1. If 7x - 5y = 13 and 2x - y = 5, then (x - y) is

A.3 B. 4 C. 5 D. 1

Answer - D
Explanation -
7x - 5y = 13 - - - - - - - equation 1
2x - y = 5 - - - - - - - - - - - - - equation 2
Subtract equation 2 from equation 1
5x - 4y = 8 - - - - - - - - - - - - - equation 3
Subtract equation 2 from equation 3
3x - 3y = 3
x - y = 1
hence option D is the correct answer
2. If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 24200, and Rs 26620 respectively, what is the rate of interest?

A.10 percent B. 5 percent C. 20 percent D. 16percent

Answer - A
Explanation -
Let the rate of interest = r
Principal amount = p
Amount = [latex]P({1 + \frac{r}{100} }^{t})[/latex]
Amount at the end of two year = 24200 = [latex]P({1 + \frac{r}{100} }^{2})[/latex] equation 1
Amount at the end of 3 year = 26620 = [latex]P({1 + \frac{r}{100} }^{3})[/latex] equation 2
Divide equation 2 by equation 1
[latex](\frac{26620}{24200}) = 1 + \frac{r}{100}[/latex]
[latex]\frac{r}{100} = \frac{26620}{24200} - 1 = \frac{2420}{24200}[/latex]
[latex]\frac{r}{100} = \frac{1}{10}[/latex]
r = 10%
so required rate of interest = 10%
3. If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is

A.xy B. xz/y C. xy/z D. xyz

Answer - D
Explanation -
First no. x Second no. = L.C.M. x H.C.F.
So, L.C.M. = xyz
4. An engineering student has to secure 25% marks to pass. He gets 47 and fails by 43 marks. What are the maximum marks of the examination?

A.385 marks B. 410 marks C. 360 marks D. 435 marks

Answer - D
Explanation -
Let the maximum marks be max.
According to the question,
Passing marks = Marks of student + 43
25% of max = 47 + 43
([latex]\frac{1}{4}[/latex]) × max = 90
max = 360 marks
5. The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

A.28, 28 B. 32, 24 C. 30, 26 D. 29, 27

Answer - C
Explanation -
Let the present age of husband = x
And present age of wife = y
x + y = 56
x = 56 - y - - - - - - equation1
(x - 10)(y - 10) = 320
xy - 10x - 10y + 100 = 320
xy - 10(x + y) = 220
xy - 10(56) = 220
xy = 780 - - - - - - - equation 2
Put value of x from equation 1 into equation 2
(56 - y)y = 780
[latex]{y}^{2}[/latex] - 56y + 780 = 0
[latex]{y}^{2}[/latex] - 30y - 26y + 780 = 0
y = 30 or 26
so we get x = 26 or 30
hence the required ages are 30 and 26
6. A shopkeeper, sold dried apricots at the rate Rs 1210 a kg and bears a loss of 12%. Now if he decides to sell it at Rs 1331 per kg, what will be the result?

A. 6.4 percent loss B. 3.2 percent gain C. 6.4 percent gain D. 3.2 percent loss

Answer - Option D
Explanation -
Old selling price = 1210 and loss percentage = 12%
Loss percentage = [latex]\frac{CP - SP}{CP}* 100[/latex]
This means (100 - 12)% of CP = 1210
88% CP = 1210 * [latex]\frac{100}{88}[/latex]
CP = 1210
CP = 1375
New selling price = 1331
Loss = CP-SP = 1375 - 1331 = 44
Loss percentage = [latex]44 * \frac{100}{1375}[/latex]
loss percentage = 3.2 %
7. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps in equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is

A. 48 : 35 B. 28 : 15 C. 24 : 20 D. 20 : 21

Answer - Option D
Explanation -
Given;
[latex]\Rightarrow \frac{Tom}{Jerry} = \frac{5}{7}[/latex]
Let jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter
then, ratio of speed of Tom and Jerry = [latex] \frac{8 * 5}{6 * 7} = \frac{20}{21}[/latex]
8. Ruchir walks at 20 km/hr and Rukma cycles at 25 km/hr towards each other. What was the distance between them when they started if they meet after 48 minutes?:

A. 54 km B. 45 km C. 36 km D. 27 KM

Answer - Option C
Explanation -
Let the distance between them at the beginning was x km
Relative speed of Rukma = 25 + 20
= 45 km/hr
Time-taken = 48 min
1 min = 1/60 hr
i.e, 48 min = [latex]\frac{48}{60}[/latex]
= 0.8 hr
We know that, distance = time-taken × speed
= 0.8 × 45
= 36 km
9. A painter can paint a fence in 24 hours. After 6 hours he takes a break. What fraction of the fence is yet to be painted?

A. 0.6 B. 0.2 C. 9.75 D. 0.8

Answer - Option C
Explanation -
Painter completed the total work in 24 hours
Work completed in one hour = [latex]\frac{1}{24}[/latex]
Work completed in six hour = 6 [latex]\frac{1}{24} = \frac{1}{4}[/latex]
remaining work = 1 - [latex]\frac{1}{4} = \frac{3}{4}[/latex]
fraction of work yet to be completed = [latex]\frac{3}{4}[/latex] = 0.75
10. [latex]\frac{sin A}{1 + cos A}[/latex] is equal to

A. cosec A - cot A B. cosec A + cot A C. [latex]\sqrt{(cosec A - cot A)}[/latex] D. [latex]\sqrt{(cosec A + cot A)}[/latex]

Answer - Option A
Explanation -
[latex]\frac{sin A}{1 + cos A}[/latex]
multiply numerator and denominator by 1-cosA
we get [latex]\frac{sin A(1 - cos A)}{1 + {cos}^{2} A} = \frac{sin A(1 - cos A)}{ {sin}^{2} A} = cosec A - cot A[/latex]

1. Find the difference between the unit digits of a and b a= [latex](437)^{253} [/latex].[latex](134)^{337} [/latex] b= [latex](437)^{253} [/latex][latex](134)^{337} [/latex].[latex](132)^{653} [/latex][latex](559)^{381} [/latex]

A. 2 B. 5 C. 4 D. 0

Answer - D
Explanation - the unit digits repeats themselves after every power multiple of 4
Now , for a unit digit of a = 7*4 = 28 = unit digit is 8
Now , for b unit digit of b = 2*9 = 18 = unit digit is 8
Therefore difference between unit digits of a and b = 8 - 8 = 0
2. Unit digit in [[latex](253)^{98} [/latex]-[latex](106)^{100} [/latex]+[latex](708)^{403} [/latex]-[latex](16)^{4} [/latex]+259]

A. 8 B. 6 C. 2 D. 4

Answer - A
Explanation - Unit digit of the expression
= [[latex](3)^{98} [/latex]-[latex](6)^{100} [/latex]+[latex](8)^{403} [/latex]-[latex](6)^{4} [/latex]+9]
= [[latex](3^{(4(24))} [/latex]-[latex](6)^{100} [/latex]+[latex](8)^{(4(100))} [/latex]-[latex](6)^{4} [/latex]+9]
= 1 * 9 - 6 + 6 * 2 - 6 + 9
= 9 - 6 + 2 - 6 + 9
= 8
3. Unit digit in [latex]264^{(5!(6!))}[/latex]* [latex]264^{(103(1024))}[/latex]= ?

A. 4 B. 6 C. 8 D. 2

Answer - A
Explanation -
[latex](5!)^{(6!)}[/latex] is an even number and [latex](103)^{(1024)}[/latex] is an odd number.
We know that,
4 even always gives 6 as last digit and 4 odd always gives 4 as last digit.
= [latex](4)^{(5!(6!))}[/latex]* [latex](4)^{(103(1024))}[/latex]
= 4even * 4odd
= 6 * 4
= 4
4. Unit digit in [latex](17)^{(256)}[/latex] is:

A. 1 B. 3 C. 7 D. 9

Answer - D
Explanation -
Unit digit of a number is the remainder when that number is divided by 10.
Rem = [latex]\frac{{17}^{256}}{10}[/latex]
= [latex]\frac{{17}^{(256)}}{10} = \frac{{({7}^{2})}^{123}}{10}[/latex]
= [latex]\frac{{49}^{(123)}}{10} = \frac{{-1}^{123}}{10}[/latex] = -1
= 10 -1
= 9(unit digit)
5. Unit digit in [latex](6474)^{(1863)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex] is:

A. 0 B. 1 C. 3 D. 2

Answer - A
Explanation -
[latex](6474)^{(1863)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex]
= [latex](6474)^{(1863/4)}[/latex]*[latex](605)^{(427)}[/latex]*[latex](281)^{(491)}[/latex]
= [latex](6474)^{(3)}[/latex]* 5* 1[(1863/4)=3]
= 4 * 5 * 1
= 0
6. Find unit digit in [latex](39)^{18(17)}[/latex] is:

A. 1 B. 4 C. 9 D. 0

Answer - A
Explanation -
([latex](18)^{(17)}[/latex]/4) = (18*18*18*...17 times)/4
= So power of 39 is completely divisible by 4.If power even and there is 9 at unit place of the base number, 1 will come at unit place of the resultant.
= [latex](39)^{(4)}[/latex] = 1
Hence the unit digit = 1
7. Find unit digit in [latex](459)^{(213)}[/latex]*[latex](545)^{(315)}[/latex]*[latex](723)^{(395)}[/latex]*[latex](634)^{(833)}[/latex] is:

A. 1 B. 2 C. 3 D. 0

Answer - D
Explanation -
[latex]\frac{213}{4}[/latex] = 1 (Remainder) = [latex]{9}^{1}[/latex] = 9 (unit digit)
[latex]\frac{315}{4}[/latex] = 3 (Remainder) = [latex]{5}^{3}[/latex] =5 (unit digit)
[latex]\frac{395}{4}[/latex] = 3 (Remainder) = [latex]{3}^{3}[/latex] = 7 (unit digit)
[latex]\frac{833}{4}[/latex] = 1 (Remainder) = [latex]{4}^{1}[/latex] = 4 (unit digit)
Requred
9 * 5 * 7 * 4 = 0 (unit digit)
8. Find unit digit in [latex]{{27}^{37}}^{35}[/latex] by 4?

A. 0 B. 1 C. 7 D. 21

Answer - C
Explanation -
7 has a cyclicity of 4 for unit digit
So [latex]{7}^{1} \rightarrow[/latex] 4
[latex]{7}^{2} \rightarrow[/latex] 9
[latex]{7}^{3} \rightarrow[/latex] 3
[latex]{7}^{4} \rightarrow[/latex] 1
ans so on
So seeing the divisibility of the power with 4
[latex]{{27}^{37}}^{35}[/latex] = [latex]\frac{(37*37*37....35 times)}{4}[/latex]
Hence in given question when we divide [latex](27)^{37(35)} {7}^{1}[/latex] = 7 as unit digit
9. Find unit digit in [latex](264)^{(102)}[/latex][latex](264)^{(103)}[/latex] is:

A. 0 B. 4 C. 6 D. 8

Answer - A
Explanation -
[latex]{(264)}^{102} + {(264)}^{103}[/latex] If the unit number is 4 in the base of the exponential number, (1)When the power is even, the unit's digit is 4.(2) If the power is odd, the unit's digit is 6.That is[latex]{(264)}^{102}[/latex][latex]{(264)}^{103}[/latex] = 6 + 4 = 10 That is, the unit number will be zero.
10. Find unit digit in [latex]{(2467)}^{153}[/latex]*[latex]{(341)}^{72}[/latex] is:

A. 7 B. 2 C. 8 D. 1

Answer - A
Explanation -
= Unit digit of [latex]{(2467)}^{153} \times {(341)}^{72}[/latex]
= Unit digit of [latex]{(7)}^{153} \times {(1)}^{72}[/latex]
= Unit digit of [latex]{{(7)^{4}}}^{38} \times {(7)}^{1} \times {(1)}^{(72)}[/latex]
= Unit digit of [latex]{(1)}^{(153)} \times 7 \times {(1)}^{72}[/latex]
So, final unit digit = 7 * 1 = 7

1. x + [latex]\frac{1}{x}[/latex], then the value of [latex]x^{7}[/latex] + [latex]\frac {1}{x^ {5}}[/latex] is

A. [latex](2)^{12}[/latex] B. 2 C. [latex](2)^{5}[/latex] D. [latex](2)^{7}[/latex]

Answer - Option B
Explanation -
x + [latex]\frac{1}{x}[/latex] = 2
[latex]x^{2}[/latex] + 1 = 2x
[latex]x^{2}[/latex] - 2x + 1 = 0
[latex](x - 1)^{2}[/latex] = 0
x - 1 =0
x = 1
therefore, [latex]x^{7}[/latex] + [latex]\frac{1}{x^{5}}[/latex] = 1 +1=2
2. If [latex](2a - 3)^{2}[/latex] + [latex](3b + 4){2}[/latex] + [latex](6c + 1){2}[/latex] = 0 then the value of [latex]\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2}}[/latex] is:

A. abc + 3 B. 6 C. 0 D. 3

Answer - Option D
Explanation -
[latex](2a - 3)^{2}[/latex] + [latex](3b + 4){2}[/latex] + [latex](6c + 1){2}[/latex]
therefore, 2a - 3 = a = [latex]\frac {3}{ 2}[/latex],
3b + 4 = 0 = b [latex]\rightarrow \frac {-4}{3}[/latex]
6c + 1 = 0 [latex]\rightarrow[/latex] c = -[latex]\frac {1}{6}[/latex]
(a + b + c) = [latex]\frac {9 - 8- 1}{6}[/latex] = 0
therefore, [latex]{a^{3} + b^{3} + c^{3} - 3abc}[/latex] = 0
[latex]\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2}}[/latex] + 3 = 0 + 3 = 3
3. What is the value of positive square root of 14 + [latex]\sqrt {6}{5}[/latex]

A. 3 + [latex]\sqrt {5}[/latex] B. 3 - [latex]\sqrt {5}[/latex] C. 5 + [latex]\sqrt {3}[/latex] D. 5 - [latex]\sqrt {3}[/latex]

Answer - Option A
Explanation -
= 14 + [latex]\sqrt {6}{5}[/latex]
= [latex]3^{2}[/latex] + [latex]\sqrt{5}^{2}[/latex] + 2 * 3* [latex]\sqrt{5}[/latex]
= (3 + [latex]\sqrt{5}^{2}[/latex])
4. A sum of money placed at compound interest doubles itself in 5 years. It will amount to eight times of itself in:

A. 15 years B. 12 years C. 10 years D. 20 years

Answer - Option A
Explanation -
A = P[latex](1 + \frac{R}{100})^{T}[/latex]
2 = 1[latex](1 + \frac{R}{100})^{5}[/latex]
Cubing both sides
[latex]2^{3}[/latex] = [latex](1 + \frac{R}{100})^{15}[/latex]
5. How many factors of 106 is/are prime number?

A. 2 B. 3 C. 5 D. 6

Answer - Option A
Explanation -
Factor of 106 = 2 × 53
Both 53 and 2 are prime
Hence there are 2 prime factor of 106
6. The population of a town is 9000. It the number of females increases by 5% and the males by 7.5%, what will be the total population after increase. The number of females currently is 3000.

A. 9600 B. 9200 C. 10500 D. 9540

Answer - Option A
Explanation -
In the village
Females = 3000
Males + 9000 - 3000 = 6000
After respective increases,
Population of Village
= 3000 * [latex]\frac {105}{100}+ {6000 * 107.5}{100}[/latex]
= 3150 + 6450 = 9600
7. Rohit buys an odd bat for Rs 1230 and spends Rs 180 on its repairs. If he sells it for Rs 1128, then what is his loss percentage?

A. 12.5 B. 15 C. 20 D. 25

Answer - Option C
Explanation -
Total cost price of bat =1230 + 180 = 1410
Selling price = 1128
Loss= 1410 - 1128 = 282
Loss percentage = [latex]\frac{282}{1410}[/latex] * 100 = 20%
8. [latex]\frac{4}{5}[/latex] part of a tank is filled with oil. After taking out 42 litres of oil the tank is 3/4 part full. What is the capacity (in litres) of the tank?

A. 420 B. 630 C. 840 D. 1680

Answer - Option C
Explanation -
Let the capacity of tank be 20x
Then initial quantity of oil in the tank= [latex]\frac{4}{5}[/latex] * 20X = 16X
Final quantity = [latex]\frac{3}{4}[/latex] * 20X = 15X
16x - 15x = 42
X = 42
So capacity of tank=20x = 20 × 42 = 840
9. To do a certain work, B would take time thrice as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. The time taken by A to complete the work separately is

A. 22 days B. 24 days C. 30 days D. 20 days

Answer - Option B
Explanation -
If B does the work in 3x days, (A + C) will do the same work in x days.
If C does that work in 2y days. (A + B) will do it in y days.
therefore, [latex]\frac{1}{x}\frac{1}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{4}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
3x = 40
x = [latex]\frac{40}{3}[/latex]
Again, [latex]\frac{1}{y}\frac{1}{2y}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{3}{2y}[/latex] = [latex]\frac{1}{10}[/latex] = y = 15
therefore(A + B+ C)'S = 1 Day's work = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A} + \frac{1}{40} + \frac{1}{30}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A}[/latex] = [latex]\frac{1}{10} - \frac{1}{40} - \frac{1}{30}[/latex]
[latex]\frac{12 - 3- 4}{120}[/latex] = [latex]\frac{5}{120}[/latex] = [latex]\frac{1}{24}[/latex]
therefore, A alone will complete the work in 24 days.
10. If sin(A - B) = Sin A Cos B - Cos A Sin B, then Sin [latex]15^{0}[/latex]will be

A. [latex]\frac{\sqrt{3} + 1}{2\sqrt{2}}[/latex] B.[latex]\frac{\sqrt{3}}{2\sqrt{2}}[/latex] C. [latex]\frac{\sqrt{3} - 1}{-\sqrt{2}}[/latex] D. [latex]\frac{\sqrt{3} - 1}{2\sqrt{2}}[/latex]

Answer - Option D
Explanation -
Let, A= [latex]45^{0}[/latex]
B= [latex]30^{0}[/latex]Sin (A-B)
=sin A. cos B-cos A. sin B Sin [latex]45^{0} - 30^{0}[/latex]
= sin[latex]15^{0}[/latex]
= [latex]\frac{1}{\sqrt{2}} * \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} * \frac{1}{2} [/latex]
[latex]\frac{\sqrt{3}}{2\sqrt{2}} * \frac {1}{2\sqrt{2}}[/latex] = [latex]\frac{\sqrt{3} - 1}{2\sqrt{2}}[/latex]

1. What is the HCF (highest common factor) of 57 and 513?

A. 10 B. 57 C. 3 D. 27

Answer - Option B
Explanation -
Factors of 57 = 3 × 19
Factors of 513 = 3 × 3 × 3 × 19
Since the common factor between the two is 3 × 19 = 57
Thus, HCF of 57 and 513 = 57
2. The two numbers are 63 and 77, HCF is 7. Find the LCM.

A. 668 B. 693 C. 674 D. 680

Answer - Option B
Explanation -
Relation b/w H.C.F, L.C.M & two no is given by
H.C.F. × L.C.M. = a × b
The numbers a = 63 and b = 77
HCF = 7
So, L.C.M. = [latex]\frac{63 * 77}{7}[/latex] = 693
3. What is the HCF (highest common factor) of 133 and 112?

A. 15 B. 7 C. 19 D. 16

Answer - Option B
Explanation -
Factor of 133 = 7× 19
Factor of 112 = 7× 16
Hence the HCF of 133 and 112 = 7
4. HCF and LCM of two numbers are 11 and 825 respectively. If one number is 275, find the other number.

A. 53 B. 45 C. 33 D. 43

Answer - Option C
Explanation -
Let the second number be n.
Formula : First Number x Second Number = LCM X HCF
275 x n = 825 x 11
n = 33
5. The sum of two numbers is 36 and their H. C. F and L. C. M are 3 and 105 respectively. The sum of the reciprocals of the two numbers will be:

A. 13 B.[latex]\frac{9}{11}[/latex] C. [latex]\frac{7}{35}[/latex] D. [latex]\frac{4}{35}[/latex]

Answer - Option D
Explanation -
HCF = 3
Numbers = 3x and 3y where x and y are co-primes.
therefore 3x + 3y = 36
x + y =12
= 3xyz = 105 and, LCM
Dividing (i) by (ii)
[latex]\frac{x + y}{xy}[/latex] =[latex]\frac{12}{35}[/latex]
[latex]\frac{1}{x}[/latex][latex]\frac{1}{y}[/latex] = [latex]\frac{12}{35}[/latex]
[latex]\frac{1}{3x}[/latex] + [latex]\frac{1}{3y}[/latex] = [latex]\frac{4}{35}[/latex]
6. A farmer has 945 cows and 2475 sheep. He farms then into flocks, keeping cows and sheep separate and having the same number of animals in each flock. If these flocks are as large as possible, then the maximum number of animals in each flock and total number of flocks required for the purpose are respectively

A. 15 and 228 B. 9 and 380 C. 45 and 76 D. 46 and 75

Answer - Option C
Explanation -
First of all we find the HCF of 945 and 2475
945 = 3 * 3 * 3 * 5 * 7
2475 = 3 * 3 * 5 * 5 * 11
HCF = 45
Maximum number of animals in each flock = 45
Required total number of flockS= [latex]\frac{945}{45}[/latex] + [latex]\frac{2475}{45}[/latex]
21 + 55 = 76
7. Find the least number which must be subtracted from 2423 so that the resultant number when divided by 15, 25 and 40 leaves remainder of 7 in each case.

A. 8 B. 12 C. 16 D. 20

Answer - Option C
Explanation -
The number which divided by 15, 25 and 40 leaves remainder 7 in each case
= n * LCM (15, 25, 40) + 7.
(where n is a positive integer)
= n *600 + 7
the closest such number is 2407.
Hence, the least number to be subtracted
= 2423 - 2407
= 16.
8. There are three classes A, B and C with 72, 96 and 144 students respectively. Find the minimum number of small groups in which they can be divided such that each group has students from one class only and each small group has equal number of students.

A. 9 B. 10 C. 11 D. 13

Answer - Option D
Explanation -
The minimum groups will be formed only when number of students in each group is largest which will be the HCF
HCF of 72, 96 and 144 = 24
Therefore, number of students in each group is 24.
Hence, the number of groups required is
= [latex]\frac{72}{24} + \frac{96}{24} + \frac{144}{24}[/latex]
= 3 + 4 + 6
= 13
9. How many numbers between 400 and 800 are divisible by each 4, 5 and 6?

A. 7 B. 8 C. 9 D. 10

Answer - Option A
Explanation -
LCM of 4, 5, 6 = 60
So 420, 480, 540, 600, 660, 720, 780
i.e, 7 no is divided between 400 - 800
10. The ratio of two numbers is 3 : 4 and their H.C.F. is 4; then their L.C.M. is:

A. 12 B. 24 C. 36 D. 48

Answer - Option D
Explanation -
Ratio of two numbers is 3 : 4
let the numbers be 3x and 4x.
HCF of (3x, 4x) = x = 4 (given)
Hence numbers are 12 and 16
The LCM of 12 and 16 = 48
Hence Option D is correct

1. The sum of the ages of husband and wife at present is 100. Ten years ago the ratio of their ages was 9:7. What is the age of the husband?

A. 45 years B. 55 years C. 65 years D. 40 years

Answer - Option B
Explanation -
Let 10 years ago, the age of husband and wife be 9x and 7x respectively.
Given: (9x + 10) + (7y + 10) = 100
[latex]\rightarrow[/latex] 16x = 100 - 20
[latex]\rightarrow[/latex] x = 5
Thus, present age of husband = 9x + 10
= 9 × 5 + 10
= 55 years
2. The ratio of present ages of Rangana and Sayed is 7:5. After 11 years the ratio of their ages will be 4:3. What is Rangana's present age?

A. 55 B. 77 C. 70 D. 96

Answer - Option B
Explanation -
The ratio of present ages of Rangana and Sayed is 7:5. Let their present ages are 7x and 5x respectively. After 11 years, the ratio of their ages = 4:3, i.e.
[latex]\frac{7x + 11}{5x + 11}[/latex] = [latex]\frac{4}{3}[/latex]
21x + 33 = 20x + 44
x = 11
Thus, the present age of Rangana = 7x = 77 years
3. The ratio of present ages of Ratnabali and Shaukat is 8:5. After 22 years the ratio of their ages will be 10:9. At present, what is Ratnabali’s age ?

A. 5 B. 14 C. 81 D. 8

Answer - Option D
Explanation -
Let the present age of Ratnabali= 8x
And present age of Shaukat= 5x
As per question
[latex]\frac{8x + 22}{5x + 22}[/latex] = [latex]\frac{10}{9}[/latex]
72x + 198 = 50x + 220
22x = 22
x = 1
hence present age of Ratnabali = 8× 1= 8 years
4. The ratio of present ages of Ranjini and Shahid is 5:4. After 13 years the ratio of their ages will be 6:5. What is Ranjini's present age?

A. 52 B. 65 C. 60 D. 32

Answer - Option B
Explanation -
Let present age of Rajini = 5x
And present age of Shahid = 4x
After 13 years,
Age of Rajini = 5x + 13
Age of Shahid = 4x + 13
Ratio of their age after 13 years is 6 : 5
[latex]\frac{(5x +13)}{(4x+13)}[/latex] = 6/5 [latex]\frac{6}{5}[/latex]
25x + 65 = 24x + 78
x = 13
Rajini’s present age = 5x = 5 × 13 = 65 years
5. The ratio of present ages of Rambha and Sarvesh is 8:5. After 7 years the ratio of their ages will be 5:4. What is Rambha's present age?

A. 5 B. 30 C. 8 D. 48

Answer - Option C
Explanation -
Let present age of Rambha and Sarvesh is 8x and 5x
After 7 years
[latex]\frac{8x + 7}{5x + 7} = \frac {5}{4}[/latex]
32x + 28 = 25x +35
7x = 7
X = 1
So Rambha present age will be 8x = 8 years
6. The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

A. 28, 28 B. 32, 24 C. 30, 26 D. 29, 27

Answer - Option C
Explanation -
Let the present age of husband = x
And present age of wife = y
x + y = 56
x = 56 - y - - - - - - equation1
(x - 10)(y - 10) = 320
xy - 10x - 10y + 100 = 320
xy - 10(x + y) = 220
xy - 10(56) = 220
xy = 780 - - - - - - - equation 2
Put value of x from equation 1 into equation 2
(56 - y)y = 780
[latex]{y}^{2}[/latex] - 56y + 780 = 0
[latex]{y}^{2}[/latex] - 30y - 26y + 780 = 0
y = 30 or 26
so we get x = 26 or 30
hence the required ages are 30 and 26
7. The ratio of present ages of Ramita and Satyajit is 9 : 7. After 9 years the ratio of their ages will be 5 : 4. What is Ramita's present age?

A. 81 B. 63 C. 56 D. 80

Answer - Option A
Explanation -
Let the present age of Ramita and Satyajit be 9x and 7x respectively.
According to the question, after 9 years the ratio of their ages will be 5:4, i.e.
[latex]\frac{9x + 9}{7x + 9} = \frac {5}{4}[/latex]
4(9x + 9) = 5(7x + 9)
36x + 36 = 35x + 45
x = 9
Thus, present age of Ramita = 9x = 9 × 9
= 81 years
8. A got married 8 years ago. A’s present age 1[latex]\frac{1}{4}[/latex]is times his age at the time of marriage. A’s son’s age is [latex]\frac{1}{10}[/latex] times his present age. His son’s age in years, is

A. 2 B. 3 C. 4 D. 5

Answer - Option C
Explanation -
Let A’s age eight years ago X years be.
X + 8 = [latex]\frac{5X}{4}[/latex]
therefore, 4x + 32 + 5x
x = 32 years
therefore, Son's present age = [latex]\frac{1}{10}[/latex](x + 8)
= [latex]\frac{1}{10}[/latex]* 40 = 4
9. Twice the age of A is 2 years more than that of B. If the product of ages of A and B is 180 years, what will be the age of A after 2 year?

A. 8 years B. 10 years C. 12 years D. 16 years

Answer - Option C
Explanation -
2A = B + 2
A * B = 180
A(2A - 2) = 180
[latex]A^{2}[/latex] - a- 90 = 0
A = 10 years
After 2 years A’s age = 10 + 2 = 12 years
10. 2 years ago, the ratio of ages of Rocky and Jacky was 6:7. 2 years hence the ratio will become 7:8, What is the present age of Rocky?

A. 28 years B. 30 years C. 24 years D. 26 years

Answer - Option D
Explanation -
Let 2 years ago, ages of Rocky and Jacky were R and J respectively. A/Q,
R : J = 6 : 7
[latex]\frac{6x + 4}{7x + 4}[/latex] = [latex]\frac{7}{8}[/latex]
48x + 32 = 49x + 28
x = 4
resent age of Rocky = R + 2 = 6X + 2 = 26

1. A is 4 years younger than B while C is 4 years younger than D but 1/5th times as old as A. If D is 8 years old, how many times as old is B as C?

A. 6 times B. 4 times C. 5 times D. 2 times

Answer - Option A
Explanation -
A = (B - 4),C = (D - 4)& C = [latex]\frac{1}{5}[/latex]A
C = 8 - 4 = 4
A = 5C = 5 * 4 = 20
B = A + 4 = 20 + 4 + 24
[latex]\frac{B}{C}[/latex] = [latex]\frac{24}{4}[/latex] = 6
B = 6C
2. 6 years ago, age of a mother was 3 times that of her daughter. If sum of their present ages is 60 years, what are the present age of mother and daughter respectively?

A. 64 and 32 B. 48 and 16 C. 42 and 18 D. 48 and 20

Answer - Option C
Explanation -
Mother’s age 6 years ago = M
Daughter’s age 6 years ago = D
[latex]\frac{M}{D}[/latex] = 60
x = [latex]\frac{48}{4} = 12[/latex]
Present age of mother = 3x + 6 = 42 years
Present age of daughter = x + 6 + 18 years
3. The ration of ages of the father and mother was 11:10 when their son was born. The ratio of ages of the father and mother will be 19:18 when the son will be twice his present age. What is the ratio of present ages of father and mother?

A. 15 : 14 B. 14 : 13 C. 16 : 15 D. 17 : 16

Answer - Option A
Explanation -
Let the age of father =11x and that of mother be 10x when there son is born
When son is twice his age, that is after 2 year the ratio of father and mother age=19:18
So, [latex]\frac{11x + 2}{10x + 2} = \frac{19}{18}[/latex]
198x - 190x = 2
X = [latex]\frac{1}{4}[/latex]
So, father age = [latex]\frac{11}{4}[/latex] and mother age = [latex]\frac{10}{4}[/latex]
So father present age = [latex]\frac{11}{4}[/latex] + 1 = [latex]\frac{15}{4}[/latex]
And mother present age = [latex]\frac{10}{4}[/latex]+ 1 = [latex]\frac{14}{4}[/latex]
So required ratio = 15 : 14
4. If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then present age of his grandson Anup is obtained. If Anup is two years younger to Madan whose age is 5 years, then what is Gagan’s present age?

A. 48 years B. 60 years C. 84 years D. 96 year

Answer - Option B
Explanation -
Madan’s age =5 years. Then Anup’s age = 5 - 2 = 3 years. Let x be the present age of
Gagan. According to question,
[latex]\frac{x-6}{18}[/latex] =3
x = 54 + 6 = 60 years
5. A person was asked to state his age in years. His reply was “take my age 3 years hence, multiply it by 3 and then subtract 3 times my age 3 years ago and you will know how old I am.” What was the age of the person?

A. 18 years B. 20 years C. 24 years D. 32 years

Answer - Option A
Explanation -
Let his present age = x years
According to question,
(x + 3) × 3- 3 × (x - 3) = x
x=18 years
6. If 10 years are subtracted from the present age of A and the remainder is divided by 6, then the present age of his grandson B is obtained. If B is 2 years younger to C whose age is 7 years, then what is A’s present age?

A. 40 years B. 55 years C. 52 years D. 65 years

Answer - Option A
Explanation -
[latex]\frac{A-10}{6}[/latex] = B, B = C - 2, C = 7 years
B = 7 - 2 = 5 years
[latex]\frac{A-10}{6}[/latex] =5
A = 40 years
7. The average age of a husband-wife and their son was 42 years. The son got married and exactly after 1 year a child was born to them. When the child became 5 years old, the average age of the family became 36 years. What was the age of bride at the time of marriage?

A. 30 years B. 27 years C. 25 years D. 22 years

Answer - Option C
Explanation -
[latex]\frac{H+W+S}{3}[/latex] = 42
H+W+S = 126
Let age of son’s bride at the time of marriage =B year. After 1 year a child was born to them and there are 5 family members now. After 5 years of his birth i.e,after 6 years of son’s marriage average age of the family
[latex]\frac{(H + 6) + (W + 6) + (S + 6) + (B + 6) + 5}{5}[/latex] = 36
(H + W + S + B) = 180 - 29 = 151
126 + B = 151
B = 151 - 126 = 25 years
8. If the ages of A and C are added to twice the age of B, the total becomes 59. If the ages of B and C are added to thrice the age of A, the total becomes 68 and if the age of A is added to thrice the age of B and thrice and age of C, the total becomes 108. What is the age of A?

A. 18 years B. 15 years C. 12 years D. 20 years

Answer - Option C
Explanation -
A + C + 2B = 59, 3A + B + C=68 and A + 3B + 3C=108
B + C = 68 - 3A and A + 3(B+C) = 108
A + 3(68 - 3A) = 108
8A = 204 - 108
A=[latex]\frac{96}{8}[/latex] = 12 years
9. Mr X has three sons namely P, Q and R. P is the eldest son of Mr. X while R is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44. What is the age of P after three years?

A. 15 years B. 13 years C. 19 years D. 17 years

Answer - Option C
Explanation -
Let the present age of P, Q and R are x, y and z (square numbers) respectively.
(x + 5) + (y + 5) + (z + 5) = 44
x + y + z = 44 – 15 = 29
Perfect square numbers are: 1, 4, 9, 16, 25 etc. The only possible values of x, y and z are 16, 9 and 4 respectively. Present age of P is 16 years (oldest). After 3 year his age would be 19 years.
10. Three years ago, the average age of Ramesh's family having 5 members was 17 years. Ramesh becomes father but the average age of his family is same today. What is the present age of baby?

A. 1 year B. 2 year C. 3 year D. 4 year

Answer - Option B
Explanation -
Three years ago total age of family having 5 members =5 × 17 = 85 years At present there are 6 members is the family but average is same. Therefore,
[latex]\frac{85 + 3 + 3 + 3 + 3 + 3 + Baby}{6}[/latex] = 17
Baby = 102 - 100 = 2 years

1. In a class of 55 students there are 34 girls. The average weight of these girls is 51 Kg and average weight of the full class is 55.2 kgs. What is the average weight of the boys of the class?

A. 62 B. 59.4 C. 56.8 D. 60

Answer - Option A
Explanation -
Average weight of the boys = [latex]\frac{sum of weights of full class - sum of weights of girls}{number of boys}[/latex]
= [latex]\frac{55.2 \times 55 - 34 \times 51}{55 - 34}[/latex]
= [latex]\frac{3036 - 1734}{21}[/latex]
= 62 kg
2. The average revenues of 11 consecutive years of a company is Rs 69 lakhs. If the average of first 6 years is Rs 64 lakhs and that of last 6 years is Rs 76 lakhs, what is the revenue for the 6th year?

A. Rs 83 lakhs B. Rs 79 lakhs C. Rs 77 lakhs D. Rs 81 lakhs

Answer - Option D
Explanation -
Revenue of the sixth year = Revenue of the first 6 years + Revenue of the last 6 years – Revenue of total 11 years
= 6 × 64 + 6 × 76 – 11 × 69
= Rs 81 lakhs
3. The mean of marks secured by 55 students in division-A of class X is 55, 45 students of division-B is 51 and that of 40 students of division-C is 49. What is the mean of marks of the students of three divisions of Class X?

A. 51.3 B. 50.6 C. 52 D. 53.4

Answer - Option C
Explanation -
Mean of all the three divisions = [latex]\frac{sum of total marks secured by all the students}{total number of students}[/latex]
= [latex]\frac{55 \times 55 + 45 \times 51 + 40 \times 49}{55 + 45 + 40}[/latex]
= [latex]\frac{7280}{140}[/latex]
= 52
4. The mean of marks secured by 55 students in division A of class X is 58, 45 students of division B is 54 and that of 75 students of division C is 52. Find the mean of marks of the students of three divisions of class X.

A. 53.7 B. 54.4 C. 53 D. 55.8

Answer - Option B
Explanation -
Total marks= mean× number of students
Total marks in three division of class X= 55 × 58 + 45 × 54 + 75 × 52
3190 + 2430 + 3900= 9520
Total number of students in class X = 55 + 45 + 75= 175
Mean marks of students of three division of class x = [latex]\frac{9520}{175}[/latex]
= 54.4
5. The mean of marks secured by 65 students in division A of class X is 54, 30 students of division B is 50 and that of 55 students of division C is 48. Find the mean of marks of the students of three divisions of Class X.

A. 50.3 B. 49.6 C. 51 D. 52.4

Answer - Option C
Explanation -
Total marks secured by class A = 65 × 54 = 3510
Total marks secured by class B = 30 × 50 = 1500
Total marks secured by class C = 55 × 48 = 2640
Average of class X = [latex]\frac{Total marks secured by class A + Total marks secured by class B Total marks secured by class C}{Total number of students in class X}[/latex]
Average of class X = [latex]\frac{3510 + 1500 + 2640}{65 + 30 + 55}[/latex]
= [latex]\frac{7650}{150}[/latex]
= 51
6. Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family remains the same today. The age of the baby today is

A. 3 years B. 2 years C. 1 years D. 1.5 years

Answer - Option A
Explanation -
Present age of 5 members = (5 * 17 + 3 * 5) = 100 Years
Present age of 5 members and a baby = 17*6 = 102 years
Age of baby = 102 - 100 = 2 years.
7. A man engaged a servant on the condition that he would pay him `90 and a turban after service of one year. He served only for nine months and received the turban and an amount of `65. The price of turban is

A. 25 B. 18.75 C. 10 D. 2.50

Answer - Option C
Explanation -
Let monthly salary = x
So,
12x = 90 + turban
9x = 65 + turban
then 3x = 25
12x = 100
= 90 + turban = 100
turban = 10
8. The costs of daily ticket of local train is Rs. 50 and Monthly Pass costs is Rs. 1125. If I buy the Monthly Pass and travel for 30 days in a month then I save?

A. 10 percent B. 25 percent C. 14 percent D. 15 percent

Answer - Option B
Explanation -
Costs of daily ticket = Rs 50
Monthly pass costs = 1125
Cost of monthly travelling = 50 × 30 = 1500
Amount saved by travelling through pass = 1500 - 1125 = 375
Required percentage =[latex]\frac{375}{1500}[/latex] * 100 = 25%
9. The average of 19 consecutive even integers is 50. The highest of these integers is

A. 88 B. 68 C. 126 D. 100

Answer - Option B
Explanation -
Given: The average of 19 consecutive even integers is 50.
Then, their sum = 19 × 50
= 950
Let the first digit be a. Common difference between these digits, d = 2
Sum of terms of AP = [latex]\frac{n}{2}[/latex][2a + (n – 1)d]
[latex]\frac{19}{2}[/latex][2a + (19 – 1)2] = 950
2a + 36 = 100
a = 32
Also, if l is the last term, then
Sum = [latex]\frac{n}{2}[/latex][a + l]
950 = [latex]\frac{19}{2}[/latex][32 + l]
l = 68
10. The average revenues of 13 consecutive years of a company is Rs 82 lakhs. If the average of first 7 years is Rs 77 lakhs and that of last 7 years is Rs 89 lakhs, find the revenue for the [latex]{7}{th}[/latex] year.

A. Rs. 98 lakhs B. Rs. 94 lakhs C. Rs. 96 lakhs D. Rs. 92 lakhs

Answer - Option C
Explanation -
Revenue of [latex]{7}{th}[/latex] year = sum of revenues of first 7 years and last 7 years – sum of revenues of 13 years
= 7 × 77 + 7 × 89 – 13 × 82
= 1162 - 1066
= Rs 96 lakhs

1. Three persons walk from place A to place B. their speed are in the ratio 4 : 3 :5. The ratio f the times taken by them to reach B will be :

A. 10 : 15 : 13 B. 2 : 3: 4 C. 15 : 20 : 12 D. 16 : 18: 15

Answer - Option C
Explanation -
Time taken is inversely proportional to relevant speeds. = [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{3}[/latex] : [latex]\frac{1}{5}[/latex]
i.e, Required ratio = [latex]\frac{1}{4} \times 60 : \frac{1}{3} \times 60 : \frac{1}{5} \times [/latex] 60
= 15 : 20 : 12
2. The ratio of boys and girls in a college is 5 : 3. If 50 boys leave the college and 50 girls join the college, the ratio becomes 9 : 7. Initial number of boys in the college was how much?

A. 300 B. 400 C. 500 D. 600

Answer - Option C
Explanation -
The ratio of boys and girls in a college is 5 : 3
If X be the number of total students in college, then number of boys = [latex]\frac{5x}{8}[/latex]
Number of girls = X - [latex]\frac{5x}{8}[/latex] = [latex]\frac{3x}{8}[/latex]
If 50 boys leave the college and 50 girls join the college,
Now number of boys = [latex]\frac{5x}{8}[/latex] - 50 = [latex]\frac{5x - 400}{8}[/latex]
Number of girls = [latex]\frac{3x}{8}[/latex] + 50 = [latex]\frac{3x + 400}{8}[/latex]
Given ratio between boys and girls = 9:7
[latex]\frac{5x - 400}{8}[/latex] : [latex]\frac{3x + 400}{8}[/latex] = 9 : 7
7 * [latex]\frac{5x - 400}{8}[/latex] = 9[latex]\frac{3x + 400}{8}[/latex]
35x - 2800 = 27x + 3600
8x = 6400
x = 800
Therefore number of boys in college = [latex]\frac{5x }{8}[/latex] = [latex]\frac{5x * 800}{8}[/latex] = 500
3. A, B and C are batsmen. The ratio of the runs scored by them in a certain match are given below : A : B = 5 : 3 and B : C = 4 : 5. In all they scored 564 runs. The number of runs scored by B is :

A. 124 B. 104 C. 114 D. 144

Answer - Option D
Explanation -
A + B + C = 564
A : B = 5 : 3 = 20 : 12 and B : C = 4 : 5 = 12 : 15
A : B : C = 20 : 12 : 15
Proportional sum of A, B and C = 20 + 12 + 15 = 47
Hence Run scored by B
= 12 * [latex]\frac {564}{47}[/latex] = 12 * 12 = 144
4. A, B, C enter into a partnership with shares in the ratio [latex]\frac {7}{2}[/latex] : [latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex] After 4 months, A increases his share by 50%. If the total profit at the end of the year was 43,200, then the B’s share in the profit is

A. 4,200 B. 4,800 C. 7,200 D. 8,000

Answer - Option D
Explanation -
A, B, C center into a partnership with shares in the ratio [latex]\frac {7}{2}[/latex] : [latex]\frac {4}{3}[/latex] : [latex]\frac {6}{5}[/latex]
After 4 months, A increases his share by 50%
Hence, A share at the end of year
= [latex]\frac {7x}{2}[/latex] * 4 + [latex]\frac {21x}{4}[/latex] * 8
14x + 42x = 56x
B' share = [latex]\frac {4x}{3}[/latex] * 12 = 16m
C, share = [latex]\frac {6x}{5}[/latex]* 12 = [latex]\frac {72x}{5}[/latex]
Now B profit = [latex]\frac {16x}{56x + 16x + \frac {72}{x}}[/latex] * 43200
= [latex]\frac {16x * 5}{432}[/latex] * 43200
= 8000 Rs.
5. In a particular type of alloy the ratio of lead to tin is 5 : 6. The amount of lead should be added to 12.1 kg of this material to make the ratio 1 : 1 is

A. 1.1 kg B. 1 kg C. 0.9 kg D.1.5 kg

Answer - Option A
Explanation -
In a particular type of alloy the ratio of lead to tin is 5 : 6
Total amount of material = 12.1 Kg
Amount of lead = 5 × [latex]\frac {12.1}{11}[/latex] = 5 × 1.1 = 5.5 Kg
Amount of tin = 6 × [latex]\frac {12.1}{11}[/latex] = 6 × 1.1 = 6.6 Kg
Difference between tin and lead = 6.6 – 5.5 = 1.1 kg
Therefore 1.1 kg of lead should be added in material to make the ratio 1 : 1.
6. If A : B = 1 : 2, B : C = 3 : 4 C : D = 6 : 9 and D : E = 12 : 16 then A : B : C : D : E is equal to

A. 1 : 3 : 6 : 12 : 16 B. 2 : 4 : 6 : 9 : 16 C. 3 : 4 : 8 : 12 : 16 D. 3 : 6 : 8 : 12 : 16

Answer - Option D
Explanation -
A : B = 1 : 2 = 3: 6 = 9 : 18
B : C = 3 : 4 = 6: 8 = 18 : 24
C : D = 6 : 6 = 24: 36
D : E = 12 : 16 = 36 : 48
therefore, A : B : C : D : E = 9 : 18 : 24: 36 : 48
= 3 : 6 : 8: 12 : 16

1. What number should be added to each of 6, 14, 18 and 38 so that the resulting numbers make a proportion?

A. 1 B. 2 C. 3 D. 4

Answer - Option B
Explanation -
From the given alternatives
[latex]\frac {6 + 2}{14 + 2}[/latex] = [latex]\frac {18 + 2}{38 + 2}[/latex]
[latex]\frac {8}{16}[/latex] = [latex]\frac {20}{40}[/latex]
[latex]\frac {1}{2}[/latex] = [latex]\frac {1}{2}[/latex]
2. If 5.5 of a = 0.65 of b, then a : b is equal to:

A. 13 : 11 B. 11 : 13 C. 13 : 110 D. 110 : 13

Answer - Option C
Explanation -
5.5 of a = 0.65 of b
a : b = 0.65 : 5.5
a : b = 13 : 110
Hence Option C is correct
3. A invests Rs. 64000 in a business: After few months B Joined him with Rs. 48000. At the end of year, the total profit was divided between them in the ratio 2 :1. After how many months did B join?

A. 8 B. 4 C. 6 D. 7

Answer - Option B
Explanation -
B entered after x months.
Ratio of profits
= 6400 * 12 : 48000(12 - X)
= 16 : (12 - X)
therefore [latex]\frac {16}{12 - x}[/latex] = [latex]\frac {2}{1}[/latex] = 24 - 2x = 16
2x = 8
x = 4 months
4. if [latex]\frac {a}{b}[/latex][latex]\frac {2}{3}[/latex] and [latex]\frac {b}{c}[/latex] = [latex]\frac {4} {5}[/latex], then (a +b) : (b + c) =?

A. 3 : 4 B. 4 : 5 C. 5 : 9 D. 20 : 27

Answer - Option D
Explanation -
a : b = 2 : 3
b : c = 4 : 5
a : b : c = 2×4 : 3×4 : 3×5
= 8 : 12 : 15
[latex]\frac {a + B}{b + C}[/latex] = [latex]\frac {8 + 12}{12 + 15}[/latex] = [latex]\frac {20}{27}[/latex]
Hence Option D is correct
5. A certain amount of money is divided among x, y and z. If x receives 25% more than y and y receives 25% less than z, they x : y : z is equal to

A. 14 : 12 : 13 B. 15 : 12 : 16 C. 10 : 9 : 12 D. 12 : 10 : 11

Answer - Option B
Explanation -
x : y = 125 : 100 = 5 : 4
y : z = 75 : 100 = 3:4
x : y : z = 15 : 12 : 16
6. A sum of 53 is divided among A, B and C in such a way that A gets Rs 7 more than what B gets and B gets Rs 8 more than what C gets. The ratio of their shares is

A. 16 : 9 : 18 B. 25 : 18 : 10 C. 18 : 25 : 10 D. 15 : 8: 30

Answer - Option B
Explanation -
Short Trick:-
C get C
B = C + 8
A = C + 8 + 7 = C + 15
So, C + C + 8 + C + 15 = 53
3C + 23 = 53
3C = 30
C gets 10, B gets 18 and A gets 25
Ratio : A : B : C =25 : 18 : 10
7. If a:b = 3:8 then find the value of (5a-3b):(2a+b).

A. [latex]\frac {9}{15}[/latex] B. [latex]\frac {14}{9}[/latex] C. -[latex]\frac {9}{14}[/latex] D. [latex]\frac {14}{8}[/latex]

Answer - Option C
Explanation -
[latex]\frac {5a - 3b}{2a + b}[/latex] = [latex]\frac {5\frac{a}{b}-3}{2 \frac {3}{8}+ 1}[/latex]
= [latex]\frac {5\frac{3}{8}-3}{2 \frac {3}{8}+ 1}[/latex]
= [latex]\frac {\frac{15}{8}-3}{\frac {6}{8}+ 1}[/latex]
= [latex]\frac{\frac{15 - 24}{8}}{\frac{6 + 8}{8}}[/latex]
= -[latex]\frac{9}{14}[/latex]
8. If x : y = 2 : 5 then (5x + 3y) : (5x - 3y) is equal to _____.

A. 5 B. 3 C. -3 D. -5

Answer - Option D
Explanation -
[latex]\frac{5x + 3y}{5x - 3y}[/latex] = [latex]\frac{5\frac{x}{y}+ 3}{5\frac{x}{y} - 3}[/latex]
= [latex]\frac{5\frac{2}{5}+ 3}{5\frac{2}{5} - 3}[/latex]
= [latex]\frac{5}{-1}[/latex]
= -5
9. A person bought some rice and wheat for Rs. 380. The ratio of weight of rice and wheat is 4 : 3 and the price of equal amount of rice and wheat is in the ratio 5 : 6. The rice was bought of worth

A. 380 B. 300 C. 200 D. 180

Answer - Option C
Explanation -
A person bought some rice and wheat for Rs. 380
Ratio of weight of rice and wheat = 4 : 3
Ratio of price of equal amount of rice and wheat = 5 : 6
= (4 * 5) : (3 * 6)
= 20 : 18
= 10 : 9
worth of rice = [latex]\frac {380}{19}[/latex] * 10 = 200 Rs
10. If 5A = 12B = 13C; FIND A : B: C

A. 60 : 65: 156 B. 65: 156: 60 C. 156 : 65 : 60 D. 13 : 12 : 5

Answer - Option C
Explanation -
It is given that, 5A = 12B = 13C
Now, using hit and trial, we can find the understated values of the variables
A = 12 × 13 = 156
B = 13 × 5 = 65
C = 5 × 12 = 60
A : B : C = 156 : 65 : 60

1. Ticket for an adult is Rs 1600 and a child is Rs 1200. 1 child goes free with two adults. If a group has 23 adults and 10 children what is the discount the group gets?

A. 25.95 percent B. 24.59 percent C. 25.77 percent D. 31.69 percent

Answer - Option B
Explanation -
Number of kids who went for free = 23/2 = 11 kids
But because there are 10 kids only, so all 10 kids will go for free
Cost of ticket of 1 kid is Rs 1200 and that of an adult = Rs 1600
Thus, total cost = 1600 × 23 + 1200 × 10 = Rs 48800
And cost of ticket only for adults = Rs
Discount = [latex]\frac {12000 * 100} {48800}[/latex]
= 24.59%
2. A student multiplied a number by 5/8 instead of 8/5. What is the percentage error in the calculation?

A. 60.94 percent B. 156 percent C. 30.47 percent D. 78 percent

Answer - Option A
Explanation -
Given that, the student multiplied a number by 5/8 instead of 8/5. Then,
Error = [latex]\frac {correct - incorrect} {correct} * 100[/latex]
= [latex]\frac {\frac{8}{5} - \frac{5}{8}} {\frac{8}{5}} * 100[/latex]
= [latex]\frac {64 - 25} {64} * 100[/latex]
= 60.94%
3. An engineering student has to secure 36% marks to pass. He gets 53 and fails by 37 marks. Find the maximum marks.

A. 275 marks B. 250 marks C. 300 marks D. 325 marks

Answer - Option B
Explanation -
According to the question,
Passing marks = 53+37 = 90
36% of maximum marks = 90
100% = [latex]\frac {90 \times 100} {36}[/latex] = 250
Maximum marks = 250
4. The price of an article is cut by 3%. To restore to its original value, the new price must be increased by

A. 3% B. 7.11% C. 3.09% D. 2.69%

Answer - Option C
Explanation -
Let initial price of the article was Rs 100, then
New price = Rs 97
To restore original price, price should be increased by [latex]\frac {3} {27} * 100[/latex] = 3.09%
5. A number is divided into two parts in such a way that 80% of [latex] {1}^ {st}[/latex] part is 3 more than 60% of 2nd part and 80% of [latex] {2}^ {nd}[/latex] part is 6 more than 90% of the [latex] {1}^ {st}[/latex] part. Then the number is

A. 125 B. 130 C. 135 D. 145

Answer - Option C
Explanation -
First part = Rs. x and second part = Rs. y
i.e, [latex]\frac {x * 80} {100}[/latex] = [latex]\frac {y * 60} {100}[/latex] + 3
[latex]\frac {4x} {5}[/latex] =[latex]\frac {3y} {5}[/latex] + 3
4x - 3y = 15 .........(i)
[latex]\frac {4y} {5}[/latex] =[latex]\frac {9x} {10}[/latex] + 6
8y - 9x = 60
By equation (i) × 8 + (ii) × 3,
32x – 24y = 120
24y – 27x = 180
5x = 300 x = 60
From equation (i)
4 × 60 – 3y = 15
3y = 240 - 15 = 225
i.e, x + y = 60 + 75 = 135
6. If population of women in a village is 90% of population of men, what is the population of men as a percentage of population of women?

A. 100% B. 105% C. 108% D. 111%

Answer - Option D
Explanation -
Required % = [latex]\frac {100 \times 100} {90}[/latex] =111 (Approx)
7.In measuring the sides of a rectangle, there is an excess of 5% on one side and 2% deficit on the other. Then the error percent in the area is

A. 3.3 B. 3.0 C. 2.9 D. 2.7

Answer - Option
Explanation -
Required percentage error = [latex]\frac {x + y + xy} {100}[/latex]
[Negative sign is used for decrease]
= 5 - 2 -[latex]\frac {5 \times 2} {100}[/latex]% = 2.9%
Required percentage error
8. For an examination it is required to get 36 % of maximum marks to pass. A student got 113 marks and failed by 85 marks. The maximum marks for th examination are :

A. 500 B. 550 C. 565 D. 620

Answer - Option B
Explanation -
If maximum marks be x, then,
[latex]\frac {36 * x} {100}[/latex] = 113 + 85 = 198
x = [latex]\frac {198 * 100} {36}[/latex] = 550
9. Due to an increase of 50% in the price of eggs, 4 eggs less are available for 24. The present rate of eggs per dozen is:

A. 24 B. 27 C. 36 D. 42

Answer - Option C
Explanation -
Ratio of price increase: 2 : 3
Ratio of decrease in quantity: 3 : 2
Difference of 1 is equal to 4
So, quantity after decrease = 8 eggs
8 eggs in 24, So Rs 3 per egg
Dozen egg = 12*3 = 36 Rs
10.If A’s income is 20% more than that of B, by how much percent is B’s income less than that of A?

A. [latex]16 \frac {4} {5}[/latex] B. [latex]16 \frac {1} {3}[/latex] C. [latex]16 \frac {2} {3}[/latex] D. [latex]16 \frac {2} {7}[/latex]

Answer - Option C
Explanation -
Required percentage = [latex]100 * \frac {x} {100 + x}[/latex]
[latex]100 * \frac {20} {100 + 20}[/latex]
[latex] \frac {50} {3}[/latex] = [latex]16 \frac {2} {3}[/latex]

1. The price of a certain television set is discounted by 10% and the reduced price is then discounted by 10%. The equivalent successive discount is ______.

A. 20% B. 19% C. 18% D. 11%

Answer - Option B
Explanation -
Let X be the Market price of a television.
First discount= 10%
= x - [latex]\frac {10X} {100}[/latex] = [latex]\frac {90X} {100}[/latex] = [latex]\frac {9X} {10}[/latex]
Hence Price after first discount
Now one more discount 0f 10%
= [latex]\frac {9X} {10}[/latex] - [latex]\frac {9X} {10}[/latex] * [latex]\frac {10} {100}[/latex]
Mow price after second discount
= [latex]\frac {9X} {10}[/latex] - [latex]\frac {9X} {100}[/latex] = [latex]\frac {81x} {100}[/latex]
= X - [latex]\frac {81x} {100}[/latex] = [latex]\frac {19x} {100}[/latex]
Hence total discount = [latex]\frac {19X} {100}[/latex] * [latex]\frac {100} {X}[/latex] = 19
Therefore total discount percentage
2. A shopkeeper gains 56 on a toy after allowing 23% discount on its marked price. If his gain is 10%, then the marked price of

A. 810 B. 800 C. 560 D. 740

Answer - Option B
Explanation -
Let ‘X’ be the market price of a toy.
Discount = 23%
Hence discount amount=[latex]\frac {23X} {100}[/latex]
Hence selling price = X - [latex]\frac {23X} {100}[/latex] = [latex]\frac {77X} {100}[/latex]
Gain = 10%
Gain amount = 56
Let C. P. = Y
Hence gain = [latex]\frac {10Y} {100}[/latex] = [latex]\frac {Y} {10}[/latex]
But given: [latex]\frac {Y} {10}[/latex] = 56 [latex]\rightarrow[/latex] Y = 560
As we know that S. P. = C. P. + Profit
[latex]\frac {77X} {100}[/latex] = 560 + 56
77X = 61600
X = [latex]\frac {61600} {77}[/latex] = [latex]\frac {5600} {7}[/latex] = 800
Hence Market price of toy = 800.
3. A fruit seller makes a profit of 20% by selling mangoes at a certain price. If he charges 1 more for each mango, he can make a profit of 40%. Find the selling price of a mango in the first case.

A. 6 B. 5 C. 5.50 D. 7

Answer - Option A
Explanation -
Let there are ‘n’ mangos and ‘x’ be the selling price of one mango.
Hence S. P. of all mangos = nx
Profit = 20%
Hence C. P. of ‘n’ mangos = [latex]\frac {100nx} {120}[/latex] = [latex]\frac {5nx} {6}[/latex]
If S.P. of each mango is increase by 1, profit = 40%
S.P. of all mangos = n(x + 1)
= [latex]\frac {5nx} {6}+ \frac{40(\frac{5nx}{6})}{100} = \frac{5nx}{6} + 2 \frac{\frac{5nx}{6}}{5}[/latex]
n(x + 1) = [latex]\frac {5nx} {6} + \frac {2nx} {6}[/latex]
n(x + 1) = [latex]\frac {7nx} {6}[/latex]
6(x + 1) = 7n
x = 6 Rs
Hence price of each mango = 6
4. A man purchased a bed sheet for Rs 450 and sold it at a gain of 10% calculated on the selling price. The selling price of the bed sheet was

A. 460 B. 475 C. 480 D. 500

Answer - Option D
Explanation -
[latex]\frac {Gain} {sp} * 100[/latex] = 10
[latex]\frac {Gain} {sp} = \frac {1} {10}[/latex]
So, CP = 10 - 1 = 9 units
SP = 10 units
The value of 9 unit is 450 Rs
So, Value of 10 units = 500 Rs
5. Aniruddha sold a bicycle at a gain of 8%. Had it been sold for 75 more, the gain would have been 14%. The cost price of the bicycle was

A. 1200 B. 1250 C. 1350 D. 1500

Answer - Option B
Explanation -
Let CP = 100 unit
Old SP = 108 unit (8% profit)
New SP = 114 unit (14% profit)
According to question,
New SP = Old SP + 75
114 unit = 108 unit + 75 Rs
6 unit = 75 Rs
1 unit = [latex]\frac {75} {6}[/latex] Rs
100 unit = [latex]\frac {75 * 100} {6}[/latex] Rs
CP = 100 unit = 1250
6. The single discount equivalent to two successive discounts of 20% and 5% is

A. 24% B. 25% C. 22% D. 23%

Answer - Option A
Explanation -
Single equivalent discount = 20 + 5 - [latex]\frac {20 * 5} {100}[/latex] = 24%
7. When a shopkeeper gives 10% discount on the list price of a toy, his gain is 20%. If he had given a discount of 20%, his percentage of gain would have been

A. [latex]\frac {20} {3}[/latex]% B. [latex]\frac {20} {3}[/latex]% C. 10% D. 15%

Answer - Option A
Explanation -
Let the MP of toy be Rs 100 Rs
SP after 10% discount = 90
120% CP = 90
CP = 75 Rs
SP after 20 % discount = 80 Rs
Gain% =[latex]\frac {5} {75}[/latex] * 100 = [latex]\frac {20} {3}[/latex]%
8. A radio dealer sold a radio at a loss of 2.5%. Had he sold it for Rs. 100 more, he would have gained [latex] 7 \frac {1} {2}[/latex]% In order to [latex] 12 \frac {1} {2}[/latex]% gain he should sell it for

A. Rs. 1080 B. Rs. 1125 C. Rs. 850 D. Rs. 925

Answer - Option B
Explanation -
Let CP = x, then
107.5% of x - 97.5% of x = 100
10% of x = 100
100% x = 1000
Desired SP = 112.5 % of CP = 1125
9. The cost price of 24 apples is the same as the selling price of 18 apples. The percentage of gain is:

A. [latex] 12 \frac {1} {2}[/latex] B. [latex] 14 \frac {2} {3}[/latex] C. [latex] 16 \frac {2} {3}[/latex] D. [latex] 33 \frac {1} {3}[/latex]

Answer - Option D
Explanation -
Let the CP of 1 apple = 1
i.e, CP of 18 apples = 18
SP of 18 apples = 24
[latex] 100 * \frac {6} {18}[/latex] = [latex] \frac {100} {3}[/latex] = 33.3%
10. A shopkeeper sells his goods at 15% discount. The marked price of an article whose selling price is 629 is:

A. ?740 B. 700 C. 730 D. 777

Answer - Option A
Explanation -
Let the marked price be x
i.e, [latex]\frac{x * 85}{100}[/latex] = 629
x = [latex]\frac{629 * 100}{85}[/latex] = 740

1. Which of the following successive discount series is the best of all for a customer?

A. 30%, 20%, 10% B. 25%, 20%, 15% C. 30%, 10%, 15% D. 25%, 15%, 10%

Answer - Option A
Explanation -
(i) Single equivalent discount for 30% and 20% = (30 + 20 - [latex]\frac{30 * 20}{100}[/latex])% = 44%
Single equivalent discount for 44% and 10% = (44 + 10 - [latex]\frac{44 * 10}{100}[/latex])% = 49.6%
(ii) Single equivalent discount for 25% and 20% = (25 + 20 - [latex]\frac{25 * 20}{100}[/latex])% = 40%
Single equivalent discount for 40% and 15% = (44 + 15 - [latex]\frac{44 * 15}{100}[/latex])% = 49%
(iii) Single equivalent discount for 30% and 10% = (30 + 10 - [latex]\frac{30 * 10}{100}[/latex])% = 37%
Single equivalent discount for 37% and 15% = (37 + 15 - [latex]\frac{37 * 15}{100}[/latex])% = 46.45%
(iv) Single equivalent discount for 25% and 15% = (25 + 15 - [latex]\frac{25 * 15}{100}[/latex])% = 36.25%
Single equivalent discount for 36.25% and 10% = (36.25 + 10 - [latex]\frac{36.25 * 10}{100}[/latex])% = 42.63%
2. The selling price of 12 articles is equal to the cost price of 15 articles. The gain per cent is

A. 6[latex]\frac{2}{3}[/latex]% B. 20% C. 25% D. 80%

Answer - Option C
Explanation -
Let cost price of each article = Rs. 1
Cost price of 12 articles = Rs. 12
Selling price of 12 articles = Rs. 15
Profit =15-12=3
Required percentage of profit = [latex]\frac{3 * 100}{12}[/latex]% = 25%
3. A vendor sells lemons at the rate of 5 for 14, gaining thereby 40%. For how much did he buy a dozen lemons ?

A. 20 B. 21 C. 24 D. 28

Answer - Option C
Explanation -
C.P. of 5 lemons
= [latex]\frac{100}{140}[/latex] = Rs. 10
C.P. of 12 lemons
[latex]\frac{10 * 12}{5}[/latex] =24
4. The percentage of profit, when an article is sold of 78, is twice than when it is sold for 69. The cost price of the articls is :

A. 49 B. 51 C. 57 D. 60

Answer - Option D
Explanation - Let the C.P. of article be Rs. x.
[latex](\frac{78 - x}{x}) * 100[/latex]
= 2 ([latex]\frac{69 - x}{x}[/latex] * 100)
78 - x = 2 * 69 - 2x
2x - x = 138 - 78
x = 60
5. Mohan sold his watch at 10% loss. If he had sold it for Rs. 45 more, he would have made 5% profit. The selling price of the watch was

A. 300 B. 900 C. 110 D. 270

Answer - Option D
Explanation -
Difference of 15% = 45 Rs
1% = 3Rs
100% = 300 Rs
So, CP = 300 Rs
SP after 10% loss = 300 - 30 = Rs 270
6. A trader purchases a watch and a wall clock for 390. He sells them making a profit of 10% on the watch and 15% on the wall clock. He earns a profit of 51.50. The difference between the original prices of the wall clock and the watch is equal to

A. 80 B. 120 C. 110 D. 100

Answer - Option C
Explanation -
If the C.P. of wrist watch be ? x, then
C.O. of wall clock = (390 - x)
[latex]\frac{x * 10}{100} + \frac{(390 - x) * 15}{100}[/latex] = 51.50
10x + 5850 - 15x = 5150
5x = 5850 - 5150
5x = 700
x = [latex]\frac{700}{5}[/latex] = 140
C.P. of wall clock = 390 – 140 = 250
Required difference = 250 – 140 = 110
7. A profit of 10% is made after giving a discount of 5% in a T.V. If the marked price of the TV is Rs. 2640.00, the cost price of the TV was:

A. 2280 B. 2296 C. 2380 D. 2396

Answer - Option A
Explanation -
MP = Rs 2640
D = 5%
SP = 2640×95/100 = Rs 2508
Profit = 10%
CP = 2508 ×100/110 = Rs 2280
Hence Option A is correct
8. A dishonest fruit vendor sells his goods at cost price but the uses a weight of 900 gm for the kg. Weight. His gain per cent is :

A. 12% B. 11 [latex]\frac{1}{9}[/latex]% C. 10 [latex]\frac{1}{9}[/latex]% D. 10%

Answer - Option B
Explanation -
Let goods cost the vendor 1 Rs / kk
i.e, He sells for Rs. 1 which cost him = 0.9
i.e, Gain% = [latex]\frac{1 - 0.9}{0.9} * 100 = \frac{100}{9} = 11 \frac{1}{9}[/latex]
9. By selling an article for 960 a man incurs a loss of 4%; what was the cost price?

A. 1000 B. 784 C. 498.4 D. 300

Answer - Option A
Explanation -
C.P.= [latex]\frac{(100 × 960)}{(100 - 4)}[/latex] = Rs.1000
10. Kamala bought a bicycle for 1,650. She had to sell it at a loss of 8%. She sold it for

A. 1,581 B. 1, 518 C. 1, 510 D. 1, 508

Answer - Option B
Explanation -
Kamala sold her Bicycle = [latex]\frac{1650 * 92}{100}[/latex] = 1518

1. Two trains, of same length, are running in parallel tracks in opposite directions with speed 65 km/hour and 85 km/hour respectively. They cross each other in 6 seconds. The length of each train is

A. 100 meters B. 115 meters C. 125 meters D. 150 meters

Answer - Option C
Explanation -
Trains are running in opposite direction hence their relative speed will be the sum of their respective speeds.
Relative speed = 65 + 85 = 150 Km/hour
Both trains are of same length (say X), if they crosses each other they travels 2X distance
Time = 6 sec = [latex]\frac{6}{3600}[/latex] hour = [latex]\frac{1}{600}[/latex] hour
Distance = speed × time
2X = 150 × [latex]\frac{1}{600}[/latex] = 0.25 km = 250m
X = 125 meter
2. A car goes 20 metres in a second. Find its speed in Km/hr.

A. 18 B. 72 C. 20 D. 36

Answer - Option B
Explanation -
Since, A car goes 20 metres in a second. Therefore,
Speed of the car = 20 m/sec
= 20 * [latex]\frac{18}{5}[/latex] = 72 km/hr
3. Two trains 100 metres and 95 metres long respectively pass each other in 27 seconds when they run Cin the same direction and in 9 seconds when they run in opposite directions. Speeds of the two trains are

A. 44 km/hr, 22 km/hr B. 52 km/hr, 26 km/hr C. 36 km/hr, 18 km/hr D. 40 km/hr, 20 km/hr

Answer - Option B
Explanation -
Let speed of first train = [latex]{s}_{1}[/latex]
Let speed of second train = [latex]{s}_{2}[/latex]
[latex]{s}_{1}[/latex] - [latex]{s}_{2}[/latex] = [latex]\frac{195}{27}[/latex] = [latex]\frac{65}{9} \rightarrow {65}{9} * {18}{5}[/latex] = 26km/hr [latex]\rightarrow[/latex] Eq (1)
similerly [latex]\rightarrow[/latex]
[latex]{s}_{1}[/latex] + [latex]{s}_{2}[/latex] = [latex]\frac{195}{9} * \frac{18}{5} [/latex] = 78 km/hr [latex]\rightarrow[/latex] Eq (2)
Add Eq (1) and Eq (2)[latex]\rightarrow[/latex]
[latex]{s}_{1}[/latex] + [latex]{s}_{2}[/latex] = 78
[latex]{s}_{1}[/latex] - [latex]{s}_{2}[/latex] = 26
Sothat, [latex]{s}_{1}[/latex] = 52 km/hr
[latex]{s}_{2}[/latex] = 26 km/hr
4. A train travelling with uniform speed crosses two bridges of lengths 300m and 240m in 21 seconds and 18 seconds respectively. The speed of the train is:

A. 72 km/hr B. 68 km/hr C. 65 km/hr D. 60 km/hr

Answer - Option A
Explanation - Let the length of the train be x metre.
[latex]\frac {x+ 300}{21}[/latex] = [latex]\frac {x + 240}{18}[/latex]
i.e, Speed of train = [latex]\frac {x+ 300}{7}[/latex] = [latex]\frac {x + 240}{6}[/latex]
7x + 1680 = 6x + 1800
7x – 6x = 1800 – 1680
x = 120
[latex]\frac {x+ 300}{21}[/latex] = [latex]\frac {420}{21}[/latex] = 20 m/sec
i.e, Speed of train = [latex](\frac {20 * 18}{5})[/latex] kmph
5. A train of length 500 feet crosses a platform of length 700 feet in 10 seconds. The speed of the train is

A. 70 ft/second B. 85 ft/second C 100 ft/second D. 120 ft/second

Answer D
Explanation
Speed of the train = [latex]\frac {(500+700)}{10}[/latex] = 120 ft/sec
6. A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.

A. 189 km B. 378 km C 270 km D. 278

Answer C
Explanation
Let the total distance covered by a man = x km
Since, he covers the first half of the journey at 60 km/h. therefore,
Time taken to cover the x/2 distance at the speed of 60 km/hr = [latex]\frac {\frac {x}{2}}{60}[/latex]
Time taken to cover rest x/2 distance at the speed of 45 km/hr = [latex]\frac {\frac {x}{2}}{45}[/latex]
But, he travels for 5 hours 15 minutes. Therefore
= [latex]\frac {x} {120}[/latex] + [latex]\frac {x}{90}[/latex] = 5 hr 15 min
= [latex]\frac {x} {120} [/latex] + [latex]\frac {x}{90}[/latex] = 5 [latex]\frac {1}{4}[/latex]
= [latex]\frac {3x + 4x} {360}[/latex] = [latex]\frac {21}{4}[/latex]
[latex]\frac {7x}{360}[/latex] = [latex]\frac {21}{4}[/latex]
X = [latex]\frac {3 * 360}{4}[/latex] = 270 km
7. A cyclist, after cycling a distance of 70 km on the second day, finds that the ratio of distances covered by him on the first two days is 4 : 5. If he travels a distance of 42 km. on the third day, then the ratio of distances travelled on the third day and the first day is:

A. 4 : 3 B. 3 : 2 C 3 : 4 D. 2 : 3

Answer C
Explanation
Let the common ratio be x
Distance covered in first day = 4x
Distance covered in second day = 5x
According to the question
5x = 70 km
So, X = 14 km
Distance covered in first day = 4x = 4×14 = 56 km
Distance covered in third day = 42 km
Required Ratio = Distance covered in third day/ Distance covered in first day
I.e, [latex]\frac {42}{56}[/latex] = 3 : 4
Hence Option C is correct
8. A train is running at a uniform speed of 60 km/hr. If the length of the train is 73 m, then the time taken by the train in crossing 77m Ling Bridge is

A. 9 sec B. 8.67 sec C 2.19 sec D. 2.51

Answer A
Explanation
Speed of the train = 60 km/hr = 60 × 5/18 = 50/3 m/sec
Length of train = 73m
Length of bridge= 77
Time taken =150 ÷ 50/3 = 9 sec
9. A man walks ‘a’ km in ‘b’ hours. The time taken to walk 200 metres is how much?

A. [latex]\frac {200 b}{a}[/latex] hours B. [latex]\frac {b}{5a}[/latex] hours C [latex]\frac {b}{a}[/latex] hours D. [latex]\frac {ab}{200}[/latex] hours

Answer B
Explanation
Speed of man = [latex]\frac {a}{b}[/latex] Km/hr
Hence time taken to walk 200 meter ([latex]\frac {1}{5}[/latex] Km) = distance/speed hrs.
[latex]\frac {1}{5}[/latex] * [latex]\frac {a}{b}[/latex] = [latex]\frac {b}{a}[/latex] hours
10. A missile travels at 1260 km/h. How many metres does it travel in one second?

A. 322 metres B. 369 metres C 384 metres D. 350 metres

Answer D
Explanation
1 km/hr = [latex]\frac {5}{18}[/latex] m/s
1260 km/hr = 1260 × = [latex]\frac {5} {18}[/latex]
= 350 m/s
Thus, in one second the missile travels 350 meters

1. A missile travels at 1422 km/h. How many meters does it travel in one second?

A. 395 meters B. 400 meters C 364 meters D.319 meters

Answer A
Explanation
Speed of missile = 1422 * [latex]\frac {5}{18}[/latex] = 395 m/s
Speed in meter per second = 1422
Distance travelled in one second = 395 x 1 = 395 meters.
Note:
To convert speed in km/h to m/s we multiply the speed in km/h by [latex]\frac {5}{18}[/latex]
2. To cover a distance of 333 km in 2 hours by a car, what should be the average speed of the car (in meter/second)?

A. 166.5 B. 46.25 C 83.25 D. 92.5

Answer B
Explanation
Speed of the car in km/hr = [latex]\frac {333}{2}[/latex]
= 166.5 km/hr
We know that, in 1km/hr = [latex]\frac {5}{18}[/latex] m/sec
Therefore, 166.5 km/hr = 166.5 × ([latex]\frac {5}{18}[/latex])
= 46.25 m/sec
3. Two trains start from stations A and B and travel towards each other at the same time at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the second train has travelled 120 km more than the first. The distance between A and B is ?

A. 1200 km B. 1440 km C 1320 km D. 990 km

Answer C
Explanation
Let the distance travelled by first train be x.
Then, the distance travelled by second train =x+120
Time taken by first train = [latex]\frac {x} {50}[/latex]
Time taken by second train= [latex]\frac {x + 120} {60}[/latex]
[latex]\frac {A}{Q}[/latex],
Distance travelled by first train=600
Distance travelled by second train = x + 120 = 720
The distance between A and B = 600 + 720 = 1320
4. 7 hrs after a goods train passed a station, another train travelling at a speed of 54 km/hr following that goods train passed through that station. If after passing the station the train overtakes the goods train in 11 hours. What is the speed of the goods train?

A. 39.6 km/hr B. 49.5 km/hr C 33 km/hr D. 26.4 km/hr

Answer C
Explanation
Distance travelled by the goods train in (11 + 7) = 18 hrs. is same as that travelled by the other train in 11 hrs. Let speed of the goods train be x, then
18x = 54 × 11
[latex]\rightarrow[/latex] x = 33 km/hr
5. Two cars travel from from city A to city B at a speed of 42 and 60 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B is :

A. 336 km B. 280 km C 420 km D. 224 km

Answer B
Explanation
Let the distance between the city A and city B be D km
As per question
[latex]\frac {D}{42}[/latex] - [latex]\frac {D}{60}[/latex] = 2
[latex]\rightarrow[/latex]18D= 2 × 42 × 60
D= 280km
6. Rakesh goes on a trip on his motorcycle and rides for 368 kms. If he rides for 5 hours at a speed of 49 km/hr, at what speed he travels for the remaining 3 hours of the journey?

A. 58 km/hr B. 41 km/hr C 54 km/hr D. 46 km/hr

Answer B
Explanation
Let speed with which he travelled for 3 hr be x, then
Distance travelled with speed 49 km/hr for 5 hr + Distance travelled in 3 hr = Total distance travelled
[latex]\rightarrow[/latex] 49 × 5 + 3 × x = 368
[latex]\rightarrow[/latex] 3x = 368 - 245
[latex]\rightarrow[/latex] 3x = 123
[latex]\rightarrow[/latex] x = 41 km/hr
7. Two boat are travelling with speed of 36 km/hr and 54 km/hr respectively towards each other. What is the distance (in metres) between the two boats one second before they collide?

A. 10 B. 15 C 25 D. 5

Answer C
Explanation
Speeds of the boats are 36km/h and 54 km/h respectively.
Then, their relative speed = 36 + 54
= 90 km/h
[latex]\rightarrow[/latex] 1 km/h = 5/18 m/s
[latex]\rightarrow[/latex] 90 km/h = 90 × 5/18
= 25 m/s
So, 1 sec before, boats were 25 metres apart.
8. Durga walks 5 km from her home to school in 60 minutes, then bicycles back to home along the same route at 15 km per hour. Her sister mriti makes the same round trip, but does so at half of Durga's average speed. How much time does Smriti spend on her round trip?

A. 120 minutes B. 40 minutes C 160 minutes D. 80 minutes

Answer C
Explanation -
Durga’s average speed = [latex](\frac{2xy}{x + y})[/latex]kmph
[latex]\frac{2 * 5 * 15}{5 + 15}[/latex]kmph
[latex]\frac{150}{20}[/latex]kmph = [latex]\frac{15}{2}[/latex]kmph
Distance of school = 5 km.
Smriti’s speed = [latex]\frac{15}{4}[/latex]kmph
[latex]\rightarrow[/latex] Required time = [latex]2(\frac{\frac{5}{15}}{4})[/latex] hours
[latex]\frac{2 * 5 * 4}{15}[/latex] = [latex]\frac{8}{3}[/latex] hours
[latex]\frac{8 * 60}{3}[/latex]minutes
= 160 minutes
9. In a 200 m sprint race, Amit can beat Rajiv by 20 m and can beat Satyendra by 30 m. If Satyendra and Rajiv participate in a 100m race, how much head start can Rajiv give to Satyendra?

A. 9.44 m B. 4.44 m C. 5.44 m D. 5.56 m

Answer - Option D
Explanation -
When Rajiv runs 180 m Satyendra can run 170 m in the same duration
Therefore when Rajiv will run 100 m Satyendra can run [latex]\frac{(170*100)}{180}[/latex] = 94.44 m
Required head start = 100-94.44 = 5.56 m
10. In what time will a 100 m long train running with a speed of 50 km/hr cross a pillar

A. 7.0 sec B. 72 sec C. 7.2 sec D. 70 sec

Answer - Option C
Explanation -
We know that, during cross a pillar, a train will cover the distance as same as its length.
Speed of the train = 50 km/hr = [latex]\frac{5}{18} * 50 = \frac{250}{18}[/latex]m/sec
The distance covered by the train = length of the train = 100 m
Therefore, required time = [latex]\frac{100}{\frac{250}{18}} = \frac{100 * 18}{250}[/latex] = 7.2 m/sec
Note:
To convert speed from km/h to m/s we multiply speed in km/h by [latex]\frac{5}{18}[/latex]

1. A man can row 30 km downstream and return to the starting point in 8 hours. If the speed of the boat in still water is four times the speed of the current, then the of the speed current is

A. 1 km/hour B. 2 km/hour C. 4 km/hour D. 3 km/hour

Answer - Option B
Explanation -
If the speed of stream = x kmph,
then speed of boat in still water = 4x kmph
Rate downstream = 4x + x = 5x kmph
Rate upstream = 4x - x = 3x kmph
i.e, [latex]\frac{30}{3x} + \frac{30}{5x}[/latex] = 8
[latex]\frac{10}{3x} + \frac{6}{x}[/latex] = 8
[latex]\frac{16}{x} [/latex] = 8
x =[latex]\frac{16}{8} [/latex] = 2 kmph
2. A boat travels 24 km upstream in 6 hours and 20 km downstream in 4 hours Then the speed of boat in still water and the speed of water current are respectively

A. 4 kmph and 3 kmph B. 4.5 kmph and 0.5 kmph C. 4 kmph and 2 kmph D. 5 kmph and 2 kmph

Answer - Option B
Explanation -
Let the speed of boat in still water = X kmph
Speed of water = Y kmph
Formula: speed = distance / time
Relative speed of boat in upstream = X – Y = 24/6 = 4 kmph …… (1)
Relative speed of boat in downstream = X+ Y = 20/4 = 5 kmph …… (2)
After solving these two equations :
X = 4.5 kmph and Y = 0.5 kmph
3. If a boat goes 100 km down ­stream in 10 hours and 75 km upstream in 15 hours, then the speed of the stream is

A. 2 km/hour B. 2.5 km/hour C. 3 km/hour D. 3.5 km/hour

Answer - Option B
Explanation -
Rate downstream = [latex]\frac{100}{10}[/latex] = 10 kmph
Rate upstream = [latex]\frac{75}{15}[/latex] = 5 kmph
Speed of current = [latex]\frac{(speed in downstream - speed in upstream)}{2}[/latex]
= [latex]\frac{1}{2}(10 - 5)[/latex]kmph
= [latex]\frac{1}{2} 5[/latex] = 2.5 kmph
4. A man rows 40 km upstream in 8 hours and a distance of 36 km downstream in 6 hour Then speed of stream is

A. 0.5 km/hr B. 1.5 km/hr C. 1 km/hr D. 3 km/hr

Answer - Option A
Explanation -
Let the speed of boat in still water = X km/hr
Speed of water = Y km/hr
Formula: speed = [latex]\frac{distance}{time}[/latex]
Relative speed of boat in upstream = X – Y = 40/8 = 5 km/hr …… (1)
Relative speed of boat in downstream = X+ Y = 36/6 = 6 km/hr …… (2)
After solving these two equations :
x = 5.5 km/hr and Y = 0.5 km/hr
Hence speed of stream = 0.5 km/ hr
5. The speed of a boat in still water is 10km/hr. It covers upstream a distance of 45 km in 6 hours. The speed (in km/hr) of the stream?

A. 2.5 B. 3 C. 3.5 D. 4

Answer - Option A
Explanation -
Let the speed of the stream be x km/hr
[latex]\frac{45}{6}[/latex] = 10 -x
60 - 6x = 45
x = [latex]\frac{15}{6}[/latex] = 2.5 km/hr
6. Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

A. 1.5 B. 3 C. 2 D. 1.75

Answer - Option B
Explanation -
Down-stream rate = [latex]\frac{9}{2}[/latex] = 4.5 kmph
Upstream rate = [latex]\frac{9}{6}[/latex] = 1.5 kmph
The speed of the boat = (4.5 – 1.5) kmph = 3 kmph
7. Rajesh rows in still water with a speed of 4.5 kmph to go to a certain place and comes back. Find his average speed for the whole journey, if the river is flowing with a speed of 1.5 kmph?

A. 2 kmph B. 4 kmph C. 6 kmph D. 8 kmph

Answer - Option B
Explanation -
Let the distance in one direction = k kms
Speed in still water = 4.5 kmph
Speed of river = 1.5 kmph
Hence, speed in upstream = 4.5 - 1.5 = 3 kmph
Speed in downstream = 4.5 + 1.5 = 6 kmph
Time taken by Rajesh to row upwards = [latex]\frac{k}{3} hrs[/latex]
Time taken by Rajesh to row downwards = [latex]\frac{k}{6} hrs[/latex]
Now, required Average speed =Total distance/Total speed
= [latex]\frac{2k}{\frac{k}{3}} + \frac{k}{6} = 2k * \frac{18}{6k} + 3k[/latex] = 4 kmph
Therefore, the average speed of the whole journey = 4kmph.
8. A boat travels 24 km upstream in 6 h and 20 km downstream in 4 h. then, the speed of a boat in still water and the speed of current are respectively.

A. 4.5 km/h and 3 km/h B. 4.5 km/h and 0.5 km/h C. 4 km/h and 2 km/h 5 km/h and 2 km/h D. 5 km/h and 2 km/h

Answer - Option B
Explanation -
Distance travelled by boat in upstream = 24 km
Time taken = 6 h
Speed of the boat in upstream = [latex]\frac{24}{6}[/latex] = 4 km/h
And distance travelled by boat in downstream = 20 km
Time taken = 4h
Speed of the boat in downstream = [latex]\frac{20}{4}[/latex] km/h = 5 km/h
Now, speed of the boat in still water = [latex]\frac{1}{2}[/latex] [ speed of the boat in upstream + speed of the boat in downstream]
= [latex]\frac{1}{2}[/latex] [4 + 5] = [latex]\frac{1}{2}[/latex] × 9 = 4.5 km/h
And speed of the current = [latex]\frac{1}{2}[/latex] [speed of the boat in downstream – speed of the boat in upstream]
= [latex]\frac{1}{2}[/latex][5 – 4] = [latex]\frac{1}{2}[/latex] × 1 = 0.5 km/h
9. P, Q and R are the three towns on the bank of a river which flows uniformly. Q is equidistant from P and R. I row from P to Q and back in 10 hours and I can row from P to R in 4 hours. Compare the speed of my boat in still water with that of the river.

A. 8 : 3 B. 5 : 3 C. 9 : 5 D. 8 : 3

Answer - Option B
Explanation -
Since time taken to cover distance from P to R= 4h
so, time taken to cover distance from P to Q= 2h
and time taken to cover distance from Q to P= 10-2= 8h
so direction of flow of current is from P to Q to R
Let the speed of the boat be v1 and the speed of the current be v2 and d be the distance between the cities.
now , = [latex]\frac{d}{{v}_{1} + {v}_{2}}[/latex] = 2 and [latex]\frac{d}{{v}_{1} - {v}_{2}}[/latex] = 8
= [latex]\frac{{v}_{1} + {v}_{2}}{{v}_{1} - {v}_{2}}[/latex]
[latex]{v}_{1} : {v}_{2}[/latex] = 5 :3
10. A man can cross a downstream river by steamer in 40 minutes and same by boat in 1 hour. If the time of crossing the river in upstream direction by steamer is 50% more than downstream time by the steamer and the time required by boat to cross the same river by boat in upstream is 50% more than the time required in downstream by boat. What is the time taken for the man to cross the river downstream by steamer and then return to same place by boat half the way and by steamer the rest of the way?

A. 185 min B. 115 min C. 170 min D. 225 min

Answer - Option B
Explanation -
Downstream (steamer) = 40min
Downstream (boat) = 60 min.
Upstream (steamer) = 60 min.
Upstream (boat) = 90min
Required time = 40 + 30 + 45 =115 min.

1. Mihir can do a piece of work in 30 hours. If he is joined by Jigya who is 50% more efficient, in what time will they together finish the work?

A. 24 hours B. 12 hours C. 6 hours D. 3 hours

Answer - Option B
Explanation -
Short Trick:-
Jigya is 50% more efficient , So their ratio is Mihir : Jigya = 2 : 3
Total work =2 * 30
Time taken by both = 60/ 5 = 12 hours
2. Abhinav is 2 times as good a workman as Balwant and therefore is able to finish a job in 36 days less than Balwant. Working together, they can do it in

A. 12 days B. 6 days C. 18 days D. 24 days

Answer - Option D
Explanation -
Let Balwant finishes his job in 36days, then Abhinav finishes the job in (x – 36) days
According to the question,
[latex]\frac{2}{x} = \frac{1}{x - 36}[/latex]
[latex]\rightarrow[/latex] 2x – 72 = x
[latex]\rightarrow[/latex] x = 72 days and (x – 36) = 36 days, i.e. Abhinav will take 36 days and Balwant will take 72 days to finish the job.
[latex]\rightarrow[/latex] if they work together, they will take
= [latex]\frac {1}{\frac {1}{36} + \frac {1}{72}}[/latex]
= [latex]\frac{1}{\frac{2 + 1}{72}}[/latex]
= 24 days
3. Pankaj has done 1/2 of a job in 12 days, Sainath completes the rest of the job in 6 days. In how many days can they together do the job?

A. 4 days B. 12 days C. 8 days D. 16 days

Answer - Option C
Explanation -
Let the total work be 24 units, then
Efficiency of Pankaj = [latex]\frac{\frac{24}{2}}{12}[/latex]
= 1 unit/day
And efficiency of Sainath = [latex]\frac{12}{6}[/latex]
= 2 units/day
If they work together their efficiency will be 3 units/day
Thus, number of days to finish the work if they work together = [latex]\frac{24}{3}[/latex]
= 8 days
4. Amar is 2 times as good a workman as Badal and therefore is able to finish a job in 15 days less than Badal. In how many days can they together complete the job?

A. 30 days B. 40 days C. 10 days D. 20 days

Answer - Option C
Explanation -
Given that, Amar is 2 times as good a workman as Badal and therefore is able to finish a job in 15 days less than Badal.
[latex]\rightarrow[/latex] Efficiency of Amar = [latex]\frac{1}{15}[/latex] units
And that of Badal = [latex]\frac{1}{30}[/latex] units
[latex]\rightarrow[/latex] number of days required to finish the job = [latex]\frac{1}{\frac{\frac{1}{15}}{\frac{1}{30}}}[/latex]
= 10 days
5. Manoj can do a piece of work in 42 hours. If he is joined by Jayashree who is 100% mor efficient, in what time they will finish the work together?

A. 7 hours B. 3.5 hours C. 14 hours D. 2.5 hours

Answer - Option C
Explanation -
Manoj can do a piece of work in 42 hour
Working efficiency of manoj = [latex]\frac{1}{42}[/latex]
Jayashree is 100% more efficient than manoj so Jayashree will take 21 hour
Efficiency of Jayashree = [latex]\frac{1}{21}[/latex]
Combined efficiency of manoj and Jayashree = [latex]\frac{1}{21} + \frac{1}{42} = \frac{3}{42} = \frac{1}{14}[/latex]
Hence time taken by both = 14 hour
6. Painter 'A' can paint a house in 10 days and 'B' can do it in 20 days. With help of 'C', they did the job in 5 days only. Then, 'C' alone can do the job in:

A. 20 days B. 10 days C. 40 days D. 50 days

Answer - Option A
Explanation -
A can paint house in 10 days
So working efficiency of A = [latex]\frac{1}{10}[/latex]
B can complete the work in 20 days
So working efficiency of B = [latex]\frac{1}{20}[/latex]
A, B and C they complete the work in 5 days
So combine efficiency of A + B + c = [latex]\frac{1}{5}[/latex]
Efficiency of C = [latex]\frac{1}{5} - \frac{1}{20} - \frac{1}{10}[/latex]
[latex]\frac{4 - 1 - 2}{20}[/latex] = [latex]\frac{1}{20}[/latex]
Efficiency of C = [latex]\frac{1}{20}[/latex]
Hence C will complete the work in 20 days
7. Mayur can complete a work in 27 hours. If he is joined by Jayantika who is 100% more efficient, in what time will they together finish the work?

A. 6 hours B. 3 hours C. 9 hours D. 10 hours

Answer - Option C
Explanation -
let total work be 27 units
Number of hours needed by Mayur to finish the work = 27
Then, efficiency of Mayur = [latex]\frac{27}{27}[/latex] = 1 unit/h
Given: Jayantika is 100% more efficient than Mayur, so
Efficiency of Jayantika = 2 units/h
Now, number of hours needed to finish the work if they work together = [latex]\frac{27}{1 + 2}[/latex]
= [latex]\frac{27}{3}[/latex] = 9 hours
8. A does 60 % of a work in 15 days. He then calls B, and they together finish the remaining work in 5 days. How long B alone would take to do the whole work ?

A. 25 days B. 20 days C. 80 days D. 24 days

Answer - Option A
Explanation -
A complete 60% work in 15 days
A will complete 100% work in = [latex]\frac{15}{60} * 100[/latex] = 25 days
A and B together complete the 40% work in 5 days
They will complete 100% work in= [latex]\frac{15}{40} * 100[/latex]
One day work of A and B together= [latex]\frac{1}{12.5} [/latex]
One day work of A = [latex]\frac{1}{25} [/latex]
One day work of B = [latex]\frac{1}{12.5} -\frac{1}{25}[/latex] = [latex]\frac{1}{25} [/latex]
Hence B complete the whole work in 25 days
9. A, B and C can complete a work in 10, 12 and 15 days respectively.All of them started working together but A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work was?

A. [latex]6 \frac{2}{3}[/latex] B. [latex]8 \frac{2}{3}[/latex] C. 6 D. 7

Answer - Option D
Explanation -
A’s 1 day work = [latex] \frac{1}{10}[/latex] ; B’s 1 day work = [latex]6 \frac{1}{12}[/latex]; A’s 1 day work = [latex]6 \frac{1}{15}[/latex]
Suppose number of days required to complete the whole work was x days.
Therefore,
[latex]\frac{(x-5)}{10} + \frac{(x-3)}{12} + \frac{(x)}{15}[/latex] = 1
[latex]\frac{(6x-30+5x-15+4x)}{60}[/latex] = 1
15x = 105
x = 7
10. A stock of food grain lasts for 21 days for village A. The same stock would last for 28 days for village B. If the food grains given to one village are used for both the villages together, the stock would last for

A. 14 days B. 12 days C. 7 days D. 49 days

Answer - Option B
Explanation -
Consumption of A for 1 day = [latex]\frac{1}{21}[/latex]
Consumption of B for 1 day= [latex]\frac{1}{28}[/latex]
Consumption of A+B=[latex]\frac{1}{21} + \frac{1}{28}[/latex] = [latex]\frac{1}{12}[/latex]
The stock would last for 12 days

1. A, B and C are three taps connected to a tank. A and B together can fill the tank in 6 hours, B and C together can fill it in 10 hours and A and C together can fill it in [latex]7 \frac{1}{2}[/latex]hours. In how much time will C alone fill the tank?

A. 10 hours B. 12 hours C. 20 hours D. 30 hours

Answer - Option D
Explanation -
A and B together can fill the tank in 6 hours
Hence part of the tank can be filled by A and B together in 1 hour = [latex]\frac{1}{6}[/latex]
[latex]\rightarrow[/latex] A+B = [latex]\frac{1}{6}[/latex] ….(1)
B and C together can fill it in 10 hours
Hence part of the tank can be filled by B and C together in 1 hour = [latex]\frac{1}{10}[/latex]
[latex]\rightarrow[/latex] B+C = 1/10 …(2)A and C together can fill it in [latex]7 \frac{1}{2}[/latex] hours
Hence part of the tank can be filled by A and C together in 1 hour = [latex] \frac{2}{15}[/latex]
[latex]\rightarrow[/latex] A+C = [latex] \frac{2}{15}[/latex] …(3)
Adding eqn. (2) and (3)
2C + (A+B)=[latex] \frac{1}{10}[/latex] + [latex] \frac{2}{15}[/latex] = 3 + [latex] \frac{4}{30}
[/latex] = [latex] \frac{7}{30}[/latex] But A + B = [latex] \frac{1}{6}[/latex]
Hence: 2C + [latex] (\frac{1}{6})[/latex] = [latex] \frac{7}{30}[/latex]
2C = [latex] \frac{7}{30} - \frac{1}{6} = 7 - \frac{5}{30} = \frac{2}{30} = \frac{1}{15}[/latex]
C = [latex]\frac{1}{30}[/latex]
Therefore C can fill 1/30 part of tank along in 1 hour or C can fill full tank in 30 hours.
2. A, B and C together earn 150 per day while A and C together earn 94 and B and C together earn 76. The daily earning of ‘C’ is

A. 6 B. 30 C. 34 D. 75

Answer - Option B
Explanation -
Let X, Y and Z be the earning per day of A, B and C respectively.
Earning of A, B and C together per day = (X + Y + Z = 150) ....(1)
Earning of A and C together per day = (X + Z = 94).....(2)
Earning of B and C together per day = (Y + Z = 76) ....(3)
Eqn. (2) + Eqn. (3) – Eqn. (1)
(X + Z) + (Y + Z) - (X + Y + Z) = 94 + 76 – 150
Z = 170 – 150 = Rs 20
Hence the daily earning of ‘C’ is 20.
3. Akash is 3 times as good a workman as Baldev and therefore is able to finish a job in 40 days . Working together, they can finish it in

A. 30 days B. 60 days C. 20 days D. 10 days

Answer - Option A
Explanation -
Let working capacity of baldev be x unit per day
then working capcity of akash = 3xunit per day
total work=40 * 3x = 120x unit
number of days taken by both working together = [latex]\frac{120x}{(x+3x)}[/latex] = 30
4. Working 8 hours a day, Anu can copy a book in 18 days. How many hours a day should she work so as to finish the work in 12 days?

A. 12 hours B. 10 hours C. 11 hours D. 13 hours

Answer - Option A
Explanation -
18 days are required to complete work working 8 hours a day
12 days are required to complete work in [latex]\frac{(8 × 18)}{12}[/latex] = 12 hours
5. B is 50% more efficient than A and A can alone do a work in 33 days. Then find thenumber of days, in which A and B, working together can finish the job.

A. 11 B. 13.2 C. 20 D. 21

Answer - Option B
Explanation -
Let B can do a work in x days.
Then, A can do a work in [latex]x + \frac{50}{100} * x = x + \frac{x}{2} = \frac{3x}{2}[/latex] days
Therefore,
[latex]\frac{3x}{2}[/latex]
x = [latex]\frac{66}{3}[/latex] = 22 days
A’s 1 day work = [latex]\frac{1}{33}[/latex]
B’s 1 day work = [latex]\frac{1}{22}[/latex]
(A +B)’s 1 day work together = [latex]\frac{1}{33} + \frac{1}{22} = \frac{5}{66}[/latex]
Thus, A+B can do a work together in [latex]\frac{66}{5}[/latex] = 13.2
6. X alone can complete a piece of work in 40 days. He worked for 8 days and left. Y alone completed the remaining work in 16 days. How long would X and Y, together take to complete the work?

A. [latex]13 \frac{1}{3}[/latex] days B. 14 days C. 15 days D. [latex]16 \frac{2}{3}[/latex] days

Answer - Option A
Explanation -
Part of the work done by X in 8 days = [latex] \frac{8}{40} = \frac{1}{5}[/latex]
Remaining work = [latex]1 - \frac{1}{5} = \frac{4}{5}[/latex]
This part of work is done by Y in 16 days.
Time taken by Y in doing 1 day = [latex] \frac{16 * 5}{4} = 20[/latex]days
Work done by X and Y in 1 day = [latex] \frac{1}{40} + \frac{1}{20} = \frac{1 + 2}{40} = \frac{3}{40}[/latex]
Hence, both together will complete the work in [latex] \frac{40}{3} = 13 \frac{1}{3}[/latex]days
7. Total weekly emoluments of the workers of a factory are ? 1534. Average weekly emolument of a worker is ? 118. The number of workers in the factory is:

A. 16 B. 14 C. 13 D. 12

Answer - Option C
Explanation -
Number of workers in the factory = [latex]\frac{1534}{118}[/latex] = 13
8. 5 men and 2 women working together can do four times as much work per hour as a man and a woman together. The work done by a man and a woman should be in the ratio :

A. 1 : 2 B. 2 : 1 C. 1 : 3 D. 4 : 1

Answer - Option B
Explanation -
Given that: work of 5 men and 2 women = 4(work of a man and a woman)
[latex]\rightarrow[/latex] work of (5 - 4) men = work of (4 - 2 ) women
[latex]\rightarrow[/latex] work of one man = work of 2 women
[latex]\rightarrow[/latex] Work of one man : work of two women = 1:1
[latex]\rightarrow[/latex] Work of one man : work of one woman = 2:1
9. A boy and a girl together fill a cistern with water. The boy pours 4 litres of water every 3 mintures and the girl pours 3 litres every 4 minutes. How much time will it take to fill 100 litres of water in the cistern?

A. 36 minutes B. 42 minutes C. 48 minutes D. 44 minutes

Answer - Option C
Explanation -
Water filled by 1 boys in 1 min = [latex]\frac{4}{3}[/latex] litres
and water filled 1 girl 1 min = [latex]\frac{3}{4}[/latex] litres
In 1 min .water filled by both = [latex]\frac{25}{12}[/latex] litres
Time taken to fill 100 litres = 100 x [latex]100 * \frac{12}{25}[/latex]= 48 min
10. A can do a piece of work in 20 days which B can do in 12 days. B worked at it for 9 days. A can finish the remaining work in

A. 5 days B. 7 days C. 11 days D. 3 days

Answer - Option A
Explanation -
A can do a piece of work in 20 days
Hence A’s work of one day = [latex]\frac{1}{20}[/latex]
B can do the same piece of work in 12 days
Hence B’s work of one day = [latex]\frac{1}{12}[/latex]
B worked at it for 9 days, therefore total work done by B = 9 × [latex]\frac{1}{12}[/latex] = [latex]\frac{3}{4}[/latex]
Remaining work = 1 –[latex]\frac{3}{4}[/latex] = [latex]\frac{1}{4}[/latex]
Now A alone do the remaining work, therefore time taken by A to do this ¼ piece of work = 20 × [latex]\frac{1}{4}[/latex] = 5 days

1. A trader had 9 Quintals of wheat. He sold a part of it at 10% profit and the rest at 20% profit, so that he made a total profit of 14 %. How much wheat did he sell at 20% profit?

A. 540 kg B. 360 kg C. 180 kg D. 720 kg

Answer - Option B
Explanation -
Given that,
Total quantity of wheat = 9 quintals
Profit on part 1 = 10% and that on part 2 = 20%

Thus, quantity of part 2 = [latex]\frac{2}{5}[/latex] × 900
= 360 kg
2. The ratio of syrup and water in a mixture is 3 : 1, then the percentage of syrup in this mixture is

A. 25% B. 66 [latex]\frac{2}{3}[/latex]% C. [latex]33 \frac{1}{3}[/latex]% D. 75%

Answer - Option D
Explanation -
The ratio of syrup and water in a mixture is 3 : 1
Let the syrup in mixture = 3x and water in the mixture = x
Then, the required percentage of the syrup in mixture = [latex] \frac{3x}{3x + x} * 100[/latex]%
= [latex] \frac{3x}{4x} * 100[/latex]%
= 75%
3. A milkman makes 20% profit by selling milk mixed with water at Rs. 9 per litre. If the cost price of 1 litre pure milk is Rs. 10, then the ratio of milk and water in the said mixture is

A. 3 : 1 B. 4 : 1 C. 3 : 1 D. 4 : 3

Answer - Option A
4. Two containers have acid and water mixed respectively in the ratio 3 : 1 and 5 : 3. To get a new mixture with ratio of acid to water as 2 : 1, the two types have to be mixed in the ratio

A. 1 : 2 B. 2 : 1 C. 2 : 3 D. 3 : 2

Answer - Option A
5. A container contains two liquids A and B in the ratio 7 : 5. When 9 litres of mixture is drawn off and the container is filled with 9 litre of pure B, the ratio of A and B becomes 7: 9. How many litres of liquid A was in the container initially?

A. 21 B. 16[latex]\frac{1}{2}[/latex] C. 36[latex]\frac{3}{4}[/latex] D. 26[latex]\frac{3}{4}[/latex]

Answer - Option A
Explanation -
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left = [latex]7X - \frac{7}{12} * 9[/latex] = [latex]7X - \frac{21}{4}[/latex] litres.
Quantity of B in mixture left = [latex]5X - \frac{5}{12} * 9[/latex] = [latex]5X - \frac{15}{4}[/latex]litres.
[latex]\frac{7x - \frac{21}{4}}{5x - {15}{4} + 9}[/latex] = [latex]\frac{7}{9}[/latex]
[latex]\frac{28x - 21}{20x + 21}[/latex] = [latex]\frac{7}{9}[/latex]
x = 3
So, the can contained 21 litres of A.
6. In two alloys A and B. the ratio of zinc to tin is 5 : 2 and 3 : 4 respectively. Seven kg of the alloy A and 21 kg of the alloy B are mixed together to form a new alloy. What will be the ratio of zinc and tin in the new alloy?

A. 2 : 1 B. 1 : 2 C. 2 : 3 D. 1 : 1

Answer - Option D
Explanation -
In alloy A ration of Zink and Tin = 5 : 2
Hence amount of Zink in 7 Kg of alloy A = 7 × [latex]\frac{5}{7}[/latex] = 5 Kg
Amount of Tin in 7 Kg of alloy A = 7 - 5 = 2 Kg
In alloy B ration of Zink and Tin = 3 : 4
Hence amount of Zink in 21 Kg of alloy B = 21 × [latex]\frac{3}{7}[/latex] = 9 Kg
Amount of Tin in 21 Kg of alloy B = 21 - 9 = 12 Kg
Seven kg of the allay A and 21 kg of the allay B are mixed together to form a new alloy
Hence in the mixture of Alloy A and B amount of Zink = 5 + 9 = 14Kg
Amount of Tin in the mixture of Alloy A and B = 2+12 = 14 Kg
Therefore ratio of Zink and Tin in the mixture of Alloy A and B = 14 : 14 = 1 : 1
7. The ratio in which a man must mix rice at Rs. 10.20 per kg and Rs. 14.40 per kg so as to make a mixture worth Rs. 12.60 per kg. is

A. 4 : 3 B. 2 : 5 C. 1.8 : 24 D. 3 : 4

Answer - Option D
Explanation -

Required ratio = 1.8 : 2.4
= 3 : 4
Hence Option D is correct
8. One type of liquid contains 20 % water and the second type of liquid contains 35 % of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is

A. 37 % B. 46 % C. 12[latex]\frac{1}{7}[/latex]% D. 24[latex]\frac{2}{7}[/latex]

Answer - Option D
Explanation -
In 10 liters of first type of liquid;
Water = [latex]\frac{1}{5}[/latex] × 10 = 2 liters
In 4 liters of second type of liquid;
Water = 4 × [latex]\frac{35}{100}[/latex] = [latex]\frac{7}{5}[/latex] liters
Total amount of water =2 + [latex]\frac{7}{5}[/latex] = [latex]\frac{17}{5}[/latex] liters
Required percentage = [latex]\frac{\frac{17}{5}}{14}[/latex] × 100
=[latex] \frac{170}{7}[/latex] = [latex]24 \frac{2}{7}[/latex]%
Hence Option D is correct
9. A container contains 60 kg of milk. From this container 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is

A. 34.24 kg B. 39.64 kg C. 43.6 kg D. 47.6 kg

Answer - Option C
Explanation -
Amount of milk left after 3 operations= 60 × [latex]{\frac{1 - 6}{60}}^{3}[/latex]
=60 × [latex]\frac{54}{60} * \frac{54}{60} * \frac{54}{60}[/latex]
= 60 × [latex]\frac{9}{10} * \frac{9}{10} * \frac{9}{10}[/latex]
= 43.74 kg
10. In a mixture of 60 litres, the ratio of milk and water is 2 : 1 How much more water must be added to make its ratio 1 : 2?

A. 40 liters B. 52 liters C. 54 liters D. 60 liters

Answer - Option D
Explanation -
In original mixture,
Milk = 40 litres
Water = 20 litres
If x litres of water is mixed,
[latex]\frac{40}{20 + x} = \frac{1}{2}[/latex]
20 + x = 40 * 2 = 80
x = 80 -20 = 60 liters

1. One type of liquid contains 25% of milk, the other contains 50% of milk. A container is filled with 6 parts of the first liquid and 4 parts of the second liquid. The percentage of milk in the mixture is:

A. 25% B. 30% C. 35% D. 33.68%

Answer - Option C
Explanation -
Using Weighted Average:
weighted average = [latex]\frac{(n{1}_{}{A}_{1} + {n}_{2}{A}_{2})} {({n}_{1} + {n}_{2})} [/latex]
Weighted Average = [latex]\frac{(6*25+4*50)}{(6+4)}[/latex]
= [latex]\frac{(150+200)}{(10)}[/latex]
= 35%
2. Ram invests Rs.9000 at the rate of 5%.How much more should he invest at the rate of 8% so that he can earn at total profit of 6%:

A. 4500 B. 5400 C. 3000 D. 4000

Answer - Option A
Explanation -
Therefore 2 = 9000
Then 1 = 4500
So he invest Rs.4500 at the rate of 6%
3. Zinc is 17 times heavy as water and silver is 8 times heavy as water.In what ratio should these be mixed to get an alloy 15 times as heavy as water?

A. 2 : 7 B. 7 : 2 C. 5 : 2 D. 9 : 7

Answer - Option B
4. Shyam buys a pen at 4% discount and a copy at 12% discount.She overall gets a discount of 10%.If the marked price of both are Rs.200/item then find the total selling price of copies, if 8 items in total are purchased?

A. 1180 B. 1182 C. 1184 D. 1056

Answer - Option D
Explanation -
8 items are purchased means no of Pens = 2
and no of copy = 6
SP of copies = [latex]\frac {6* 200* 88}{100}[/latex] = 1056
5. An alloy contains only zinc and copper. One such alloy weighing 25 gm contains zinc and copper in the ratio of 2 : 3 by weight. If 20 gm of zinc is added then find what amount of copper has to be removed from the alloy such that the final alloy has zinc and copper in the ratio of 4 : 1 by weight?

A. 3.5 gm B. 5.5 gm C. 7.5 gm D. 4.8 gm

Answer - Option C
Explanation -
In [latex]{1}^{st}[/latex] alloy zinc= [latex]\frac{2}{5}[/latex] ×25=10gm
Copper= [latex]\frac{3}{5}[/latex] × 25=15gm
Let copper be removed by x gm
Then ATQ
[latex]\frac{10 + 20}{15 - x} = \frac{4}{1}[/latex]
X=[latex]\frac{30}{4}[/latex]gm = 7.5gm
6. A 24 liters cylinder contains a mixture of oxygen and nitrogen, the volume of oxygen being 36% of total volume. A few liters of the mixture is released and an equal amount of nitrogen is added. Then the same amount of the mixture as before is released and replaced by nitrogen for the second time .As a result the oxygen content becomes 4% of the total volume. How many liters of mixture is released each time?

A. 12 B. 15 C. 16 D. 20

Answer - Option C
Explanation -
[latex]\frac{Remaining oxygen}{original oxygen} = 1 - \frac{Vol. of each time released}{total vol of vessel}[/latex]
[latex]\frac{4 %}{36 %} = (1 - {\frac{x}{24}}^{2})[/latex]
X= 16 liter
Therefore amount of mixture released each time =16 liter
7. A man purchased two chairs in Rs.1200; he sells the first chair at [latex]\frac{4}{5}[/latex] of its cost price and the second chair at [latex]\frac{5}{4}[/latex] of its cost price. If during the whole transaction he earns a profit of Rs.120.Find the cost price of cheaper chair?

A. 350 B. 400 C. 200 D. 150

Answer - Option B
Explanation -
[latex]{1}^{st}[/latex] chair = [latex]\frac{4}{5}[/latex] * 100 = 20%
[latex]{2}^{nd}[/latex] chair = [latex]\frac{5}{4}[/latex] * 100 = +25%
Profit% = [latex]{1}^{st}[/latex] chair = [latex]\frac{120}{12oo}[/latex] * 100 = 10%
3 = Rs. 1200
1 = Rs. 400 = cost price of [latex]{1}^{st}[/latex] chair
2 = Rs. 800 = cost price of [latex]{2}^{nd}[/latex] chair
8. A goldsmith has two qualities of gold, one of 22 carats and another of 16 carats purity. In what proportion should he mix both to make a jewelry of 18 carat purity?

A. 1 : 2 B. 2 : 1 C. 3 : 1 D. 3 : 2

Answer - Option A
Explanation -
Let the ratio of 22 carat gold be a.
[latex]\rightarrow[/latex] Ratio of 16 carat gold = 1 – a
Given, 22 carat gold and 16 carat gold are mixed to obtain 18 carat gold.
[latex]\rightarrow[/latex] 22 × a + 16 × (1 – a) = 18
[latex]\rightarrow[/latex] 22a + 16 – 16a = 18
[latex]\rightarrow[/latex] 6a = 2
[latex]\rightarrow[/latex] a = [latex]\frac{1}{3}[/latex]
Ratio of 16 carat gold = 1 – a = [latex]\frac{2}{3}[/latex]
Ratio of 22 carat gold and 16 carat gold in the mixture = [latex]\frac{\frac{1}{3}}{\frac{2}{3}}[/latex] = 1 : 2
9. The volume of a mixture of X and Y in the ratio 16:25 is 155L more than the volume of mixture of Y and Z in the ratio 5:17 and the quantity of Z in mixture is 102L. If two mixture are mixed find the new ratio between Z and Y.

A. [latex]\frac{205}{102}[/latex] B. [latex]\frac{102}{205}[/latex] C. [latex]\frac{255}{102}[/latex] D. [latex]\frac{107}{102}[/latex]

Answer - Option C
Explanation -
Let the quantity of Y and Z mixture = a
[latex]\frac{17a}{22}[/latex] = 102
a = 132 L
Quantity of Y = 132-102 = 30 L
Volume of mixture = 132 + 155= 287 L
Quantity of
Y = 25 * [latex]\frac{287}{41}[/latex] = 175 L
X = 287 - 175= 112 L
Total Y = 175 + 30 = 205 L
Total Z = 102 L
Ratio = [latex]\frac{Z}{Y} = \frac{102}{205L}[/latex]
10. Three containers whose volumes are in the ratio of 2:3:4 are full of mixture of spirit and water. In the first container the ratio of spirit and water is 4:1, in [latex]{2}^{nd}[/latex]container the ratio is 11:4 and in the [latex]{3}^{rd}[/latex] container ratio is 7:3. All the three mixtures are mixed in a big container. The ratio of spirit and water in the resultant mixture is?

A. 2 : 9 B. 5 : 7 C. 11 : 4 D. 17 : 8

Answer - Option C
Explanation -
Let the volume of the containers be 200 L, 300 L and 400 L respectively
So spirit in [latex]{1}^{ST}[/latex] container= [latex]\frac{4}{5}[/latex] * 200 = 160 L
Water =40 L
Spirit in [latex]{2}^{nd}[/latex] container= [latex]\frac{11}{15}[/latex] * 300 = 220 L
Water =80 L
Spirit in 3rd container= [latex]\frac{7}{10}[/latex] * 400=280 L
Water =120 L
Spirit in new mixture= 660 L
ater in new mixture= 240 L
So, ratio= 660:240= 11:4

1. There is 40% increase in an amount in 10 years at simple interest. What will be the compound interest of Rs. 30000 after 3 years at the same rate?

A. Rs 3745. 92 B. 7491. 84 C. 9364. 8 D. 5618. 88

Answer - Option A
Explanation -
There is 40% increase in an amount in 10 years, it mean yearly increase will be [latex]\frac{40}{10}[/latex]% = 4 %
Hence yearly rate of interest = 4 %
Principal amount=30000
Time =3 years
Compound interest= p[latex]{(1 + \frac {r}{100})}^{t}[/latex] - 1
30000 [latex]{(1 + \frac {4}{100})}^{3}[/latex] - 30000
33745.92 - 30000 = 3745.92
So required compound interest=3745.92
2. A sum fetched a total simple interest of Rs. 8100 at the rate of 6% per year in 9 years. What is the sum?

A. 15ooo B. 18000 C. 12000 D. 9000

Answer - Option
Explanation -
Let the sum be P
We know that SI = [latex]\frac{P * R * T}{100}[/latex]
Here SI =8100
r-6%
t=9 years
hence 8100 * 100 = P * 6 * 9
P=15000
Hence required sum is 15000
3. John invested a sum of money at an annual simple interest rate of 10%. At the end of four years, the amount invested plus interest earned was Rs 770. The amount invested was how much?

A. Rs. 650 B. Rs. 350 C. Rs. 550 D. Rs. 500

Answer - Option C
Explanation -
P+ SI = Rs 770
(770 – P) = [latex]\frac{P * 10 * 4}{100}[/latex]
P = Rs 550
4. A sum becomes 2,916 in 2 years at 8% per annum compound interest. The simple interest at 9% per annum for 3 years on the same amount will be

A. ?600 B. ?675 C. ?650 D. ?625

Answer - Option B
Explanation -
A = P[latex]{(1 + \frac{R}{100})}^{T}[/latex]
2916 = x [latex]{(1 + \frac{8}{100})}^{2}[/latex]
2916 = x [latex]{(\frac{27}{25})}^{2}[/latex]
x = x [latex]\frac{2916 * 25 * 25}{27 * 27}[/latex]
= ?2500
i.e, S.I = [latex]\frac{P * R * T}{100}[/latex] = ?675
5. A sum was lent at simple interest at a certain rate for 2 years. Had it been lent at 3% more rate, it would have fetched Rs. 300 more. The original sum of money was:

A. Rs. 5000 B. Rs. 6000 C. Rs. 7000 D. Rs. 4000

Answer - Option A
Explanation -
For 2 years we got Rs 300 more
So for 1 year we must get Rs 150 extra
Assume we deposited Rs 100 in bank. If bank gives us 3% extra rate it gives us Rs 3 extra
To get Rs 150 extra we need to invest 150/3 = 50 times of Rs 100 that is Rs 5000
Hence Option A is correct
6. Simple interest on a certain amount is [latex]\frac{9}{16}[/latex] of the principle. If the numbers representing the rate of interest in present and time in years be equal, then time, for which the principle is lent out, is:

A. 5 [latex]\frac{1}{2}[/latex] years B. 6 [latex]\frac{1}{2}[/latex] years C. 7 years D. 7 [latex]\frac{1}{2}[/latex] years

Answer - Option D
Explanation -
Let sum = x, then, S.I. = [latex]\frac{9}{16}[/latex]x
Let rate = R% and time = R years.
i.e, [latex]\frac{x * R * R}{100} = \frac{9x}{16}[/latex]
[latex] {R}^{2} = \frac{900}{16} = R = \frac{30}{4}[/latex]
Hence, time = 7 [latex]\frac{1}{2}[/latex] years
7. A certain sum of money lent out at simple interest amounts to ?1380 in 3 years and ?1500 in 5 years. Find the rate per cent, per annum.

A. 3 B. 3.5 C. 4 D. 5

Answer - Option D
Explanation -
S.I. for 2 years = 1500 – 1380 = 120
S.I. for 1 year = 120/2 = 60
S.I. for 3 years = 60*3 = 180
Amount after 3 years = 1380
Principal = 1380 - 180 = 1200
And interest per year is already known to be 60.
S.I. = PRT/100
60 = [latex]\frac{1200*R*1}{100}[/latex]
R = [latex]\frac{6000}{1200}[/latex]
R = 5%
8. A person deposited Rs. 500 for 4 years and Rs. 600 for 3 years at the same rate of simple interest in bank. Altogether he received Rs. 190 as interest. The rate of simple interest per annum was

A. 4% B. 5% C. 2% D. 3%

Answer - Option B
Explanation -
90 = [latex]\frac{500 * 4 * r}{100} + \frac{600 * 3 * r}{100}[/latex]
=>20r + 18r = 190
=>38r = 190
=>r = [latex]\frac{190}{38}[/latex] = 5%
Hence Option B is correct
9. If x, y, z are three sums of money such that y is the simple Interest on x and z is the simple interest on y for the same time and at the same rate of interest, then the relationship between x,y & z will be?

A. [latex]{z}^{2}[/latex] = xy B. xyz = 1 C. [latex]{x}^{2}[/latex] = yz D. [latex]{y}^{2}[/latex] = zx

Answer - Option D
Explanation -
SI = [latex]\frac{P * T * R}{100}[/latex]
Y = [latex]\frac{x * T * R}{100}[/latex]
Z = [latex]\frac{y * T * R}{100}[/latex]
[latex]\frac{y}{z} = \frac{x}{y}[/latex]
[latex]{y}^{2}[/latex] = 2x
10. The difference between the simple interests received from two different banks on ? 500 for 2 years is ? 2.50. The difference between their per annum rates of interest is:

A. 0.10% B. 0.25% C. 0.50% D. 1.00%

Answer - Option B
Explanation -
[latex]\frac{500 * 2 * {R}_{1}}{100} - \frac{500 * 2 * {R}_{2}}{100}[/latex] = 2.5
[latex]10({R}_{1} - {R}_{2})[/latex] = 2.5
[latex]{R}_{1} - {R}_{2} = \frac{2.5}{10}[/latex] = 0.25

1. A sum of money was invested at a certain rate of simple interest for 2 years. Had it been invested at 1 % higher rate, it would have fetched 24 more interest. The sum of money is _____.

A. 1200 B. 1050 C. 1000 D. 9600

Answer - Option A
Explanation -
At 1% additional rate of interest, the interest is Rs 24.
[latex]\frac{P * 1 * 2}{100} = 24[/latex]
P = [latex]\frac{2400}{2}[/latex]
= Rs. 1200
2. In how many years will a sum of money double itself at 12% per annum at simple interest?

A. 8 yrs. 6 months B. 6 yrs. 9 months C. 8 yrs. 4 months D. 7 yrs. 6 months

Answer - Option
Explanation -
In case of simple interest, money is going to double, it means the interest will be equal to principal.
Present rate of interest = 12%
t= time period
p= [latex]\frac{P * 12 * t}{100} = 24[/latex]
t=[latex]\frac{100}{12} = 24[/latex]= 8 years 4 months
3. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is 31, the sum is

A. 500 B. 750 C. 1000 D. 1250

Answer - Option C
Explanation -
Given: rate of interest ‘r’ = 10% per annum
Time period ‘T’ = 3 year
Let Principle amount = X
Simple interest = [latex]\frac{X * T * r}{100} = \frac{30X}{100} = 0.3X[/latex]
Compound interest = [latex]X ({ 1 + \frac{ r}{100}})^{T} - X = X ({ 1 + \frac{10}{100}})^{3} - X[/latex]
[latex]{\frac {110}{100}}^{3} - X = 1.331X - X = 0.331X[/latex]
Given : difference between compound interest and simple interest = ?31
0.331X – 0.30X = 31
0.031X = 31
X = [latex]\frac{31}{0.031}[/latex] = 1000
Hence sum = 1000
4. In how many years will the simple interest on a sum of money be equal to the principal at the rate of [latex]16 \frac{2}{3}[/latex]% per annum ?

A. 4 B. 5 C. 6 D. 8

Answer - Option C
Explanation -
100 = 100 [latex]\frac{50}{3} * \frac{x}{100} [/latex]
x = 6 years
5. Deepak lent Rs. 8800 to Jaichand for 13 years and Rs 5500 to Kapil for 12 years on simple interest at the same rate of interest and received Rs 14430 in all from both of them as interest. The rate of interest per annum is __________.

A. 8.5 percent B. 9 percent C. 9.5 percent D. 8 percent

Answer - Option D
Explanation -
Let rate of interest be = R
SI = [latex]\frac{PTR}{100}[/latex]
Simple interest for Deepak = [latex]\frac{(8800 × 13 × R)}{100}[/latex]
Simple interest for Kapil = [latex]\frac{(5500 × 12 × R)}{100}[/latex]
14430 =[latex]\frac{(8800 × 13 × R)}{100} + \frac{(5500 × 12 × R)}{100}[/latex]
14430 = 1144R + 660R
14430 = 1804 R
R = [latex]\frac{14430}{1804}[/latex] = 7.99 ~ 8%
6. Deepinder lent Rs 8200 to Jairaj for 16 years and Rs 4900 to Karna for 15 years on simple interest at the same rate of interest and received Rs 19446.5 in all from both of them as interest. The rate of interest per annum is:

A. 10% B. 10.5% C. 9.5% D. 11%

Answer - Option C
Explanation -
Let the rate of interest be r
We know that Simple interest = [latex]\frac{Prt}{100}[/latex]
Total simple interest = 19446.5
= 19446.5 = [latex]\frac{8200 * 16 * r}{100} = \frac{4900 * 15 * r}{100}[/latex]
1944650 = 131200r + 73500r
204700r = 1944650
R = 9.5
7. If a sum of money amounts to ? 12,900 and ? 14,250 at the end of 4th year and 5th year respectively at a certain rate of simple interest, then the rate of interest is

A. 10% B. 12% C. 18% D. 20%

Answer - Option C
Explanation -
SI Rate = [latex]\frac{(14250 - 12900) * 100}{12900 * 5 - 14250 * 4}[/latex]% = 18%
8. P borrowed a sum of money form Q at simple interest the rate of interest is 10% p.a. for the first 2 years and 12% p.a. for the next 3 years and 15% p.a. thereafter. If paid 5332 as interest after 7 years, then find the sum (in)

A. 5200 B. 6200 C. 4200 D. 3200

Answer - Option B
Explanation -
Let the sum be p
[latex]\frac {p * 2 * 10}{100} + \frac {p * 3 * 12}{100} + \frac {p * 2 * 15}{100}[/latex]= 5332
[latex]\frac {20p}{100} + \frac{36p}{100} + \frac {30p}{100}[/latex]= 5332
[latex]\frac {86p}{100}[/latex] = 5332
P = 5332 × [latex]\frac {100}{86p}[/latex] = Rs. 6200
9. Ram earned Rs 4800 as simple interest on a certain sum of money in four years at 8% rate per annum. What would be the compound interest Ram will receive if he invest thrice the sum and at the same rate of interest in two years?

A. 2031 B. 4546 C. 8488 D. None of these

Answer - Option D
Explanation -
Principal= [latex]\frac{4800 * 100}{8 * 4}[/latex] = RS 15000
thrice the sum is invested ,
so Principal becomes = 15000*3 = 45000
Compound Interest = 45000[latex](\frac{1 + 8{100}) 2 - 1[/latex]
Compound Interest= 45000 [latex](\frac{729}{625}) - 1[/latex]
= 45000* [latex]\frac{104}{625}[/latex]
= Rs 7488
10. A man borrowed some money from a private organization at 8% simple interest per annum. He lent 75% of his money to another person at 10% compound interest per annum. If the difference between the CI on 3/4 th amount and the SI on whole amount, that was borrowed is Rs 561.5 in 4 years. Then Find the total amount Borrowed?

A. 30000 B. 23000 C. 20000 D. 35000

Answer - Option C
Explanation -
CI in 1 year= 10%
CI in 2 years= 21%
For 4 years @ 10%= 21+21+ [latex]\frac{21*21}{100}[/latex] = 46.41%
SI in 1 year= 8%
SI in 4 years= 32%
Now,
46.41% of 3/4P – 32% of P = 561.5
= 46.41% of 3P – 32 *4P%= 561.5*4
= 139.23%P – 128P%= 2246
= 11.23P%= 2246
P= [latex]\frac{2246*100*100}{1123}[/latex] = Rs 20000

1. On what sum of money will the difference between the simple interest and the compound interest for 2 years at 8% per annum be equal to 8?

A. Rs. 1200 B. Rs. 1300 C. Rs. 1250 D. Rs. 1350

Answer - Option C
Explanation -
Difference between CI and SI for two years = [latex]P{(\frac{R}{100})}^{2}[/latex]P(R/100)2
Difference = [latex]\frac{P * 8* 8}{10000}[/latex]
8 = [latex]\frac{P * 8* 8}{10000}[/latex]
P = [latex]\frac{10000}{8}[/latex] = Rs. 1250
2. The principal that amounts to Rs.4913 in 3 years at [latex]6 \frac{1}{4}[/latex] % per annum compound interest compounded annually, is :

A. Rs. 4076 B. Rs. 4085 C. Rs. 3096 D. 4098

Answer - Option D
Explanation -
Principle = Rs. [latex]6 \frac{4913}{{(1 + \frac {25}{4 * 100})}^{3}}[/latex] = Rs. [latex](4913 * \frac{16}{17} * \frac{16}{17}* \frac{16}{17})[/latex] = Rs. 4096
3. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ? 31, the sum is

A. 500 B. 750 C. 1000 D. 1250

Answer - Option C
Explanation -
Given: rate of interest ‘r’ = 10% per annum
Time period ‘T’ = 3 year
Let Principle amount = X
Simple interest = [latex]\frac{X * T * r}{100} = \frac{30X}{100} = 0.3X[/latex]
Compound interest =
[latex]X{1 + \frac{r}{100}}^{T} - X = X {1 + \frac{10}{100}}^{3} - X[/latex]
[latex]X{\frac{110}{100}}^{T} - X = 1.331X -X = 0.331X[/latex]
Given : difference between compound interest and simple interest = ?31
[latex]\rightarrow[/latex] 0.331X – 0.30X = 31
[latex]\rightarrow[/latex] 0.031X = 31
[latex]\rightarrow[/latex] X = [latex]\frac{31}{0.031}[/latex] = 1000
Hence sum = 1000
4. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs.1. the sum (in Rs.) is:

A. 625 B. 630 C. 640 D. 650

Answer - Option A
Explanation -
Let the sum be Rs.x. Then,
C.I. = [latex]x{(1 + \frac{4}{100})}^{2} -x = (\frac{676}{625}x - x) = \frac{51}{625}x[/latex]
S.I. = [latex](\frac{x * 4 * 2}{100}) = \frac{2x}{25}[/latex]
i.e, [latex](\frac{51x }{100}) - \frac{2x}{25}[/latex] = 1
or x = 625
5. At some rate per annum, the compound interest for 2 years on Rs. 1500 is ? 449.40 The rate of interest per annum is :

A. 10% B. 12% C. 14% D. 15%

Answer - Option C
Explanation -
A = P [latex]x{(1 + \frac{R}{100})}^{T} [/latex]
= 1500 + 449.40
= 1500 [latex]x{(1 + \frac{R}{100})}^{2} [/latex]
= [latex]\frac{1949.4}{15000} = {(1 + \frac{R}{100})}^{2}[/latex]
= [latex]{\frac{57}{50}}^{2} = {(1 + \frac{R}{100})}^{2}[/latex]
= 1 + [latex] \frac{R}{100} = \frac{57}{50} [/latex]
[latex] \frac{R}{100} = \frac{57}{50} [/latex] R = 14
6. The compound interest on 6,400 for 2 years at[latex]7 \frac{1}{2}[/latex] per annum is (in Rs)

A. 1016 B. 996 C. 976 D. 966

Answer - Option B
Explanation -
C.I.= 6400 [latex]{(\frac{1 + 15}{200})}^{2} - 1[/latex] = 6400 × 0.155625 = Rs.996
7. A sum of money becomes 1.331 times in 3 years as compound interest. The rate of interest is

A. 8% B. 7.5% C. 10% D. 50%

Answer - Option C
Explanation -
If principal = Rs 1000 ; amount = Rs 1331
[latex]\rightarrow[/latex] A = [latex]P{(1 + \frac{r}{100})}^{T} [/latex]
[latex]\rightarrow[/latex] [latex]\frac{1331}{1000} = {(1 + \frac{r}{100})}^{3}[/latex]
[latex]\rightarrow[/latex] [latex]{\frac{1331}{1000}}^{3} = {(1 + \frac{r}{100})}^{3}[/latex]
[latex]\rightarrow[/latex] 1 + [latex]\frac{r}{100} = \frac{1}{10}[/latex]
[latex]\rightarrow[/latex] [latex]\frac{r}{100} = \frac{1}{10}[/latex]
[latex]\rightarrow[/latex] r = [latex]\frac{1}{10} * 100[/latex] = 10%
Hence Option C is correct
8. Simple interest on a certain sum of money for 3 years at 18 % per annum is half the compound interest on Rs. 9000 for 2 years at 10 % per annum. The sum placed on simple interest is :

A. Rs 3500 B. Rs 875 C. Rs 1750 D. Rs 1400

Answer - Option C
Explanation -
Compound interest= [latex]P{(1 + \frac{r}{100})}^{2} - P [/latex]
Compound interest at 9000 for two year= [latex]9000 {(1 + \frac{10}{100})}^{2} - 9000 [/latex]
CI = 9000 * 1.1 * 1.1 - 9000
CI= 10890 - 9000 = 1890
Simple interest = [latex]\frac{CI}{2} = \frac{1890}{2} = 945[/latex]
SI= [latex]\frac {P * r * t}{100}[/latex]
945 = [latex]\frac {P * 18 * 3}{100}[/latex]
P = [latex]\frac {945 * 100}{18 * 3} = 1750[/latex]
Sum placed on simple interest= 1750
9. If the amount received at the end of [latex]{2}^{nd}[/latex] and [latex]{3}^{rd}[/latex] year at compound interest on a certain Principal is Rs 30250, and Rs 33275 respectively, what is the rate of interest?

A. 10% B. 5% C. 20% D. 16%

Answer - Option A
Explanation -
Let the rate of interest be r% and sum be p, then
We have,
P + CI of 3 years = Rs 30250 ----------(i)
P + CI of 2 years = Rs 33275 ----------(ii)
Subtracting (ii) from (i), we get
CI of [latex]{3}^{rd}[/latex] year = 33275 – 30250
= Rs 3025
Thus, the CI calculated in the [latex]{3}^{rd}[/latex] year which is Rs.3025 is basically the amount of interest on
the amount generated after 2 years, which is Rs 30250
? Rate of interest = [latex]\frac{3025 * 100}{30250}[/latex]
= 10%
10. There is 40% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 20000 after 3 years at the same rate?

A. Rs 3152.5 B. Rs. 6305 C. Rs 7881.25 D. Rs 4728.75

Answer - Option A
Explanation -
As there is 40% increase in an amount in 8 years,
So yearly increase in amount = [latex]\frac{40}{8}[/latex] = 5% = rate of interest
Compound interest = [latex]P{(1 + \frac{r}{100})}^{t} - P [/latex]
Compound interest = [latex]20000 {(1 + \frac{5}{100})}^{3} - 20000 [/latex]
= 20000 * [latex]\frac{21}{20} * \frac{21}{20} * \frac{21}{20} - 20000[/latex]
= 23152.5 - 20000
Compound interest = 3152.5

1. A man borrowed some money from a private organization at 8% simple interest per annum. He lent 75% of his money to another person at 10% compound interest per annum. If the difference between the CI on 3/4 th amount and the SI on whole amount, that was borrowed is Rs 561.5 in 4 years. Then Find the total amount Borrowed?

A. 30000 B. 23000 C. 20000 D. 35000

Answer - Option C
Explanation -
CI in 1 year = 10%
CI in 2 years = 21%
For 4 years @ 10%= 21+21+[latex]\frac{21*21}{100}[/latex] = 46.41%
SI in 1 year = 8%
SI in 4 years = 32%
Now,
46.41% of [latex]\frac{3}{4P}[/latex] – 32% of P = 561.5
= 46.41% of 3P – 32 *4P% = 561.5*4
= 139.23%P – 128P% = 2246
= 11.23P% = 2246
P = 2246 * 100 * [latex]\frac{100}{1123}[/latex] = Rs 20000
2. The compound interest on Rs.2400 at 10% per annum, for a certain period of time is Rs.504. Find the time in years.

A. 15 B. 2.5 C. 2 D. 3

Answer - Option B
Explanation -
CI = P([latex]{(1 + \frac{R}{100})}^{t} -1[/latex])
Where, P is principal, t is time in years and r is rate% p.a.
Given, P = Rs. 2400, rate = 10% and CI = Rs. 504
504 = 2400([latex]{(1 + \frac{10}{100})}^{t} -1[/latex])
[latex]\rightarrow[/latex] 0.21 =[latex]{(1.1)}^{t}[/latex] – 1
[latex]\rightarrow[/latex] 1.21 = [latex]{(1.1)}^{t}[/latex]
[latex]\rightarrow[/latex] [latex]{(1.1)}^{2} = {(1.1)}^{t}[/latex]
[latex]\rightarrow[/latex] t = 2 years
3. if the difference between compound interest at 8 %p.a. and simple interest at 13/2 if the difference between compound interest at 8p.a. and simple interest at 13/2 p.a. on a certain sum of money for 2 years is 1820, then find the sum (in rs.)?p.a. on a certain sum of money for 2 years is 1820, then find the sum (in rs.)?

A. 50000 B. 54000 C. 25000 D. 41000

Answer - Option A
Explanation -
Let, P = 100
C.L = 100 [latex]{(1.08)}^{2}[/latex] – 100 = 16.64
S.L = 2 × [latex]\frac{13}{200}[/latex] × 100 = 13
When the difference is 3.64, the sum is 100
When the difference is 1820 the sum is
[latex]\frac{100}{3.64}[/latex] × 1820 = 50000
Alternative method.
P([latex]{(1 + \frac{8}{100})}^{2}[/latex] – 1) - [latex]\frac{P * 2 * 13}{100 * 2}[/latex]= 1820
P = 50, 000
4. 2000 was lent at compound interest, interest being compounded annually for 3 years. The respective rates of interest for the first, the second and the third years are 10% pa 20% pa and 30% pa had the sum of 2000 been lent at 20% pa simple interest for 3 years, how much more/less interest would have been realized?

A. 116 more B. 116 less C. 232 less D. 232 more

Answer - Option C
Explanation -
The value that the sum of 2000 would have amounted to at the end of 3 years
= 2000 (1 + [latex]\frac{10}{100}[/latex]) (1 +[latex]\frac{20}{100}[/latex] ) (1 + [latex]\frac{30}{100}[/latex] ) = 2000 (1.716)
The value that sum would have amounted to the end of 3 years had it been lent at 20% pa simple interest =2000
(1 + [latex]\frac{3(20)}{100}[/latex]) = 2000 (1.6)
The difference of the interests in the two cases = The difference of the amounts in the two cases = 2000(1.716 – 1.6) = 2000 (0.116) = 232
The sum amounts to a higher value when lent at compound interest than when lent at simple interest
Had the sum been lent at 20% pa simple interest for 3 years, the interest realized would have been 232 less.
5. Mr. Girish deposits an amount of Rs 96,000 to obtain compound interest at the rate of 10 p.c.p.a. for 3 years. What total amount will Mr. Girish get at the end of 3 years?

A. Rs 1, 24,776 B. Rs 1, 25,776 C. Rs 1, 26,776 D. Rs 1, 27,776

Answer - Option D
Explanation -
C.I. = 96000 × [latex]\frac{33.10}{100}[/latex] = 31776
So amount = 96000 + 31776
= 127776
6. What would be the compound interest accrued on an amount of Rs. 14,800 @13.5 p.c.p.a.at the end of two years? (Rounded off to two digits after decimal)

A. Rs 4, 136.87 B. Rs 4, 306.81 C. Rs 3, 032.18 D. Rs 4, 265.73

Answer - Option D
Explanation -
CI = 14800(1+[latex]{(\frac{13.5}{100})}^{2}[/latex] -1)
= 14800[1.288225 - 1] = 14800 × 0.288225
= Rs 4265.7
7. What would be the compound interest obtained on an amount of ? 1,210 at the rate of 6 p.c.p.a. after a year?

A. 70.5 B. 74.6 C. 73.8 D. 72.6

Answer - Option D
Explanation -
C.I. for one year
= [latex]\frac{PTR}{100}[/latex]
= [latex]\frac{1200 * 6 * 1}{100}[/latex]
= Rs. 72.6
8. The compound interest accrued in two years on a principal of ?15,800 is ?7716.72. What is the rate of interest pcpa?

A. 22% B. 16 % C. 18% D. Cannot be determined

Answer - Option A
Explanation -
Amount = P[latex]({1 + \frac{r}{100}}^{2}[/latex]
15800 + 7716.72 = 15800 [latex]{(1 + \frac{r}{100}})^{2}[/latex]
or, 23516.72 = 15800 [latex]{(1 + \frac{r}{100}})^{2}[/latex]
[latex]\frac{23516.72}{15800} = {(1 + \frac{r}{100}})^{2}[/latex]
[latex]1.4884 = {(1 + \frac{r}{100}})^{2}[/latex]
Taking square root of both the sides,
[latex]1.22 = 1 + \frac{r}{100}[/latex]
122 = 100 + r
or, r = 22%
9. The simple interest accrued on a sum of certain principle is Rs 7,200 in six years at the rate of 12 p.c.p.a. What would be the compound interest accrued on that principle at the rate of 5 p.c.p.a. in 2 years?

A. Rs. 1000 B. Rs. 1025 C. Rs. 1050 D. Rs. 1075

Answer - Option B
Explanation -
Required difference = 456-272 = 184
Let the principal be Rs x . Then,
[latex]\rightarrow \frac{x * 12 * 6}{100} = Rs. 7200 \rightarrow x = Rs 10000[/latex]
Required CI = [latex] 10000 (1 + {(\frac{5}{100})}^{2} - 1)[/latex] = Rs. 1025
10. What will be compound interest on a sum of Rs. 1550 at the rate of 10 p.c.p.a. after 3 years?

A. Rs. 520.55 B. Rs. 513.05 C. Rs. 625.50 D. Rs. 605.05

Answer - Option B
Explanation -
A = 1550 [latex]{(1 + \frac{10}{100})}^{3}[/latex]
= 1550 [latex]{(\frac{11}{10})}^{3}[/latex]
= 1550 * [latex]\frac{10}{100} * \frac{10}{100} * \frac{10}{100}[/latex]
= [latex]\frac{1550 * 1331}{1000} = 2065.05[/latex]
Interest = 2063.5 – 1550 = 513.05 Rs.

1. If [latex]{cos}^{2}\alpha + {cos}^{2}\beta = 2[/latex] then the value of [latex]{tan}^{3}\alpha + {sin}^{3}\beta[/latex] is

A. -1 B. 0 C. 1 D. [latex]\frac{1}{\sqrt{3}}[/latex]

Answer - Option B
Explanation -
[latex]{cos}^{2}\alpha + {cos}^{2}\beta = 2[/latex]
The maximum value of cosx cannot be more than 1.
[latex]{cos}^{2}\alpha = 1 = {cos}^{2}\beta[/latex]
[latex]\alpha = n\pi; \beta = m\pi [/latex]{n = 0, 1, 2,......; m = 0. 1,2,.....}
[latex]{tan}^{3}\alpha + {sin}^{3}\beta = {tan}^{3}n \pi + {sin}^{5}m \pi = 0 + 0 = 0[/latex]
2. If A= tan 11º tan 29º, B = 2 cot 61º cot 79º, then:

A. A = 2B B. A = -2B C. 2A = B D. 2A = -B

Answer - Option C
Explanation -
As we know that; tan (90 - x) = cotx
A = If A = tan 11º tan 29º,
B = 2 cot 61° cot 79º = 2tan(90 - 61º)tan(90 - 79º)= 2 tan 11º tan 29º
[latex]\Rightarrow [/latex] B = 2A
3. The value of tan 10° tan 15° tan 75° tan 80° is

A. 0 B. 1 C. -1 D. 2

Answer - Option B
Explanation -
We know that tan (90 - x) = cotx
Hence tan10° = tan (90 - 80) = cot 80°
& tan15 = tan (90 - 15) = cot 75°
Therefore
tan 10° tan 15° tan 75° tan 80°
= cot80° cot75° tan 75° tan 80° = 1
as cotx = [latex]\frac{1}{tanx }[/latex]
[latex]\Rightarrow [/latex] cotx.tanx= 1
4. If [latex]{sec}^{2}\theta + {tan}^{2}\theta= 7, then the value of \theta when 0° \leq \theta \leq 90°[/latex]

A. 60° B. 30° C. 0° D. 90°

Answer - Option A
Explanation -
[latex]{sec}^{2}\theta + {tan}^{2}\theta = 7[/latex]
[latex]1 + {tan}^{2}\theta + {tan}^{2}\theta = 7[/latex]
[latex]2{tan}^{2}\theta = 7 - 1 = 6[/latex]
[latex]{tan}^{2}\theta= \frac {6}{3} = 3[/latex]
[latex]tan\theta = \sqrt {3} [/latex]
[latex]\theta = 60°[/latex]
5. If [latex]{tan}^{2}\alpha = 1 + 2{tan}^{2}\beta (\alpha, \beta[/latex] are positive acute angles), then [latex]\sqrt{2}cos \alpha - cos\beta [/latex] is equal to

A. 0 B. [latex]\sqrt{2}[/latex] C. 1 D. -1

Answer - Option A
Explanation -
As [latex]{sec}^{2}x = 1 + 2{tan}^{2}x [/latex]
[latex]{tan}^{2}\alpha = 1 + 2{tan}^{2}\beta[/latex]
[latex]\Rightarrow {sec}^{2}\alpha - 1 = 1 + 2({sec}^{2}\beta - 1) [/latex]
[latex]\Rightarrow {sec}^{2}\alpha - 1 = 1 + 2{sec}^{2}\beta - 2 [/latex]
[latex]\Rightarrow {sec}^{2}\alpha = 2{sec}^{2}\beta [/latex]
[latex]\frac {1}{{cos}^{2}\alpha} = \frac {2}{{cos}^{2}\beta}[/latex]
[latex]\Rightarrow {cos}^{2}\beta = 2{cos}^{2}\alpha [/latex]
[latex]\Rightarrow 2{cos}^{2}\alpha - {cos}^{2}\beta = 0 [/latex]
[latex]\Rightarrow (\sqrt{2}cos\alpha - cos\beta)(\sqrt{2}cos\alpha + cos\beta) = 0 [/latex]
[latex]\Rightarrow (\sqrt{2}cos\alpha - cos\beta) = 0 [/latex]
6. If [latex]2({cos}^{2}\theta - {sin}^{2}\theta) = 1 (\theta[/latex] s a positive acute angle), then cot? is equal to

A. -[latex]\sqrt{3} [/latex] B.[latex]\frac {1}{\sqrt{3}}[/latex] C. 1 D. [latex]\sqrt{3} [/latex]

Answer - Option D
Explanation -
[latex]2({cos}^{2}\theta - {sin}^{2}\theta) = 1 [/latex]
[latex]\Rightarrow{cos}^{2}\theta - (1 - {cos}^{2}\theta )= \frac {1}{2} [/latex]
[latex]\Rightarrow 2{cos}^{2}\theta = 1 + \frac {1}{2} = \frac {3}{2}[/latex]
[latex]\Rightarrow 2{cos}^{2}\theta = \frac {3}{4}[/latex]
[latex]\Rightarrow 2{sec}^{2}\theta = \frac {4}{3}[/latex]
[latex]\Rightarrow 1 + {tan}^{2}\theta = \frac {4}{3}[/latex]
[latex]\Rightarrow tan\theta = \sqrt {3} = cot\theta = \sqrt {3}[/latex]
7. he numerical value of [latex]\frac{5}{{sec}^{2}\theta} + \frac{2}{1 + {cot}^{2}\theta } + 3 {sin}^{2}\theta[/latex] is:

A. 5 B. 2 C. 3 D. 4

Answer - Option A
Explanation -
[latex]\frac{5}{{sec}^{2}\theta} + \frac{2}{1 + {cot}^{2}\theta } + 3 {sin}^{2}\theta = 5 {cos}^{2}\theta + 2 {sin}^{2}\theta + 3 {sin}^{2}\theta = 5[/latex]
8. If [latex] tan\theta = \frac {4}{3}[/latex], then the value of [latex]\frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta}[/latex] is

A. 0.5 B. -0.5 C. 3.0 D. -3.0

Answer - Option C
Explanation -
[latex] tan\theta = \frac {4}{3}[/latex](Given)
i.e, [latex]\frac{3sin\theta + 2cos\theta}{3sin\theta - 2cos\theta} = \frac{3tan\theta + 2}{3tan\theta - 2}[/latex]
[latex]\frac{3 * \frac{4}{3} + 2}{3 * \frac{4}{3} - 2} = \frac{4 + 2}{4 - 2} = 3[/latex]
9. The numerical value of [latex]\frac{9}{{cosec}^{2}\theta} + 4{cos}^{2}\theta + \frac{5}{1 + {tan}^{2}\theta}[/latex] will be?

A. 7 B. 9 C. 4 D. 5

Answer - Option B
Explanation -
[latex]\frac{9}{{cosec}^{2}\theta} + 4{cos}^{2}\theta + \frac{5}{1 + {tan}^{2}\theta}[/latex]
= [latex]9{sin}^{2}\theta + 4{cos}^{2}\theta + \frac{5}{{sec}^{2}\theta}[/latex]
= [latex]9{sin}^{2}\theta + 4{cos}^{2}\theta + 5{cos}^{2}\theta [/latex]
= [latex]9{sin}^{2}\theta + 9{cos}^{2}\theta [/latex]
= [latex]9({sin}^{2}\theta + {cos}^{2}\theta) [/latex]
= 9 * 1 = 9
10. If [latex]\frac{2sin\theta - cos\theta}{cos\theta + sin\theta } = 1[/latex] then value of cot ? is :

A. [latex]\frac{1}{2}[/latex] B. [latex]\frac{1}{3}[/latex] C. 3 D. 2

Answer - Option A
Explanation -
[latex]\frac{2sin\theta - cos\theta}{cos\theta - sin\theta} = 1 \Rightarrow 2sin\theta - cos\theta = cos\theta + sin\theta[/latex]
[latex]sin\theta = 2cos\theta \Rightarrow cot\theta = \frac{1}{2} (\frac{cos\theta}{sin\theta} = cot\theta)[/latex]

1. The distance between two pillars of length 16 metres and 9 metres is x metres. If two angles of elevation of their respective top from the bottom of the other are complementary to each other, then the value of x (in metres) is

A. 15 B. 16 C. 12 D. 9

Answer - Option C
Explanation -
If the angles are complementary then the distance between two pillars of length 'a' and 'b' is vab.
In this question,
x = [latex]\sqrt{(16*9)} = \sqrt{144}[/latex] = 12
2. The length of the shadow of a vertical tower on level ground increases by 10 meters when the altitude of the sun changes from 45 ° to 30 °. Then the height of the tower is

A. [latex]5 \sqrt{3} [/latex] m B. [latex]10 (\sqrt{3} + 1)[/latex] m C. [latex]5 (\sqrt{3} + 1)[/latex] m D. [latex]10 \sqrt{3} [/latex] m

Answer - Option C
Explanation -

The ratio of sides in these triangles are shown in red colour.
[latex] (\sqrt{3} - 1)[/latex] unit = 10m
1 unit = [latex]\frac{10} {(\sqrt{3} - 1)}[/latex] m
h = 1 unit = [latex]\frac{10 (\sqrt{3} - 1)} {2}[/latex] m
h = [latex]5 (\sqrt{3} + 1)[/latex] m
3. A man standing at a point C is watching the top of a tower, which makes an angle of elevation of 30 °. The man walks some distance towards the tower and then his angle of elevation of the top of the tower is 60 °. If the height of the tower is 30 m, then the distance he moves is

A. 22 m B. [latex]22 \sqrt{3} [/latex] m C. 20 m D. [latex]20 \sqrt{3} [/latex] m

Answer - Option D
Explanation -

B 60º D 30º C
AB = Tower = 30 m
CD = X m
ACB = 30º
ADB = 60º
From ABD
tan 60º = [latex]\frac{AB}{BD}[/latex]
[latex]\sqrt {3} = \frac{30}{BD}[/latex]
BD [latex] = \frac{30}{\sqrt {3}} = 10\sqrt {3}[/latex]m
From ? ABC
tan 300 = [latex]\frac{AB}{BD}[/latex]
[latex] = \frac{1}{\sqrt {3}} = \frac{30}{10\sqrt {3} + x}[/latex]
[latex]x = 30\sqrt {3} - 10\sqrt {3} [/latex]
[latex]x = 20\sqrt {3}[/latex] m
4. If a pole of 12 m height casts a shadow of [latex]4 \sqrt{3} [/latex] m long on the ground, then the sun’s angle of elevation at that instant is

A. 30º B. 60º C. 45º D. 90º

Answer - Option B
Explanation -

Tan C = [latex] = \frac{AB}{BC} = \frac{12}{4\sqrt {3}} = \frac {3}{\sqrt {3}} = \sqrt {3}[/latex]
or, C = 60º
5. From two points on the ground and lying on a straight line through the foot of a pillar, the two angles of elevation of the top of the pillar are complementary to each other. If the distances of the two points from the foot of the pillar are 12 meters and 27 meters and the two points lie on the same side of the pillar, then the height (in meters) of the pillar is

A. 12 B. 18 C. 15 D. 16

Answer - Option B
Explanation -

Let, ACB = [latex]\theta[/latex]
i.e, ACB = 90 - [latex]\theta[/latex]
BC = 12 meter
BD = 27 meter
AB = Pillar = h meter
From [latex]\triangle[/latex]ABC,
[latex]tan \theta = \frac{AB}{BC} = \frac{h}{12}[/latex] .....(i)
From ABC,
[latex]tan (90 - \theta) = \frac{AB}{BD} \Rightarrow cot \theta \frac{h}{27}[/latex] .....(ii)
i.e, [latex]tan \theta .cot \theta = = \frac{h}{12} * \frac{h}{27}[/latex]
[latex]{h}^{2}[/latex] = 12 * 27
h = [latex]\sqrt{12 * 27} [/latex]
= [latex]\sqrt{2 * 2 * 3 * 3 * 3 * 3} [/latex]
= 2 * 3 * 3
6. The shadow of a tower standing on a level plane is found to be 40m longer when the sun's altitude is 45 °, than when it is 60 °. The height of the tower is

A. [latex]30 (3 + \sqrt{3}) [/latex] m B. [latex]40 (3 + \sqrt{3}) [/latex] m C. [latex]20 (3 + \sqrt{3}) [/latex] m D. [latex]10 (3 + \sqrt{3}) [/latex] m

Answer - Option C
Explanation -

Let, ACB = 60º; BC = Xmeters
CD = 40 m, AB = Tower = h metre
From [latex]\triangle[/latex] ABC
[latex]tan 60º = \frac{AB}{BC} \Rightarrow \sqrt {3} \frac{h}{x}[/latex]
[latex] h = \sqrt {3}x[/latex]
From [latex]\triangle[/latex] ABD
[latex]tan 45º = \frac{AB}{BD} [/latex]
[latex] \Rightarrow h = x + 40 = \frac{h}{\sqrt {3}} + 40[/latex]
[latex] \Rightarrow h = \frac{h}{\sqrt {3}} = 40[/latex]
[latex] \Rightarrow \frac{40\sqrt {3} }{\sqrt {3} - 1} + 40[/latex]
[latex] = \frac{40\sqrt {3} (\sqrt {3} + 1)}{(\sqrt {3} - 1)(\sqrt {3} + 1)} + 40[/latex]
[latex] = \frac{40\sqrt {3} (\sqrt {3} - 1)}{3 - 1} + 40[/latex]
= [latex]20 (3 + \sqrt{3}) [/latex] m
7. A ladder Leaning against a house has angle of elevation 60º and the foot of the ladder is 6.5 metres from the house. The length of the ladder is

A. [latex]\frac{13}{\sqrt {3}} = 40[/latex] m B. 13 m C. 15 m D. 3.25 m

Answer - Option B
Explanation -

AC = Ladder, AB = House
In [latex]\triangle[/latex] ABC
cos 60º = [latex]\frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{6.5}{AC} [/latex]
[latex] \Rightarrow [/latex] = 2 * 6.5
= 13 m
8. The angle of elevation of the sun when the length of the shadow of a pole is equal to its height is :

A. 30º B. 45º C. 60º D. 90º

Answer - Option B
Explanation -

AB = BC
[latex] tan \theta = \frac{AB}{BC}[/latex]
[latex] \Rightarrow tan \theta =[/latex] tan 45º
[latex] \Rightarrow \theta =[/latex] 45º
9. The cliff of a mountain is 180 m high and the angles of depression of two ships on the either side of cliff are 30º and 60º respectively. What is the distance between the two ships? ([latex]\sqrt{3}[/latex] = 1.732)

A. 400 m B. 400 [latex]\sqrt{3} [/latex] m C. 415.69 m D. 398.62 m

Answer - Option C
Explanation -

AC = Cliff = 180 m
In [latex]\triangle[/latex] ABC
tan 60º = [latex]\frac{AD}{BD}[/latex]
[latex]\Rightarrow \sqrt {3} = \frac{180}{BD}[/latex]
[latex]BD = \frac{180}{\sqrt {3}} = 60\sqrt {3}[/latex] m
From [latex]\triangle[/latex] ACD,
tan 30º = [latex]\frac{AD}{CD}[/latex]
[latex]\frac{1}{\sqrt {3}} = \frac{180}{CD}[/latex]
CD= 180 [latex]\sqrt{3} [/latex]metre
[latex]\triangle[/latex] BC = BD + DC
= 60 [latex]\sqrt{3} [/latex] + 180 [latex]\sqrt{3} [/latex]
= 240 [latex]\sqrt{3} [/latex]metre
=(240 x 1.732) metre
= 415.69 metre
10. The angles of elevation of the top of a tree 220 meters high from two points lie on the same plane are 30 ° and 45 °. What is the distance (in metres) between the two points?

A. 193 .22 B. 144.04 C. 176.12 D. 161.05

Answer - Option D

1. The distance between the tops of two building 38 metres and 58 metres high is 52 metres. What will be the distance (in metres) between two buildings?

A. 46 B. 42 C. 44 D. 48

Answer - Option D
Explanation -

[latex]{52}^{2} = {x}^{2} + {h}^{2}[/latex]
2704 = [latex]{x}^{2} + {20}^{2} = {x}^{2} + 400[/latex]
[latex]{x}^{2}[/latex] = 2704 - 400 = 2304
x = 48 meter
2. The angle of elevation of an aeroplane from a point on the ground is 60º. After flying for 30º seconds, the angle of elevation changes to. If the aeroplane is flying at a height of 4500m, then what is the speed (in m/s) of aeroplane?

A. 50 [latex]\sqrt{3} [/latex] B. 100 [latex]\sqrt{3} [/latex] C. 200 [latex]\sqrt{3} [/latex] D. 300 [latex]\sqrt{3} [/latex]

Answer - Option B
Explanation -

AB=ED= height of aeroplane= 4500m
In triangle ABC
Tan60° = [latex]\frac{AB}{BC}[/latex]
[latex]\sqrt{3} = \frac{4500}{BC}[/latex]
BC=1500 [latex]\sqrt{3} [/latex] m
In triangle CED
In triangle CED
Tan30 ° = [latex]\frac{ED}{DC}[/latex]
[latex]\frac{1}{\sqrt{3}} = \frac{4500}{BC}[/latex]
DC=4500 [latex]\sqrt{3} [/latex] m
Distance covered in 30 second = BC = 4500 [latex]\sqrt{3} m-1500 \sqrt{3} = 3000 \sqrt{3} [/latex] m
Speed of aeroplane= [latex]\frac {3000 \sqrt{3}}{30} [/latex] m
3. A kite is flying in the sky. The length of string between a point on the ground and kite is 420 m. The angle of elevation of string with the ground is 30º Assuming that there is no slack in the string, then what is the height (in metres) of the kite?

A. 210 B. 140 [latex]\sqrt{3} [/latex] C. 210 [latex]\sqrt{3} [/latex] D. 150

Answer - Option A
Explanation -

Here length of string= AC=420m
In triangle ABC , AB= height of kite
So sin30 ° = [latex]\frac{AB}{AC}[/latex]
[latex]\frac{1}{2} = \frac{AB}{420}[/latex]
AB=210 m
So height of kite is 210m
4. If the angle of elevation of sun changes from 45º to 60º, the length of the shadow of a pillar decreases by 20 meters. The height of the pillar is:

A. [latex]20\sqrt{3}(\sqrt{3} + 1) [/latex] B. [latex]20\sqrt{3}(\sqrt{3} - 1) [/latex] C. [latex]10\sqrt{3}(\sqrt{3} - 1) [/latex] D. [latex]10\sqrt{3}(\sqrt{3} + 1) [/latex]

Answer - Option D
Explanation -
Let h be the height of the pillar. Then in [latex]\triangle[/latex]ABC
Tan 45° = [latex]\frac{h}{BD + 20}[/latex]
h = BD + 20
In [latex]\triangle[/latex]ABd,
Tan 60° = [latex]\frac{h}{BD} = \sqrt{3}[/latex]
h = BD [latex]\sqrt{3} = (h -20) \sqrt{3}[/latex]
h = [latex]h \sqrt{3} - 20 \sqrt{3}[/latex]
h = [latex]\frac{20 \sqrt{3}}{\sqrt{3} - 1} = 10\sqrt{3}(\sqrt{3} + 1) [/latex]
5. From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

A. 149 m B. 156 m C. 173 m D. 200 m

Answer - Option C
Explanation -

[latex]\frac{AB}{AP} = Tan 30° = \frac{1}{\sqrt{3}}[/latex]
AP = AB [latex]\sqrt{3}[/latex]
AP = 100 [latex]\sqrt{3}[/latex] = 173 m
6. The angle of elevation of the sun, when the length of the shadow of a tree [latex]\sqrt{3} [/latex] times the height of the tree, is:

A. 30º B. 45º C. 60º D. 90º

Answer - Option A
Explanation -
Let AB be the tree and AC be its shadow.

Let [latex]\triangle[/latex] ACB = [latex]\theta, \frac{AC}{AB} = \sqrt{3} \rightarrow cot \theta = \sqrt{3}[/latex]
[latex]\theta = 30º[/latex]
7. A tree is cut partially and made to fall on ground. The tree however does not fall completely and is still attached to its cut part. The tree top touches the ground at a point 10m from foot of the tree making an angle of 30 °. What is the length of the tree?

A. 10 [latex]\sqrt{3}[/latex] m B. [latex]\frac{10}{\sqrt{3}}[/latex] m C. 10 m D. [latex]\frac{10}{\sqrt{2}}[/latex]

Answer - Option A
Explanation -
AC = [latex]\frac{10}{COS 30 °} = 20\sqrt{3} m[/latex]
AB = 10 tan 30 ° = [latex]10 \sqrt{3} m[/latex]
Length of the tree = AC + BC = [latex]\frac{30}{\sqrt{3}} = 10\sqrt{3} m[/latex]
8. At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is [latex]\frac{1}{5}[/latex]. On walking 138 meters towards the monument the secant of the angle of elevation is found to be [latex]\frac{\sqrt{193}}{12}[/latex]. The height of the monument (in meter) is:

A. 42 B. 49 C. 35 D. 56

Answer - Option A
Explanation -

tan ? D = [latex]\frac{AB}{BD} = \frac{1}{5}[/latex] Now,
sec ? C = [latex]\frac{AB}{BD} = \frac{\sqrt{193}}{12}[/latex]
AB = [latex]\sqrt{{AC}^{2} - {BC}^{2}} = \sqrt{193 - 144} = 7[/latex] Perpendicular AB in both cases must be same, therefore
[latex]\frac{AB}{BD} = \frac {1}{5} * \frac{7}{7} = \frac {7}{35}[/latex]
BD - BC = 35r - 12r = 23r = 138
AB = 7r = 7 * [latex]\frac {138}{23} = 42 meter[/latex]
9. The angle of elevation of the top of a tower from a point A on the ground is 30 °. On moving a distance of 20 meters towards the foot of the tower to a point B, the angle of elevation increases to 60 °. The height of the tower is:

A. [latex]\sqrt{3} [/latex] m B. [latex]5 \sqrt{3} [/latex] m C. [latex]10 \sqrt{3} [/latex] m D. [latex]20 \sqrt{3} [/latex] m

Answer - Option C
Explanation -

tan 30 °= [latex]\frac{1}{\sqrt{3}} = \frac{h}{20 + x} \Rightarrow x = h \sqrt{3} - 20[/latex] Now,
tan 60 ° = [latex]\sqrt{6} = \frac{h}{x}\Rightarrow x = \frac {h}{\sqrt{3}}[/latex]
[latex]h \sqrt{3} - 20 = \frac {h}{\sqrt{3}}[/latex]
3h - h = 20 [latex] \sqrt{3}[/latex]
h = 10 [latex] \sqrt{3}[/latex]
10. The shadow of a tower is [latex]\sqrt{3} [/latex] times its height. Then the angle of elevation of the top of the tower is

A. 45° B. 30° C. 60° D. 90°

Answer - Option B
Explanation -

[latex]tan \theta = \frac {h}{h \sqrt{3}} = \frac {1}{ \sqrt{3}} = tan30°[/latex]
[latex]\theta = 30°[/latex]

1. A circle is touching the side BC of [latex]\triangle[/latex]ABC at P and is also touching AB and AC produced at Q and R respectively. If AQ = 6 cm, then perimeter of [latex]\triangle[/latex] ABC is

A. 6 cm B. 10 cm C. 12 cm D. 18 cm

Answer - Option C
Explanation -
Given: A circle is touching the side BC of [latex]\triangle[/latex] ABC at P and is also touching AB and AC produced at Q and R respectively. If AQ = 6 cm

By the property of triangle
AQ = AR = 6 cm; BQ = BP and CR= CP
AQ +AR = AB + BQ + AC + CR
Hence 12 = AB + AC + BP + CP = AB + AC + BC
Hence AB + AC + BC = 12 cm
Perimeter of triangle = 12 cm
2. [latex]\triangle[/latex] ABC is similar to [latex]\triangle[/latex] PQR. Length of AB is 18 cm and length of the corresponding side PQ is 12 cm. If area of [latex]\triangle[/latex] ABC is 324 sq cm, what is the area of [latex]\triangle[/latex] PQR?

A. 72 sq cm B. 144 sq cm C. 36. sq cm D. 487.5 sq cm

Answer - Option B
Explanation -
Since, [latex]\triangle[/latex]ABC is similar to [latex]\triangle[/latex]PQR. And, AB = 18 cm and PQ = 12 cm
[latex]\frac{ar (\triangle ABC)}{ar (\triangle PQR)} = \frac{{AB}^{2}}{{PQ}^{2}}[/latex]
[latex]\frac{324}{ar (\triangle PQR)} = \frac{{18}^{2}}{{12}^{2}}[/latex]
[latex]\frac{324}{ar (\triangle PQR)} = \frac{324}{144}[/latex]
ar (\triangle PQR) = 144[latex]{cm}^{2}[/latex]
3. E is the mid-point of the median AD of [latex]\triangle[/latex] ABC. BE is joined and produced to meet AC at F. F divides AC in the ratio :

A. 3 : 2 B. 1 : 2 C. 2 : 3 D. 2 : 1

Answer - Option B
Explanation -
Since, E is the mid-point of the median AD of [latex]\triangle[/latex]ABC. BE is joined and produced to meet AC at F. here, BD = DC and AE = ED

In [latex]\triangle[/latex] ADG
EF [latex]\parallel[/latex] DG
According to the theorem,
[latex]\frac{AF}{FG} = \frac{1}{1}[/latex]
FG = AF
FG = GC = AF
In [latex]\triangle[/latex] BCF
DG [latex]\parallel[/latex] BF and D is mid point of BC so G is also the mid point of FC.
FG = GC = AF
So [latex]\frac{AF}{FC} = \frac{1}{2}[/latex]
4. For a triangle, base is [latex]6 \sqrt{3}[/latex]cm and two base angles are 30 ° and 60. Then height of the triangle is

A. [latex]3 \sqrt{3}[/latex] cm B. 4.5 cm C. [latex]4 \sqrt{3}[/latex] cm D. [latex]2\sqrt{3}[/latex] cm

Answer - Option B
Explanation -

sin 30º = [latex]\frac{AC}{BC}[/latex]
[latex]\Rightarrow \frac{AC}{6 \sqrt{3}} \Rightarrow AC = 3 \sqrt{3}[/latex]
sin 60º = [latex]\frac{ACD}{AC}[/latex]
[latex]\Rightarrow \frac {\sqrt{3}} {2} \frac{AD}{3 \sqrt{3}} [/latex]
[latex]\Rightarrow AD = \frac{3\sqrt{3} * \sqrt{3}}{2} = 4.5 cm[/latex]
5. A straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E respectively. If the area of ?ABE be 36 sq. cm then the area of ?ACD is

A. 18 sq. cm B. 36 sq. cm C. 10 sq. cm D. 16 sq. cm

Answer - Option B
Explanation -

[latex]\triangle[/latex] DBC and [latex]\triangle[/latex] EBC lie on the same base and between same parallel lines.
[latex]\triangle[/latex] DBC = [latex]\triangle[/latex] BEC
[latex]\triangle[/latex] ABC – [latex]\triangle[/latex] ADC = [latex]\triangle[/latex] ABC – [latex]\triangle[/latex] ABE
[latex]\triangle[/latex] ADC = [latex]\triangle[/latex]ABE = 36 sq cm
6. The perpendiculars, drawn from the vertices to the opposite sides of a triangle, meet at the point whose name is

A. Incenter B. Circumcentre C. Centroid D. Orthocentre

Answer - Option D
Explanation -

O = Ortho - centre
7. The ratio of length of each equal side and the third side of an isosceles triangle is 3 : 4. If the area of the triangle is [latex]18 \sqrt{5}[/latex] square unit, the third side is

A. 16 unit B. [latex]5 \sqrt{10}[/latex] unit C. [latex]8 \sqrt{2}[/latex] unit D. 12 unit

Answer - Option D
Explanation -
Sides = 3x, 3x and 4x
Semi perimeter = [latex]\frac{3x + 3x + 4x}{2} = 5x[/latex]
area of triangle = [latex]\sqrt{5x(5x - 3x)(5x - 3x)(5x - 4x)}[/latex]
[latex]\sqrt{5x * 2x * 2x * x}[/latex]
[latex]20{x}^{4} = 18 \sqrt{5} * 18 \sqrt{5}[/latex]
[latex]20{x}^{4} = 81[/latex]
[latex]{x}^{2} = 9[/latex]
[latex]x = \sqrt {5} = 3[/latex]
Third side = 4x = 4 * 3 = 12 units
8. The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ?PSR is

A. 30° B. 15° C. 60° D. 45°

Answer - Option A
Explanation -
Side PR= QR= RS
Hence in triangle PRS ; PR = RS ; so ?P = ?S= x
PRS = 180 PRQ = 180 - 60 – 120
Now in triangle PRS = PRS + ? P + ?S= 180
120 + x + x = 180
x = 30° = ?PSR
9. ABC is an isosceles triangle With AB = AC. The side BA is produced to D such that AB = AD If ?ABC = 30 °, then BCD is equal to

A. 45° B. 90° C. 30° D. 60°

Answer - Option B
Explanation -
AB = AC = AD
Angle ABC = Angle ACB = 30°
Angle BAC = 180 – 60 = 120°
Angle DAC = 180 – 120 = 60°
Angle ADC + Angle ACD = 120
Angle ACD = 120/2 = 60°
Angle BCD = Angle ACB + Angle ACD = 30 + 60 = 90°
10. The in-radius of an equilateral triangle is of length 3 cm. Then the length of each of its medians is

A. 12 cm B. [latex]\frac{9}{2}[/latex] C. 4 cm D. 9 cm

Answer - Option D
Explanation -

Since, in equilateral triangle in-centre, centroid, circumference, and ortho-centre lie on the same point.
Therefore, centroid divides the median in the ratio 2:1
i.e, Length of each median = (2 + 1) × in-radius
= 3 × 3 = 9 cm
Thus, length of each of its medians is 9 cm.
hence, option D is correct.

1. The diagonal of a square is [latex]10 \sqrt{2}[/latex]cm. Find its perimeter?

A. 160 cm B. 80 cm C. 20 cm D. 40 cm

Answer - Option D
Explanation -
Diagonal of a square = a [latex]\sqrt{2}[/latex] = 10 [latex]\sqrt{2}[/latex] cm
a = 10 cm
Perimeter of the square = 4a
= 4 × 10
= 40 cm
2. The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, then the distance between the longer sides is

A. 10 cm B. 12 cm C. 16 cm D. 9 cm

Answer - Option D
Explanation -

We have,
Area of parallelogram = base × height
27 * 12 = 9 cm
x = [latex]\frac{27 * 12}{36} = 9 cm[/latex]
Hence, option D is correct.
3. The ratio of the areas of the circum circle and the in circle of a square is?

A. 2 : 1 B. [latex]\sqrt{2}[/latex] : 1 C. [latex]\sqrt{2} : \sqrt{3}[/latex] D. [latex]\sqrt{3} : 1[/latex]

Answer - Option A
Explanation -

Let ABCD be a square with sides ‘2a’.
AB = BC = CD = DA = 2a
AC be the diagonal of square, hence
[latex]AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{4{a}^{2} + 4{a}^{2}} = 2 \sqrt{2}a[/latex]
AO = OC = [latex]\sqrt{2} a[/latex]
AO is the radius of circum Circles of square and OP is the radius of in Circles of same square.
OP = [latex] \frac{1}{2}[/latex](AB) = a
Hence area of circum Circles = [latex]\pi (2{a}^{2}) = 2\pi {a}^{2}[/latex]
Area of in Circles = [latex]\pi ({a}^{2}) = \pi {a}^{2}[/latex]
Hence required ratio (ratio of the areas of the circum Circles and the in Circles of a square) = [latex]2 \pi {a}^{2} : \pi {a}^{2} = 2 : 1[/latex]
4. A quadrilateral is inscribed in a circle. If the opposite angles of the quadrilateral are equal and length of its adjacent sides are 6 cm and 8 cm, what is the area of the circle?

A. 64 [latex]\pi[/latex] sq cm B. 25 [latex]\pi[/latex] sq cm C. 36 [latex]\pi[/latex] sq cm D. 49 [latex]\pi[/latex] sq cm

Answer - Option B
Explanation -
Length of diagonal of the quadrilateral = [latex]\sqrt{{8}^{2} + {6}^{2}}[/latex]
= 10 cm
i.e, Radius of the circle = [latex]\frac{10}{2}[/latex]
= 5 cm
i.e, Area of the circle =[latex]\pi {r}^{2} [/latex]
= 25[latex]\pi[/latex] sq. cm
5. A parallelogram ABCD has sides AB = 24 cm and AD = 16 cm. The distance between the sides AB and DC is 10 cm. Find the distance between the sides AD and BC.

A. 16 cm B. 18 cm C. 15 cm D. 26 cm

Answer - Option C
Explanation -

Area of the parallelogram = Base × Height
= 24 * 10 = 240 sq. cm
If the required distance becm, then
240 = 16 * x
x = [latex]\frac{240}{16} = 15[/latex] cm.
6. ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. I DP = [latex]\frac{1}{2}[/latex] AB, then the ratio of the lengths of BQ and AB is

A. 2 : 1 B. 1 : 2 C. 1 : 1 D. 3 : 1

Answer - Option A
Explanation -

Angle P is common in triangle DPC and APQ. Apart from that DC || AB hence other two angles are also same. triangle DPC and APQ are similar.
Let AB = BC = CD = DA = 2 and PD = 1 to satisfy the question condition.
Now using similarity
[latex]\frac{PD}{PA} = \frac{DC}{AQ}[/latex]
[latex]\frac{1}{3} = \frac{2}{AQ}[/latex]
AQ = 6
BQ = AQ - AB = 6 - 2 = 4
BQ : AB = 4 : 2 = 2 : 1
7. A quadrilateral ABCD circumscribes a circle and AB = 6 cm. CD = 5 cm and AD = 7 cm. The length of side BC is

A. 4 cm B. 5 cm C. 3 cm D.

Answer - Option A
Explanation -
Let ABCD is a quadrilateral in which a circle is inscribed
Given AB = 6cm; CD = 5cm

By the property of circle, sum of opposite sides are equal
AB + CD = BC + AD
6 + 5 = BC + 7
BC = 11 - 7 = 4cm
8. The area of a rhombus is 216 cm and the length of its one diagonal is 24 cm. The perimeter (in cm) of the rhombus is

A. 52 B. 60 C. 120 D. 100

Answer - Option B
Explanation -
Let ABCD be a rhombus, all sides of rhombus equal and its diagonal cuts each other at right angle. Let AC = 24cm

Area of rhombus = 1/2 (product of diagonals) = [latex]\frac{1}{2}[/latex] (AC × BD)
216 = [latex]\frac{1}{2}[/latex] (24 × BD)
24 × BD = 512
BD = 18cm
Now in right angled Triangles AOB
[latex]{AB}^{2} = {AO}^{2} + {BO}^{2}; AO[/latex]
= [latex]\frac{1}{2}[/latex](AC)
= 12cm; BO [latex]\frac{1}{2}[/latex] (BD)
= 9cm
[latex]{AB}^{2} = {12}^{2} + {9}^{2}[/latex]
= 144 + 81 = 225
AB = 15cm
Hence perimeter of rhombus = 4 × side
= 4 × 15 = 60cm
9. Two adjacent sides of a parallelogram are of lengths 15 cm and 18 cm. If the distance between two smaller sides is 12 cm, then the distance between two bigger sides is

A. 8 cm B. 10 cm C. 12 cm D. 15 cm

Answer - Option B
Explanation -
Adjacent sides of parallelogram = 15 cm and 18 cm
Distance between shorter sides = 12 cm
Area of parallelogram = b × h
= 15 × 12 cm2 = 180 [latex]{cm}^{2}[/latex]

Again, area of parallelogram = b × h
180 = 18 × h
h = [latex]\frac{180}{18}[/latex]
h = 10 cm
Therefore, the distance between its longer side = 10 cm.
10. Q is a point in the interior of a rectangle ABCD. If QA = 3 cm, QB = 4 cm and QC = 5 cm, then the length of QD in centimetre is

A. [latex]3 \sqrt{2}[/latex] B. [latex]5 \sqrt{2}[/latex] C. [latex]\sqrt{34}[/latex] D. [latex]\sqrt{41}[/latex]

Answer - Option A
Explanation -

Hint: Use pythagorous theore in all 4 small rectangles to derive this formula.
[latex]{QD}^{2} + {QB}^{2}= {QA}^{2} + {QC}^{2}[/latex]
[latex]{QD}^{2} + {16}^{2}= 9 + 15[/latex]
[latex]{QD}^{2} = 34 - 16 = 18[/latex]
[latex]{QD}^{2} = \sqrt{18} = 3 \sqrt{2}[/latex]cm

1. What is the area of the sector whose central angle is 90 ° and radius of the circle is 14 cm?

A. 308 sq cm B. 77 sq cm C. 154 sq cm D. 231 sq cm

Answer - Option C
Explanation -
Area of circle = [latex]\pi {r}^{2}[/latex]
As the central angle is 90 ° which is one fourth of 360 °. It means the sector is quadrant and area of
quadrant will be one fourth of the area of complete cirle
Hence required area = [latex]\frac{1}{4} * \pi {r}^{2} = \frac{1}{4} * \frac{22}{7} * 14 * 14 = 154 sq cm[/latex]
2. A gear 12 cm in diameter is turning a gear 18 cm in diameter. When the smaller gear has 42 revolutions, how many has the larger one made?

A. 28 B. 20 C. 15 D. 24

Answer - Option A
Explanation -
Smaller gear perimeter = 12 [latex]\pi[/latex]
Larger gear perimeter = 18 [latex]\pi[/latex]
No of revolutions of larger = [latex]\frac{44 * 12 * \pi}{18 \pi}[/latex]
3. Two circles touch each other externally at point A and PQ is a direct common tangent which touches the circles at P and Q respectively. Then ?PAQ =

A. 45º B. 90º C. 80º D. 100º

Answer - Option B
Explanation -
Two circles with centres [latex]{O}_{1}[/latex] and [latex]{O}_{2}[/latex] touch each other externally at point A and PQ is a direct common tangent which touches the circles at P and Q respectively.
Let ?PAQ = [latex]\alphaº [/latex]
and [latex]?{O}_{1}AP = ?{O}_{1}PA = {\Thetaº}_{1}[/latex]
and [latex]?{O}_{2}AQ = ?{O}_{2}QA = {\Thetaº}_{2}[/latex]
then, in [latex]\triangle APQ -[/latex]
?APQ + ?AQP + ?PAQ = 180º
(90º - [latex]{\Thetaº}_{1}) + (90º - {\Thetaº}_{2}[/latex]) + [latex]\alphaº = 180[/latex]
[latex]\alphaº = {\Thetaº}_{1} + {\Thetaº}_{2}[/latex] ...(i)
The straight line [latex]{O}_{1}A?{O}_{2}[/latex] joining the centres of the circles.
[latex]{\Thetaº}_{1} + {\Thetaº}_{2} + \alphaº = 180º[/latex]
[latex]\alphaº + \alphaº = 180º[/latex]
?PAQ = [latex]\alphaº[/latex]
= [latex]\frac{1}{2} * 180º = 90º [/latex]
Hence, option D is correct.
4. A 7 m wide road runs outside around a circular park, whose circumference is 176m.The area of the road is: [use [latex]\pi \frac{22}{7}[/latex]]

A. 1386 [latex]{m}^{2}[/latex] B. 1472 [latex]{m}^{2}[/latex] C. 1512 [latex]{m}^{2}[/latex] D. 1760 [latex]{m}^{2}[/latex]

Answer - Option A
Explanation -
If the radius of the circular park be metre, then
[latex]2 \pi r = 176[/latex]
[latex]2 * \frac{22}{7} * r = 176[/latex]
[latex]\frac{176 * 7}{2 * 22} = 28[/latex] metre
Radius of the park with road = 28 + 7 = 35 metre
Area of the road = [latex]\frac{22}{7} ({35}^{2} - {28}^{2}) {m}^{2}[/latex]
= [latex]\frac{22}{7} * 63 * 7 = 1386 {m}^{2}[/latex]
5. Two circles [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] touch each other internally at P. Two lines PCA and PDB meet the circles [latex]{C}_{1}[/latex] in C, D and [latex]{C}_{2}[/latex] in A, B respectively. If [latex]\triangle[/latex] BDC = 120º, then the value of [latex]\triangle[/latex] ABP is equal to

A. 60º B. 80º C. 100º D. 120º

Answer - Option A
Explanation -

[latex]\triangle PC{O}_{1} \simeq \triangle PA{O}_{2} and \triangle PD{O}_{1} \simeq \triangle PB{O}_{2} [/latex]
So [latex]/frac {PC}{PA} = \frac{r}{R}[/latex]
[latex]\frac{r}{R} = \frac{PD}{PB}[/latex]
So [latex]\frac{PC}{PA} = \frac{PD}{PB}[/latex]
Now

In [latex]\triangle[/latex] PCD and [latex]\triangle[/latex] PAB ;
P is common and PC/PA = PD/PB So [latex]\triangle[/latex] PCD = [latex]\triangle[/latex] PAB So CD II AB.
So, ABP + BDC =180 or,
ABP + 120 = 180 or, ABP = 60
6. Find the radius of the circle if the length of its arc is 33 cm and the corresponding central angle is 90 °?

A. 21 cm B. 14 cm C. 7 cm D. 28 cm

Answer - Option A
Explanation -
We know that,
Angle = [latex]\frac{arc}{radius}[/latex]
Here, angle = 90 ° = [latex]\frac{\pi}{2}[/latex]
And arc = 33 cm
i.e, Radius = [latex]\frac{33}{\frac{\pi}{2}}[/latex]
= 21 cm
7. In circular measure, the value of the angle 11 °15’ is

A. [latex]\frac{\pi}{16}[/latex] B. [latex]\frac{\pi}{8}[/latex] C. [latex]\frac{\pi}{4}[/latex] D. [latex]\frac{\pi}{12}[/latex]

Answer - Option A
Explanation -
[latex]11 + \frac{15}{60} = \frac{45}{4}[/latex]degrees
[latex]\frac{45}{4} \times \frac{\pi}{180} = \frac{\pi}{15} [/latex] radians
8. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ?APB is

A. 45º B. 90º C. 30º D. 60º

Answer - Option D
Explanation -

OA = OB = r
OP = 2r
sin APO = [latex]\frac {OA}{OP} = \frac {r}{2r} = \frac {1}{2}[/latex]
?APO = 30º
i.e, ?APB = 60º
9. In a circle of radius 21 cm, an arc subtends an angle of 72 ° at the centre. The length of the arc is

A. 21.6 cm B. 26.4 cm C. 13.2 cm D. 19.8 cm

Answer - Option B
Explanation -
[latex]\theta = 72º = 72 \times \frac {\pi}{180} radians[/latex]
[latex]= \frac {2\pi}{5}[/latex] radians i.e, [latex]\theta = \frac {s}{r}[/latex]
s = [latex]= \theta r = \frac {2 \pi}{5} \times 21[/latex]
[latex]= \frac {2}{5} \times \frac {22}{7} \times 21[/latex]
[latex]= \frac {132}{5} = 26.4 cm[/latex]
10. If the four equal circles of radius 3 cm touch each other externally, then the area of the region bounded by the four circles is

A. 4 (9 - [latex]\pi[/latex]) sq .cm B. 9 (4 - [latex]\pi[/latex]) sq .cm C. 5 (6 - [latex]\pi[/latex]) sq .cm D. 6 (5 - [latex]\pi[/latex]) sq .cm

Answer - Option B
Explanation -

Area of squard abed = 2r [latex]\times[/latex]
= 4 [latex]{r}^{2}[/latex]
= 4 [latex]\times[/latex] 9 = 36 sq.cm
Area of shaded part = Area of square abed - Area of the circle
[latex]{(2r)}^{2} - \pi {r}^{2}[/latex]
= [latex](4 - \pi) {r}^{2}[/latex]
= [latex]9(4 - \pi)[/latex]sq cm (r = 3 cm, given)

1. The cross section of a canal is in the shape of an isosceles trapezium which is 2 m wide at the bottom and 3 m wide at the top. If the depth of the canal is 1 m and it is 100 m long, what is the maximum capacity of this canal?

A. 500 cubic mts B. 250 cubic mts C. 750 cubic mts D. 1000 cubic mts

Answer - Option B
Explanation -
Area of isosceles trapezium = [latex]\frac{1}{2} (a + b) \times h[/latex]
Here a = 2, b = 3 and h = 1
Area trapezium = [latex]\frac{1}{2} (2 + 3) \times 1 = 2.5 sqm[/latex]
Required volume = area × length of canal
[latex]\Rightarrow 2.5 \times 100 = 250 [/latex]cubic meter
2. Volume of a cylinder is 770 cubic cm. If circumference of its base is 22 cm, what will be the curved surface area of the cylinder? (Take p = 22/7)

A. 440 sq cms B. 880 sq cms C. 220 sq cms D. 660 sq cms

Answer - Option A
Explanation -
Let the radius of base = r and height of cylinder = h
Circumference of base = 2p r = 22cm, here r is the radius of base
[latex]2 \times \frac{22}{7} \times r = 22[/latex]
r = 3.5 cm
volume of cylinder = [latex]\pi \frac{r}{2} h[/latex]
[latex]\Rightarrow 770 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h[/latex]
[latex]\Rightarrow[/latex] h = 20cm
curved surface area of cylinder = 2prh
CSA = [latex]2 \times \frac{22}{7} \times \frac{7}{2} \times 20 = 440 sqcm[/latex]
3. The total surface area of a hemisphere is 1848 sq cm, what is its radius? (Take n = 22/7)

A. 28 cms B. 7 cms C. 21 cms D. 14 cms

Answer - Option D
Explanation -
Total surface area of a hemisphere = [latex]3 \pi {r}^{2}[/latex]
[latex]\Rightarrow \times \frac{22}{7} \times {r}^{2}[/latex]
[latex]\Rightarrow[/latex] r = 14 cms
4. A pentagonal prism has 15 edges. How many vertices does it have?

A. 12 B. 10 C. 15 D. 20

Answer - Option B
Explanation -
In geometry, the pentagonal prism is a prism with a pentagonal base. It is a type of heptahedron.

According to Euler's formula:
F + V - E = 2
F = Number of faces (7)
E = Edges (15 given)
V = Vertices
F + V - E = 2
7 + V - 15 = 2
V = 2 + 15 - 7 = 10
Thus, B is correct.
5. Sum of lengths of all edges of a cube is 84 cm, find its volume?

A. 686 cubic cms B. 343 cubic cms C. 171.5 cubic cms D. 514.5 cubic cms

Answer - Option B
Explanation -
Given: sum of all edges of the cube, 12 a = 84 cm
[latex]\Rightarrow[/latex] a = 7 cm
Then, its volume = [latex]{a}^{3}[/latex]
= [latex]{7}^{3}[/latex]
= 343 cubic cm
6. The total surface area of a hemisphere is 41.58 sq cm. Find its curved surface area.

A. 27.72 sq cm B. 55. 44 sq cm C. 9.24 sq cm D. 13.86 sq cm

Answer - Option A
Explanation -
Total surface area = 3 × p × [latex]{r}^{2}[/latex]
Curved surface area = 2 × p × [latex]{r}^{2} = 41.58 × (\frac{2}{3}) = 27.72 {cm}^{2}[/latex]
7. A cube and a sphere have equal surface areas. The ratio of their volumes will be _____.

A. p : 6 B. [latex]\sqrt {\pi} : \sqrt{6}[/latex] C. [latex]\sqrt {6} : \sqrt {\pi}[/latex] D. 6 : p

Answer - Option B
Explanation -
Let the length of the side of the cube to be 'l' units & radius of the sphere to be 'r' units.
Then,
Surface area of the cube = 6 [latex]{I}^{2}[/latex]
Surface area of sphere = 4p [latex]{R}^{2}[/latex]
[latex]\Rightarrow \frac{{I}^{2}}{{R}^{2}} = \frac{4 \pi}{6} [/latex]
i.e, [latex]\frac{Volume of the cube}{Volume of the sphere}[/latex]
= [latex]\frac{{I}^{3}}{\frac{4}{3}} \pi {r}^{3}[/latex]
= [latex]\frac{3}{4 \pi} \times \frac{8 \pi \times \sqrt {\pi}}{6 \times \sqrt {\pi}}[/latex]
= [latex]\sqrt {\pi} : \sqrt {6}[/latex]
8. If the radii of the circular ends of a truncated conical bucket which is 45 cm high be 28 cm and 7 cm, then the capacity of the bucket in cubic centimeter is

A. 45810 B. 45810 C. 48150 D. 48051

Answer - Option A
Explanation -
Volume of bucket = [latex]\frac{1}{3} \pi h ({{r}_{1}}^{2} + {{r}_{2}}^{2} + {r}_{1} {r}_{2})[/latex]
= [latex]\frac{1}{3} \times \frac{22}{7} \times 45 ({28}^{2} + {7}^{2} + 28 \times 7)[/latex]
= [latex]\frac{1}{3} \times \frac{22}{7} \times 45 \times 1029[/latex]
= 4810 cu. cm
9. A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical part is 63 m then the total area of the canvas required to make the tent is how much?(Take p = 22/7)

A. 11385 [latex]{m}^{2}[/latex] B. 10395 [latex]{m}^{2}[/latex] C. 9900 [latex]{m}^{2}[/latex] D. 990 [latex]{m}^{2}[/latex]

Answer - Option A
Explanation -
Total area of the canvas = 2prh + prl
= p r (2h + l)
= [latex]\frac{22}{7} \times \frac{105}{2} (2 \times 3 + 63)[/latex]
= [latex]\frac{22}{7} \times \frac{105}{2} \times 69[/latex]
= [latex]11 \times 15 \times 69 = 11385 sq.m[/latex]
10. If the ratio of the diameters of two right circular cones of equal height be 3 : 4, then the ratio of their volumes will be

A. 3 : 4 B. 9 : 16 C. 16 : 9 D. 27 : 64

Answer - Option B
Explanation -
Ratio in the diameter of two right circular cone = 3 : 4
Hence ratio in radius is same as ratio in diameter = 3 : 4
Hence radius of first cone = [latex]\frac{3X}{7}[/latex]
Radius of second cone = [latex]\frac{4X}{7}[/latex]
They have equal height h (say)
Volume of cone = [latex]\frac{1}{3} \pi {r}^{2} h[/latex]
Now ratio in their volume
= [latex]\frac{1}{3} \pi {(\frac {3X}{7})}^{2} h : \frac{1}{3} \pi {(\frac {4X}{7})}^{2} h[/latex]
= [latex]{3}^{2} : {4}^{2} \Rightarrow 9 : 16[/latex]

1. Let O be the in-centre of a triangle ABC and D be a point on the side BC of [latex]\triangle[/latex] ABC, such that OD ? BC. If ?BOD = 15 °, then ?ABC =

A. 75 ° B. 45° C. 150° D. 90°

Answer Option C
Explanation :

BO is the internal bisector of ?B. ?ODB = 90°; ?BOD = 15°, ?OBD = 180° - 90° - 15° = 75°
[latex]\triangle[/latex] ABC = 2 * 75° = 150°
2. D is any point on side AC of ABC. IF P, Q, X, Y are the midpoints of AB, BC, AD and DC respectively, then the ratio of PX and QY is

A. 1 : 2 B. 1 : 1 C. 2 : 1 D. 2 : 3

Answer Option B
Explanation :

PX [latex]\parallel[/latex] BD and PX = [latex]\frac{1}{2}[/latex] BD
QY [latex]\parallel[/latex] BD and QY = [latex]\frac{1}{2}[/latex] BD
i.e, PX : QY = 1:1
3. The sides of a triangle are in the ratio [latex]\frac{1}{3}[/latex] : [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{5}[/latex] and its perimeter is 94 cm. The length of the smallest side of the triangle is:

A. 18 cm B. 24 cm C. 22.5 cm D. 27 cm

Answer Option B
Explanation :
[latex]\frac{1}{3}[/latex] : [latex]\frac{1}{4}[/latex] : [latex]\frac{1}{5}[/latex] = [latex]\frac{1}{3} \times 60[/latex] : [latex]\frac{1}{4} \times 60[/latex] : [latex]\frac{1}{5} \times 60[/latex]
20 : 15: 12
i.e, 20x + 15 + 12x = 94
47x = 94
= [latex]\frac{94}{47} = 2[/latex]
i.e, The smallest side = 12x
= 12 × 2 = 24 cm
4. If in a [latex]\triangle[/latex]ABC, the medians CD and BE intersect each other at O, then the ratio of the areas of ?ODE and ?ABC is

A. 1 : 6 B. 6 : 1 C. 1 : 12 D. 12 : 1

Answer Option C
5. If the length of the side of an equilateral triangle is 8 cm, what is its area?

A. 32 [latex]\sqrt{3}[/latex] sq cm B. 16 sq cm C. 16 [latex]\sqrt{3}[/latex] sq cm D. 32 sq cm

Answer Option C
Explanation :
Given the side of the equilateral triangle, a = 8 cm
[latex]\Rightarrow [/latex] Area = [latex](\frac{\sqrt{3}}{4}) {a}^{2}[/latex]
[latex]\Rightarrow [/latex] Area = [latex](\frac{\sqrt{3}}{4}) {8}^{2} = 16 \sqrt{3} sq cm[/latex]
6. ABC is an isosceles right angled triangle with ?B = 90 °. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of areas of ?ABE and ?ACD is

A. 1 : 3 B. 2 : 3 C. 1 : 2 D. 1 : [latex]\sqrt{2}[/latex]

Answer Option C
Explanation :

AB = x units
BC = x units
AC = [latex]\sqrt{2}[/latex] x units
[latex]\frac{\Delta ABE}{\Delta ACD}[/latex] = [latex]\frac{\sqrt{3} \frac{{x}^{2}}{4}}{\sqrt{3} \frac{({\sqrt{3}2)}^{2}}{4}}[/latex] = [latex]\frac {1}{2}[/latex]
7. If a, b and c are the sides of a triangle and [latex]{a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca[/latex] then the triangle is

A. right-angled B. obtuse-angled C. equilateral D. isosceles

Answer Option C
Explanation :
[latex]{a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca[/latex]
[latex]2{a}^{2} + 2{b}^{2} + 2{c}^{2} = 2ab + 2bc + 2ca[/latex]
[latex]{a}^{2} - 2ab + {b}^{2} + {b}^{2} - 2bc + {c}^{2} + {c}^{2} - 2ca + {a}^{2} = 0[/latex]
[latex]{a -b}^{2} + {b -c}^{2} + {c -a}^{2} = 0[/latex]
a - b = 0; a = b
b - c = 0; b = c
c - a = 0; c = a
i.e, a = b= c
8. ABCD is a trapezium in which AB || DC and AB = 2 CD. The diagonals AC and BD meet at O. the ratio of areas of triangles AOB and COD is

A. 1 : 2 B. 1 : [latex]\sqrt{2}[/latex] C. 4 : 1 D. 1 : 4

Answer Option C
Explanation :
AB || DC (AC is a transversal)
Angle DCA = Angle CAB(alternate interior angles)
In Triangle AOB and Triangle COD
Angle OAB= Angle OCD ( Proved)
Angle AOB= Angle COD (Vertically opposite angles)
By AA of similarity, Triangle AOB ~ Triangle COD
Now using Area theorem of similar triangle
[latex]\frac{Area(AOB)}{Area(COD)} = \frac{{AB}^{2}}{{CD}^{2}} = \frac{{(2CD)}^{2}}{{CD}^{2}}[/latex] (AB 2CD)
[latex]\rightarrow \frac{Area(AOB)}{Area(COD)} = \frac{4}{1} = \frac{4}{1}[/latex]
Hence option C is correct
9. AD is the Median of a triangle ABC and O is the centroid such that AO = 10 cm. Length of OD (in cm) is

A. 2 B. 4 C. 5 D. 7

Answer Option C
Explanation :
AD is median of the triangle.
OA : OD = 2 : 1
So, OD = [latex]\frac{10}{2}[/latex] = 5cm
10. The points D and E are taken on the sides AB and AC of [latex]\triangle ABC [/latex] such that AD = [latex]\frac{1}{3}[/latex] AB, AE = [latex]\frac{1}{3}[/latex] AC. If the length of BC is 15 cm, then the length of DE is:

A. 10 cm B. 8 cm C. 6 cm D. 5 cm

Answer Option D
Explanation :
Let ABC be a Triangles, D and E are taken on the sides AB and AC
Given: AD = [latex]\frac{1}{3}[/latex] AB, AE = [latex]\frac{1}{3}[/latex] AC
BC = 15cm

In [latex]\triangle[/latex] ABC and [latex]\triangle[/latex] ADE
[latex]\triangle[/latex] BAC= [latex]\triangle[/latex] DAE
[latex]\frac{AD}{AB} = \frac{1}{3} ; \frac{AE}{AC} = \frac{1}{3}[/latex]
Hence these two triangles are similar.
Hence;
[latex]\frac{AD}{AB} \frac{AE}{AC} = \frac{DE}{BC} = \frac{1}{3}[/latex]
Hence DE = [latex] \frac{1}{3}[/latex] (BC) = [latex] \frac{15}{3}[/latex] = 5

1. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 9 cm. Determine the corresponding side of the second triangle.

A. 13.5 cm B. 6 cm C. 15 cm D. 5 cm

Answer - Option B
Explanation -
If the required side be x cm, then
[latex]\frac {30}{20} =\frac {9}{x}[/latex]
3 = 9 * 2
x = [latex]\frac {9 * 2}{3}[/latex] = 6 cm
2.The area of similar triangles ABC and DEF are 20 [latex]{cm}^{2}[/latex] and 45 [latex]{cm}^{2}[/latex] respectively. If AB = 5 cm, then DE is equal to :

A. 6.5 cm B. 7.5 cm C. 8.5 cm D. 5.5 cm

Answer - Option B
Explanation -
[latex]\frac {\triangle ABC}{\triangle DEF} =\frac {{AB}^{2}}{{DE}^{2}}[/latex]
[latex]\frac {20}{45} =\frac {25}{{DE}^{2}}[/latex]
[latex]{DE}^{2} = \frac {45 * 25}{20} = \frac {225}{4}[/latex]
i.e, DE = [latex] \frac {15}{2} = 7.5 cm[/latex]
3. If area of an equilateral triangle is a and height is b then the value of [latex] \frac {{b}^{2}}{a} [/latex] will be how much?

A. 3 B. [latex] \frac {1}{3} [/latex] C. [latex] \sqrt {3}[/latex] D. [latex] \frac {1}{\sqrt {3}}[/latex]

Answer - Option C
Explanation -

AD = b
BD = DC = x
tan 60 = [latex] \frac {AD}{BD}[/latex]
[latex] \sqrt {3} = \frac {b}{x}[/latex]
x = [latex] \frac {b}{\sqrt {3}}[/latex]
i.e, BC = 2x = [latex] \frac {2b}{\sqrt {3}}[/latex]
i.e, Area of the triangle
[latex] \frac {1}{2} * BC * AD[/latex]
a = [latex] \frac {1}{2} * \frac {2b}{\sqrt {3}} * b[/latex]
[latex]\frac {{b}^{2}}{a} = \sqrt {3}[/latex]
4. If the orthocenter and the centroid of a triangle are the same, then the triangle is:

A. Scalene B. Right angled C. Equilateral D. Obtuse angled

Answer - Option C
Explanation -

Here, AB = BC = CA
So, the concerned triangle is equilateral triangle.
5. What is the area of a largest triangle that can be inscribed in a circle of radius r?

A. [latex]\frac{1}{3} \pi {r}^{2}[/latex] B. [latex]\frac{4}{\sqrt{3}} {r}^{2}[/latex] C. [latex]\frac{3 \sqrt{3}}{4} {r}^{2}[/latex] D. [latex]\frac{\sqrt{3}}{2} {r}^{2}[/latex]

Answer - Option C
Explanation -
The only possible triangle with maximum area is equilateral triangle. Since all the centers in an equilateral triangle coincide.
So, O is circumcenter as well as centroid. Now, AP is a medium and O is centroid. Therefore,
[latex]\frac{AO}{OP} = \frac{2}{1}[/latex]
[latex]\frac{r}{OP} = \frac{2}{1}[/latex]
[latex]OP = \frac{r}{2}[/latex]
BC = 2BP = 2 [latex]\sqrt{{BO}^{2} - {OP}^{2}}[/latex]
= 2 [latex]\sqrt {{r}^{2} - {(\frac{r}{2})}^{2}} = \sqrt {3}r[/latex]
Area of [latex]\triangle[/latex] ABC
[latex] = \frac{ \sqrt{3}}{4} {BC}^{2} = \frac{ \sqrt{3}}{4} {(\sqrt{3 r})}^{2} = \frac{3 \sqrt{3}}{4} {r}^{2}[/latex]
6. [latex]\triangle[/latex] ABC is a right triangle. [latex]\angle ACB = 90°, CD \perp AB, BC = a, AC = B[/latex] and CD = P, then:

A. [latex]{a}^{2} + {b}^{2} = {p}^{2}[/latex] B. [latex]\frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} = \frac{1}{{p}^{2}}[/latex] C. ab = [latex]{p}^{2}[/latex] D. a + b = p

Answer - Option B

1. The length of aside of a rhombus is 10 m and one of its diagonal is 12 m. The length of the other diagonal is

A. 15 m B. 18 m C. 16 m D. Can't be determined

Answer - Option C
Explanation -
Let one diagonal = x ; and second diagonal = y
Side of rhombus= 10
[latex]{x}^{2} + {y}^{2} = {4a}^{2} [/latex]
[latex]{12}^{2} + {y}^{2} = 4{10}^{2} [/latex]
[latex]{y}^{2} [/latex] = 256
y = 16 m
2. The length and breadth of a rectangle are increased by 20% and 25% respectively. The increase in the area of the resulting rectangle will be:

A. 60 % B. 50% C. 40% D. 30%

Answer - Option B
Explanation -
Effect on area of rectangle = (20 + 25 + [latex]\frac{20 * 25}{100} [/latex] )% = 50%
3. AB is a chord to a circle and PAT is the tangent to the circle at A. If BAT = 75o and BAC = 45o, C being a point on the circle, then ABC is equal to

A. 40 B. 45 C. 60 D. 70

Answer - Option C
Explanation -

If a line touches a circle and from the point of contact a chord is drawn, the angles which this chord makes with given line are equal respectively to the angles formed in the corresponding alternate segments.
i.e, ACB = BAT = 75°
ABC = 180° - 45° - 75° = 60°
4. If a chord of a circle of radius 5 cm is a tangent to a circle of radius 3 cm, both the circles being concentric, then the length of the chord is

A. 10 cm B. 12. 5 cm C. 8 cm D. 7 cm

Answer - Option C
Explanation -
Let [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] are the circles of radius 5 cm and 4cm respectively with centre O.

PQ be a chord of circle [latex]{C}_{1}[/latex], PQ is tangent at the point R of circle [latex]{C}_{2}[/latex].
As we know that tangent makes an angle of 90° with the radius of respective circle.
Hence PQ makes 90° with OR.
Therefore Triangles ORQ is a right angle Triangles.
Hence by Pythagoras theorem:
[latex]{RQ}^{2} = {OP}^{2} - {OR}^{2} [/latex]
= 25 – 9 = 16
RQ = 4cm
Hence PQ = 2RQ = 2 × 4 = 8cm
5. The length of two chords AB and AC of a circles are 8 cm and 6 cm and ?BAC = 90°, then the radius of circle is

A. 25 cm B. 20 cm C. 4 cm D. 5 cm

Answer - Option D
Explanation -
A circle with centre O has two chords AB = 8cm and Ac = 6cm

Given: = 900
Therefore Triangles ABC is right angle Triangles.
By Pythagoras theorem
[latex]{BC}^{2} = {AB}^{2} - {AC}^{2} [/latex]
[latex]{BC}^{2}[/latex] = 64 + 36
[latex]{BC}^{2}[/latex] =100
BC = 10 cm
BC is the diameter of circle, hence radius of circle = [latex]\frac{diameter}{2}[/latex]
Radius = [latex]\frac{10}{2}[/latex] = 5cm
6. The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle be a tangent to the smaller circle, then the length of that chord is

A. 24 cm B. 12 30 cm D. 18 cm

Answer - Option A
Explanation -

Let [latex]{C}_{1} and {C}_{2}[/latex]are the circles of radius 15 cm and 9 cm respectively with centre O.
PQ be a chord of circle [latex]{C}_{1}[/latex]. PQ is tangent at the point R of circle [latex]{C}_{2}[/latex].
We know that tangent makes an angle of 90° with the radius of respective circle. Hence, PQ makes 90° with OR.
Therefore triangle ORQ is a right angle triangle.
Hence, by Pythagoras theorem:
[latex]{RQ}^{2} = {OQ}^{2} - {OR}^{2} [/latex]
= 225 – 81 = 144
RQ = 12cm
Hence, PQ = 2 RQ = 2 × 12 = 24 cm
7. A chord 12 cm long is drawn in a circle of diameter 20 cm. The distance of the chord from the centre is

A. 8 cm B. 6 cm C. 10 cm D. 16 cm

Answer - Option A
Explanation -
OA = radius= 10 cm
CE= RD = 6 cm
i.e, OE = [latex]\sqrt{{10}^{2} - {6}^{2}}[/latex]
= [latex]\sqrt{100 - 36}[/latex]
= [latex]\sqrt{64}[/latex] = 8 cm
8. In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?

A. 1 : 3 B. 7 : 1 C. 3 : 1 D. 2 : 3

Answer - Option C
Explanation -

Now, Using section formula
[latex]\frac{3m - n} {m + n} = x[/latex]
[latex]\frac{4m - 12n} {m + n} = 0[/latex]
4m - 12n =0
4m = 12 n
[latex]\frac{m} {n} = \frac{3} {1}[/latex]
Therefore line is divided in the ratio 3 : 1
9. The graph of the equation intersects the x-axis at the point

A. (2, 0) B. (5, 0) C. (4, 5) D. (0, 5)

Answer - Option B
Explanation -
When a straight line cuts x-axis, the coordinates of point of intersection = (x, 0),
i.e. y = 0
Putting y = 0 in 4x – 5y = 20
4x = 20 x = 5
Point of intersection = (5, 0)
Note: Putting y = 0 in 4x – 5y = 20, point of intersection on x-axis = (5, 0)
Putting x = 0 in 4x – 5y = 20, point of intersection on y-axis = (0, -4)
Look at the graph of the equation:

10. Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then the length of PB is

A. 5 cm B. 7.35 cm C. 6 cm D. 4 cm

Answer - Option B
Explanation -

AB = 6 cm; CD = 3cm
PD = 5 cm; PB= ?
PA * PB = PC * PD
(PB - 6)PB = 2 * 5
[latex]{PB}^{2}[/latex] - 6PB = 2 * 5
PB = [latex]\frac {6 ± \sqrt {36 + 40}}{2}[/latex]
= [latex]\frac {6 ± \sqrt {76}}{2}[/latex]

1. In the adjoining figure, sides AB and AC of a [latex]\triangle[/latex]ABC are produced to P and Q respectively. The bisectors of PBC and QCB intersect at O. then BOC is equal to:

A. 90 - [latex]\frac{1}{2}[/latex] BAC B. [latex]\frac{1}{2}[/latex]( PBC + QCB) C. 90 + [latex]\frac{1}{2}[/latex] BAC D. None of these

Answer - Option A
Explanation -
1 = 90° - [latex]\frac{1}{2} ?3[/latex]
2 = 90° - [latex]\frac{1}{2} ?4[/latex]
Now in BOC
1 + 2 + BOC = 180°
BOC = 180° - (2 + 1)
= 180° - [90° - [latex]\frac{1}{2} 4 + 90° - \frac{1}{2} 3 [/latex]]
Or, BOC = [latex]\frac{1}{2}(180° - A)[/latex]
So, A + 3 = 180°
BOC = 90° - [latex]\frac{1}{2} A[/latex]
2. Two circle of radii 5 cm and 3 cm intersects at two points and the distance between their centres is 4 cm. find the length of common line segment.

A. 8 cm B. 4 cm C. 3 cm D. 2 cm

Answer - Option B
Explanation -
Given [latex]{C}_{1}{C}_{2}[/latex] = 4 cm
[latex]{C}_{1}P[/latex] = 3 cm (Radius)
= 5 cm (Radius)
Figure according to the question.

Common line segment PQ = 1 + 3 = 4cm
Difficult level : Esay
3, Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. find the distance between their centres.

A. 6 cm B. 12 cm C. 13.29 cm D. 15 cm

Answer - Option C
Explanation -

Let O and [latex]{O}^{1}[/latex] be the centre of the circle of radii 10 cm and 8 cm and PQ = 12c m.
PL = [latex]\frac{1}{2} PQ = 6 cm[/latex]
In right triangle OLP, we have
[latex]{OP}^{2} = {OL}^{2} + {OP}^{1}{LP}^{2}[/latex]
OL = [latex]\sqrt{{10}^{2} - {6}^{2}} = \sqrt{64} = 8 cm[/latex] and in right triangle , we have
[latex]{O}^{1}{L}^{2} + {LP}^{2} = {O}^{1}{P}^{2}[/latex]
[latex]{O}^{1}L = \sqrt{{8}^{2} - {6}^{2}} = \sqrt{28} = 5 .29 cm[/latex]
[latex]O{O}^{1} = OL + LQ = (8 + 5.29) = 13.29 cm[/latex]
4. In the adjoining figure BAC = 60° and BC = a, AC = b and AB = c, then

A. [latex]{a}^{2} = {b}^{2} + {c}^{2}[/latex] B. [latex]{a}^{2} = {b}^{2} + {c}^{2} - bc[/latex] C. [latex]{a}^{2} = {b}^{2} + {c}^{2} + bc[/latex] D. [latex]{a}^{2} = {b}^{2} + 2bc[/latex]

Answer - Option B
Explanation -
cos A = [latex]\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc} [/latex]
[latex]\frac{1}{2} = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc}[/latex]
[latex]{a}^{2} = {b}^{2} + {c}^{2} - bc[/latex]
5. An equilateral triangle of side 6 cm has its corners cut off to form a regular hexagon. Area (in [latex]{cm}^{2}[/latex]) of this regular hexagon will be

A. 6[latex]\sqrt{3}[/latex] B. 3[latex]\sqrt{6}[/latex] C. 10[latex]\sqrt{2}[/latex] D. 5[latex]\sqrt{3}[/latex]

Answer - Option A
Explanation -
Side of the regular hexagon = [latex]\frac{(1*6)}{3}[/latex] = 2
Area of the hexagon = [latex]\frac{(3\sqrt{3} * {a}^{2})}{2}[/latex]

[latex]\frac{3 \sqrt{3}}{2} * 2* 2 = 6 \sqrt{3}[/latex]
6, The length of a chord of a circle is equal to the radius of the circle. The angle which this chord subtends in the major segment of the circle is equal to

A. 30 B. 45 C. 60 D. 90

Answer - Option A
Explanation -

AO = OB = AB
i.e, AOB = 600
i.e, ACB = 300
7. In [latex]\triangle[/latex] PQR, right-angled at Q, PR= 3 [latex]\sqrt{5}[/latex]cm. If the difference between the lengths of the perpendicular sides is 3cm, then find the value of sinP + sinR?

A. 3 [latex]\sqrt{5}[/latex] cm B. [latex]\frac{3}{\sqrt{5}}[/latex] C. [latex]\sqrt{5}[/latex] D. 3 cm

Answer - Option B
Explanation -

Let lengths of perpendicular sides be ‘x’ and ‘x+3’. So using Pythagoras Theorem,
we get [latex]{x}^{2} + {x + 3}^{2} = {(3\sqrt{5})}^{2}[/latex]. So, ‘x’ = 3cm. So,
sin P + sin R = [latex]\frac{QR}{PR} + \frac{PQ}{PR} +\frac{3}{3 \sqrt{5}} + \frac{6}{3 \sqrt{5}} + \frac{3}{ \sqrt{5}}[/latex]
8. At the centers two circles, two Arcs of equal length subtend angles of 60° and 75° respectively, the ratio of the radii of the two circles is –

A. 5 : 2 B. 5 : 4 C. 3 : 4 D. 2 : 1

Answer - Option B
Explanation -
Radius of the first circle is and second circle is
[latex]{q}_{1} = 60 and {q}_{2} = 75[/latex](given)
[latex]{arc}_{1} = 2\pi{r}_{1} * \frac{{Q}_{1}}{360} = 2\pi{r}_{1} * \frac {60}{360}[/latex]
[latex]{arc}_{2} = 2\pi{r}_{2} * \frac{{Q}_{2}}{360} = 2\pi{r}_{2} * \frac {75}{360}[/latex]
[latex]{arc}_{1} = {arc}_{2}[/latex]
[latex] 2\pi{r}_{2} * \frac {60}{360}\frac{{Q}_{2}}{360} = 2\pi{r}_{2} * \frac {75}{360}[/latex]
[latex] \frac {{r}_{1}}{{r}_{2}} = \frac {5}{4}[/latex]
= 5 : 4
9. The radius of two concentric circles are 9 cm and 15 cm. If the chord of the greater circle be a tangent to the smaller circle, then the length of that chord is

A. 24 cm B. 12 cm C. 30 cm D. 18 cm

Answer - Option A
Explanation -

BO = OC = 15 cm. OD = 9 cm. BD = [latex] \sqrt{{15}^{2} - {9}^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12[/latex] cm.
BC = 2 × 12 = 24 cm.
10. In the adjoining figure a parallelogram ABCD is shown. AB = 24 cm and AO = BO = 13 cm. Find BC :

A. 8 cm B. 10 cm C. 11 cm D. None of these

Answer - Option B
Explanation -
AO = BO = 13 cm

[latex]\rightarrow[/latex] AC = BD = 26 CM
Now , since the diagonals are equal, it means the given figure is actually a rectangle.
[latex]{BC}^{2} = {AC}^{2} - {AB}^{2}[/latex]
[latex]{BC}^{2} = {26}^{2} - {24}^{2}[/latex]
[latex]{BC}^{2}[/latex] = 100
[latex]\rightarrow[/latex] BC = 10 cm

1. Chord PQ is the perpendicular bisector of radius OA of circle with center O (A is a point on the edge of the circle). If the length of Arc PAQ =[latex]\frac {2\pi}{3}[/latex]. what is the length of chord PQ?

A. 2 B. [latex]\sqrt{3}[/latex] C. 2[latex]\sqrt{3}[/latex] D. 1

Answer - Option B
Explanation -

PQ is perpendicular bisector of OA.
OP = OQ = PA = AQ
OPAQ is a rhombus.
2 PAQ = Reflex POQ
(The angle subtended at the centre by an arc is twice to that at the circumference)
2 PAQ = 360° - POQ
3 PAQ = 360°
(? PAQ = ? POQ)
? PAQ = 120° = ? POQ = [latex]\frac {\pi 2}{3}[/latex]
Again redius (r)= [latex]\frac {t}{0} = \frac {\frac {\pi 2}{3}}{\frac {\pi 2}{3}} = 1[/latex]
From ? OPB
OP = 1 unit
? POB = 60°
? sin 60° = [latex]\frac {PB}{OP}[/latex]
? PB = [latex]\frac {\sqrt{3}}{2}[/latex]
? PQ = 2 × = [latex]\sqrt{3}[/latex] unit
2. What can be the maximum number of common tangent which can be drawn to two non-intersecting circles?

A. 2 B. 4 C. 3 D. 6

Answer - Option B
Explanation -

4 tangents.
3. There is a circular garden of radius 21 metres. A path of width 3.5 metres is constructed just outside the garden. What is the area (in [latex]{metres}^{2}[/latex]) of the path?

A. 500.5 B. 575.6 C. 521.2 D. 560.7

Answer - Option A
Explanation -

Area of path AB = p[latex]{R}^{2}[/latex] - p[latex]{r}^{2}[/latex]
= p([latex]{24.5}^{2} - {21}^{2}[/latex])
= p(600.25-441)
= [latex]\frac{22}{7}[/latex]*159.25 = 500.5[latex]{metres}^{2}[/latex]
4. Two circles touch each other at point X. A common tangent touch them at two distinct points Y and Z. If another tangent passing through X cut YZ at A and XA= 16 cm, then what is the value (in cm) of YZ?

A. 18 B. 24 C. 16 D. 32

Answer - Option D
Explanation -

We know that XA = YA = ZA
Required
YZ = AY + AZ = 2 × XA = 2 × 16 = 32
5. The sum of radii of the two circles is 91 cm and the difference between their area is 2002 [latex]{cm}^{2}[/latex]. What is the radius (in cm) of the larger circle?

A. 56 B. 42 C. 63 D. 49

Answer - Option D
Explanation -
[latex]{R}_{1} + {R}_{2}[/latex]= 91 ………(i)
[latex]\pi{{R}_{1}}^{2} - \pi{{R}_{2}}^{2}[/latex] = 2002
[latex]{{R}_{1}}^{2} - {{R}_{2}}^{2}[/latex] = 2002 * [latex]\frac {7}{22}[/latex]
[latex]({R}_{1} + {R}_{2}) ({R}_{1} - {R}_{2})[/latex]=91×7
[latex]({R}_{1} - {R}_{2})[/latex] = 7 …(ii)
Solve eqn. (i) & (ii)
[latex]{R}_{1}[/latex]= 49 (Larger Circle)
[latex]{R}_{2}[/latex]= 42 (Smaller Circle) .
6. Two circle of same radius 5 cm intersect each other at A and B. If AB = 8 cm, then distance between their centers is:

A. 6 cm B. 8 cm C. 10 cm D. 4 cm

Answer - Option A
Explanation -
AB = 8 cm, OA = AP = 5 cm. OP and AB are perpendicular bisector of each other.
OP = 2 * OC = [latex]\sqrt{{OA}^{2} - {AC}^{2}} = \sqrt{25 - 16} = 6 cm[/latex]
7. In the given figure, ?ONY = 50° and ?OMY = 15° Then the value of ?MON is

A. 30° B. 40° C. 20° D. 70°

Answer - Option D
Explanation -
Since OM = OY = ON = radius of circle.
?OMY = ?OMY = 15°
?ONY = ?OYN = 50°
?MON = 180° - 2 * 15° = 150°
?NOY = 180° - 2 * 50° = 80°
?MON = ?MOY - ?NOY = 150° - 80° = 70°
8. A, B and C are the three points on a circle such that the angles subtended by the chords AB and AC at the center O are 90 ° and 110 ° respectively. ?BAC is equal to

A. 70° B. 80° C. 90° D. 100°

Answer - Option B
Explanation -

We know that,
?BOA + ?AOC + ?BOC = 360
90° + 110° + ?BOC = 360°
?BOC = 360° - 200° = 160°
Note: The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
?BAC = [latex]\frac{1}{2}[/latex] ?BOC = [latex]\frac{1}{2}[/latex] * 160° = 80°
9. In the given figure ?AOC = 120° Find the ?CBE where O is the center:

A. 60° B. 100° C. 120° D. 150°

Answer - Option C
Explanation -
?CBA = [latex]\frac{1}{2}[/latex] ?COA = [latex]\frac{1}{2}[/latex] * 120 = 60°
?CBE = 180° - 60° = 120°
10. ABC is a right angled at C. What is the in radius of the in circle shown in the figure?

A. 9cm B. 4 cm C. Can't be determined D. None of these

Answer - Option B
Explanation -
In radius of the right-angled triangle ABC = [latex]\frac{AC + BC - AB}{2} = \frac{9 + 40 - 41}{2} = 4cm[/latex]

1. At least two pairs of consecutive angles are congruent in a _________.

A. Parallelogram B. Isosceles trapezium C. Rhombus D. Kite

Answer - Option B
Explanation -
At least two pairs of consecutive angles are congruent in an isosceles trapezium
2. A solid has 12 vertices and 30 edges. How many faces does it have?

A. 22 B. 24 C. 26 D. 20

Answer - Option D
Explanation -
We know that F + V - E = 2
Here F = number of faces, V = number of vertices and E = number of edges
F + 12 - 30 = 2
F = 2 + 18 = 20
So required number of faces = 20
3. If the sum of the measures of all the inter for angles of polygon is 1440° , find the number of sides of the polygon.

A. 8 B. 12 C. 10 D. 14

Answer - Option C
Explanation -
Sum of interior angles of a polygon = (N - 2) × 180 °
[latex]\rightarrow[/latex] 1440 ° = (N - 2) × 180 °
[latex]\rightarrow[/latex] [latex]\frac {1440 °}{180 °} [/latex] = (N - 2)
[latex]\rightarrow[/latex] 8 + 2 = N
[latex]\rightarrow[/latex] N = 10
Number of sides of the polygon = 10
4. Dodecahedron has 12 faces and 20 vertices . How many edges does it have?

A. 30 B. 16 C. 12 D. 24

Answer - Option A
Explanation -
According to Euler's formula
V + F = E +2
where,
V= number of vertices
E= Number of edges
F= number of faces
20 + 12 = E + 2
E = 30
5. The measure of each interior angle of a regular hexagon is how much?

A. 60° B. 45° C. 120° D. 100°

Answer - Option C
Explanation -
We have each interior angle of a polygon = [latex]\frac {(n - 2)180°}{n}[/latex]
[latex]\frac {(6 - 2)180°}{6}[/latex](n = 6 of hexagon)
[latex]180° * \frac {4}{6} = 120°[/latex]
6. If the interior angle of a regular polygon is double the measure of exterior angle, then the number of slides of the polygon is

A. 6 B. 8 C. 10 D. 12

Answer - Option A
Explanation -
Sum of interior +exterior angles = 180°
interior angle = 2*exterior angle
3 * exterior angle = 180
exterior angle = 60
60 = [latex]\frac {360}{n}[/latex]
n= [latex]\frac {360}{60}[/latex]
n = 6
7. The sum of interior angles of a regular polygon is 1440°. The number of sides of the polygon is how much?

A. 10 B. 12 C. 6 D. 8

Answer - Option A
Explanation -
If the number of sides regular polygon be n, then
(2n – 4) × 90 = 1440
2n - 4 = [latex]\frac {1440}{90}[/latex] = 16
n =10
8. The area of an isosceles trapezium is 176 [latex]{cm}^{2}[/latex]cm2 and the height is [latex]\frac {2}{{11}^{th}}[/latex] of the sum of its parallel sides. If the ratio of the length of the parallel sides be 4 : 7, then the length of a diagonal is

A. [latex]\sqrt{168} cm[/latex] B. 2 [latex]\sqrt{147} cm[/latex] C. [latex]\sqrt{137} cm[/latex] D. 2 [latex]\sqrt{137} cm[/latex]

Answer - Option D
Explanation -
Let ABCD be a trapezium
Given : the ratio of the length of the parallel sides be 4 : 7
Let length of parallel sides are 4x and 7x respectively.
height is [latex]\frac {2}{{11}^{th}}[/latex] of the sum of its parallel sides

hence height = [latex]\frac {2}{11}[/latex] * (11x) = 2x
area of trapezium = [latex]\frac {1}{2}[/latex] (sum of parallel sides)×height
area = [latex]\frac {1}{2}[/latex] (11x)×2x = 11x2
given: The area of an isosceles trapezium is 176 cm2
hence, 11x2 = 176 [latex]\rightarrrow[/latex] x2 = 16 [latex]\rightarrrow[/latex] x = 4cm
hence sides of trapezium = 16, 28 cm
height = 8cm
Diagonal of trapezium ABCD is AC.
QC = AB+[latex]\frac {(CD – AB)}{2}[/latex]
QC = 16 + [latex]\frac {12}{2}[/latex]= 16+6 = 22cm
In triangle AQC
AC2 = AQ2 + QC2
AC2 = 82+222 [latex]\rightarrow[/latex] 64+ 484 = 548
AC = [latex]\sqrt{548} = 2 \sqrt{137} cm[/latex]
9. If the difference between an exterior angle of a regular polygon of n sides and an exterior angle of another regular polygon of (n + 1) sides is equal to 5 °, then the value of n is

A. 5 B. 6 C. 7 D. 8

Answer - Option D
Explanation -
Exterior angle of regular polygon = 360o/ number of sides
ATQ
[latex]\frac {360°}{n} - \frac {360°}{n + 1}[/latex] = 5
360 (n+1 – n ) = 5n(n+1)
[latex]\rightarrow[/latex] 72 = [latex]{n}^{2}[/latex] + n
[latex]\rightarrow[/latex] [latex]{n}^{2}[/latex] + n – 72 = 0
[latex]\rightarrow[/latex] (n – 8)(n + 9)= 0
[latex]\rightarrow[/latex] n = 8, -9
Hence number of sides = 8
10. The number of sides in two regular polygons are in the ratio 5 : 4 and the difference between each interior angle of the polygons is 6 °. Then the numbers of sides are

A.15, 12 B. 5, 6 C. 10, 8 D. 20, 16

Answer - Option A
Explanation -
If the difference between Interior angles = 6o then the difference between Exterior angles will also be 6o. Since sides are in the ratio 5:4, take them 5n and 4n.
[latex]\frac {360°}{4n} - \frac {360°}{5n}[/latex] = 6
[latex]\frac {90}{n} - \frac {72}{n}[/latex] = 6
18 = 6n
n = 3
So sides are 5n, 4n = 15, 12

1. If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is:

A. 6 B. 8 C. 10 D. 12

Answer - Option B
Explanation -
Sum of the interior angles of a regular polygon of n sides
= (2n -4) * 90 °
i.e, (2n -4) * 90 ° = 1080°
(2n -4) = 1080° ÷ 90 ° = 12
i.e, 2n = 12 + 4 = 16
n = [latex] \frac{16}{2} = 8[/latex]
2. If each interior angle of a regular polygon is 150 ° the number of sides of the polygon is

A. 8 B. 10 C. 15 D. None of these

Answer - Option D
Explanation -
The relationship between angle and side of a polygon can be given as:
Angle = [latex] \frac{(2 × number of sides - 4) × 90°}{number of sides} [/latex]
Hence 150 = [latex] \frac{(2n - 4) ×90}{n}[/latex]
[latex]\rightarrow[/latex] 150n = 180n – 360
[latex]\rightarrow[/latex] 30n = 360
[latex]\rightarrow[/latex] n = 12
Hence number of sides = 12
3. If the measure of each interior angle of a regular Polygon be 144° the number of sides of the Polygon is

A. 10 B. 20 C. 24 D. 36

Answer - Option A
Explanation -
Interior angle = 144
Exterior anlge = 180 - 144 = 36
36 = [latex] \frac{360}{n}[/latex]
n = [latex] \frac{360}{36}[/latex]
n = 10
4. The number of diagonals of a polygon is 54. The number of sides of the polygon is how much?

A. 7 B. 9 C. 12 D. 15

Answer - Option C
Explanation -
Number of diagonals of a polygon = [latex] \frac{n(n-3)}{2}[/latex]
54 = [latex] \frac{n(n-3)}{2}[/latex]
108 = n(n-3)
12*9 = n(n-3)
n = 12
5. In the given figure, O is the center of the circle, PQR = 100° and STR = 105°. What is the value (in degrees) of OSP?

A. 95 B. 45 C. 75 D. 65

Answer - Option D
Explanation -

Angle PQR=100 °
Angle STR=105 °
Refex angle of SOR= 2 × STR = 2 × 105 = 210 °
Angle SOR = 360 - 210 = 150
Angle PQR = 100
So angle formed by chord PR at the centre must be double of PQR
Hence angle POR = 2 × PQR = 200 °
Angle SOP+angle SOR = 200 °
Angle
So angle SOP = 180 - 150 = 50 °
In tringle SOP OS = OP = radius of angle
So angle OSP = angle OPS = x
2x + 50 = 180
X = 65 °
So angle SOP = 65°
6, The diagonal of a cube is [latex]\sqrt {192}[/latex] Its volume (in [latex]{cm}^{3}[/latex]) will be

A. 216 B. 432 C. 512 D. 624

Answer - Option C
Explanation -
Diagonal of a cube is = [latex]\sqrt {3}[/latex] x = [latex]\sqrt {192}[/latex]
x= [latex]\sqrt {\frac {192}{3}}[/latex] = 8cm
Now, volume of the cube = [latex]{a}^{3}[/latex] = ([latex]{8}^{3}[/latex]) = 512 [latex]{cm}^{3}[/latex]
7. Find the ratio of internal angle to external angle in a regular polygon which has 54 diagonals.

A. 3 : 4 B. 5 : 3 C. 5 : 1 D. 6 : 2

Answer - Option C
Explanation -
We know that,
no of diaonals = [latex]\frac {n(n - 3)}{2}[/latex]
54 = [latex]\frac {n(n - 3)}{2}[/latex]
n (n - 3) = 108
[latex]{n}^{2} - 3n - 108 = 0[/latex]
[latex]{n}^{2} - 12n + 9n - 108 = 0[/latex]
n(n - 12) + 9(n - 12) = 0
(n - 12)(n + 9) = 0
n = 12
Each interior angle
[latex]\frac {(n - 2)180}{n}[/latex]
[latex]\frac {10 * 180}{12} = 150[/latex]
Each exterior angle = 360/12 = 30°
Required
Interior angle : Exterior angle = 5 : 1
8. What is the area (in sq cm) of a rectangle if its diagonal is 26 cm and one of its sides is 10 cm?

A. 120 B. 240 C. 360 D. 480

Answer - Option B
Explanation -
Given: diagonal of a rectangle = 26 cm and length, l = 10 cm
Then, breadth, b = [latex]\sqrt {{26}^{2} - {10}^{2}}[/latex]
= [latex]\sqrt {576}[/latex]
= 24 cm
Now, area of the rectangle = l × b
= 10 × 24
240 [latex]{cm}^{2}[/latex]
9. If ABCDEF is a regular hexagon, then what is the value (in degrees) of ADB?

A. 15 B. 30 C. 45 D. 60

Answer - Option B
Explanation -
Since ABCDEF is a regular hexagon
Measure of each interior angle = [latex]\frac {(n - 2)180}{n}[/latex] = [latex]\frac {(6 - 2)180}{6}[/latex] = 120
ADB = [latex]\frac {120}{4}[/latex] = 30
10. An equilateral triangle of area 300 [latex]{cm}^{2}[/latex] is cut from its three vertices's to form a regular hexagon. Area of hexagon is what percent of the area of triangle?

A. 66.66% B. 33.33% C. 83.33% D. 56.41%

Answer - Option A
Explanation -
Area of [latex]\triangle[/latex] = [latex]\frac {\sqrt {3}}{4}{a}^{2} = 300[/latex]
[latex]{a}^{2} = \frac {4}{\sqrt {3}} * 300[/latex]
a = 20[latex]{\sqrt {3}}{4}[/latex]
Side of Hexagon = [latex]\frac {20}{3}\sqrt 4{3}[/latex]
Area of Hexagon = [latex]6 * \frac {\sqrt {3}}{4} {\frac {20}{3}\sqrt 4{3}}^{2}[/latex]
= 6 * [latex]\frac {\sqrt {3}}{4}\frac {400}{9}\sqrt {3}[/latex]
= 200
Required
% = [latex]\frac {200}{300} * 100[/latex]
= 66.66%

1. If [latex]{2}^{x} = {3}^{y} = {6}^{-z} [/latex] then [latex]\frac {1}{x} + \frac {1}{y} + \frac {1}{z}[/latex] is equal to

A. 0 B. 1 C. [latex]\frac {3}{2}[/latex] D. [latex]\frac {3}{2}[/latex]

Answer Option A
Explanation :
[latex]{2}^{x} = {3}^{y} = {6}^{-z} [/latex] = k
[latex]2 = {k}^{\frac {1}{x}}: 3 = {k}^{\frac {1}{y}}; 6 = {k}^{- \frac {1}{z}} [/latex]
2 * 3 = 6
[latex]{k}^{\frac {1}{x}} \times {k}^{\frac {1}{y}}; = {k}^{- \frac {1}{z}} [/latex]
[latex]{k}^{\frac {1}{x} + \frac {1}{y}} = {k}^{- \frac {1}{z}} [/latex]
[latex]\frac {1}{x} + \frac {1}{y} = - \frac {1}{z}[/latex]
[latex]\frac {1}{x} + \frac {1}{y} + \frac {1}{z}[/latex] = 0
2. If the selling price is Rs 1680 after getting a discount of 16%, what was the marked price?

A. Rs 2000 B. Rs 1948.8 C. Rs 1411.2 D. Rs 1448

Answer Option A
Explanation :
We know that,
MP = SP [latex]\times \frac {100}{100 - discount%}[/latex]
= 1680 [latex]\times \frac {100}{84}[/latex]
= Rs 2000
3. A number consists of two digits and the digit in the ten’s place exceeds the digit in the units place by 5. If 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is

A. 11 B. 7 C. 9 D. 13

Answer Option C
Explanation :
Let unit’s digit be x.
Ten’s digit = x + 5
Number = 10 (x + 5) + x = 11x+ 50
Again,
= 11x + 50 - 5 (2x + 5) = 10x + x + 5
[latex]\Rightarrow [/latex] 11x + 50 - 10x - 25 = 11x + 5
[latex]\Rightarrow [/latex] 10x = 20 [latex]\Rightarrow [/latex] x = 2
i.e, Required sum
= 2x + 5 = 2 [latex]\times [/latex] 2 + 5 = 9
4. A discount of 20% on one article is the same as a discount of25% on another article. The cost price of the two articles can be:

A. Rs. 90 and Rs. 40 B. Rs. 50 and Rs. 40 C. Rs. 40 and Rs. 60 D. Rs. 50 and Rs. 90

Answer Option B
Explanation :
20% of x = 25% of y
[latex]\Rightarrow \frac {x}{y} = \frac {25}{20} = \frac {5}{4}[/latex]
So ,x and y must be in the ratio of 5 : 4.
Rs. 50 and Rs. 40
5. The bus fare between two cities is increased in the ratio 1:2. Find the increase in the fare, if the original fare is Rs. 175.

A. Rs 350 B. Rs 70 C. Rs 140 D. Rs 175

Answer Option D
Explanation :
Let the fare before and after increase be x and 2x respectively, then
Increase in fare = 2x – x = x
i.e, Increase in fare = x = Rs 175
6. If the diameter of a circle is increased by 8% then its area is increased by:

A. 16.64% B. 6.64% C. 16% D. 16.46%

Answer Option A
Explanation :
8 + 8 + [latex] \frac {8 \times 8}{100}[/latex] = 16.64%
7. The cross section of a canal is in the shape of an isosceles trapezium which is 4 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and it is 120 m long, what is the maximum capacity of this canal?

A. 2160 cubic mts B. 3240 cubic mts C. 4329 cubic mts D. 1080 cubic mts

Answer Option D
Explanation :
Width of canal = 4m
Length of canal = 120m
Depth of canal = 2m
Crosse sectional area of canal = [latex] \frac {1}{2} 120 (4 + 5)[/latex] = 540sqcm
Maximum capacity of canal= cross sectional area × height = 540× 2 = 1080 cubic mts
8. A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical shaped bottles of diameter 5 cm and height 8 cm. The number of bottles required to empty the bowl is

A. 45 B. 40 C. 50 D. 60

Answer Option A
Explanation :
Number of bottles = [latex] \frac {Volume of hemisphere}{Volume of cylinder} = \frac { \frac {2}{3}\pi {R}^{3}}{\pi {r}^{2} h }[/latex]
[latex] \frac {\frac {2}{3} \times 15 \times 15 \times 15}{{(2.5)}^{2}\times 8}[/latex] = 45
9. Marked price of an item is Rs 200. On purchase of 1 item discount is 22%, on purchase of 4 items discount is 33%. Rabia buys 5 items, what is the effective discount?

A. 35 percent B. 30.8 percent C. 34 percent D. 20.4 percent

Answer Option B
Explanation :
Marked price of 4 items = 800
Discount on 4 items= 33 %
Then dicount= 33 % 800 = 264
Discount on one item = 22 % 200 = 44
Total discount on 5 items = 264 + 44 = 308
Effective discount = 308 [latex]\times \frac {100}{1000}[/latex] = 30.8
10. The average of seven consecutive positive integers is 26. The smallest of these integers is:

A. 21 B. 23 C. 25 D. 26

Answer Option B
Explanation :
x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 26 * 7
x = 182 - 21 = 161
x = [latex] \frac {161}{7}[/latex] = 23

1. The distance between the centres of two equal circles each of radius 3 cm is 10 cm. The length of a transverse tangent is

A. 4 cm B. 6 cm C. 8 cm D. 10 cm

Answer Option C
Explanation :
Transverse tangent formula
PQ = [latex]\sqrt {{d}^{2} - {(r + r)}^{2}} = \sqrt {{10}^{2} - {(3 + 3)}^{2}}[/latex]
[latex]\sqrt {64}[/latex] = 8
2.If [latex]{x}^{2} + {y}^{2} - 4x - 4y + 8 = 0,[/latex] then the value of x - y is

A. 4 B. - 4 C. 0 D. 8

Answer Option C
Explanation :
[latex]{x}^{2} + {y}^{2} - 4x - 4y + 8 = 0,[/latex]
[latex]\Rightarrow {x}^{2} -4x + 4 + {y}^{2} - 4y + 4 = 0,[/latex]
[latex]\rightarrow[/latex][latex]{(x - 2)}^{2} + {(y - 2)}^{2} = 0,[/latex]
[latex]\rightarrow[/latex](x - 2) = 0 and (y - 2) = 0
[latex]\rightarrow[/latex] = 2, y = 2
Now (x - y) = (2 - 2) = 0
3. Two trains start from station A and B and travel towards each other at speeds of 16 miles/hour and 21 miles/hour respectively. At the time of their meeting, the second train has travelled 60 miles more than the first. The distance between A and B (in miles) is:

A. 444 B. 496 C. 333 D. 540

Answer Option A
Explanation :
Here speeds of two trains = 16 miles/hour and 21 miles/hour respectively.
So I can say in one-hour second train runs 5 miles more than the first train.
Now at the time of their meeting, the second train has travelled 60 miles more than the first. So I can say trains meet after 12 hours([latex] \frac {60}{5}[/latex])
Hence distance between A & B = (16 + 21)(12) = 444 miles
Hence Option A is correct
4. By selling a fan for Rs. 600, a man loses 10%. To make a gain of 20%, the selling price of the fan should be

A. 900 B. 1000 C. 700 D. 800

Answer Option D
Explanation :
CP of fan = [latex] \frac {600 \times 100}{90}[/latex]
Required SP = [latex] \frac {600 \times 100}{90} \times \frac {120}{100}[/latex] = Rs 800
Hence Option D is correct
5. Find the value of 3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3} + 3} + \frac {1}{\sqrt {3} - 3}[/latex]

A. 3 B. 4 C. 45 D. 9

Answer Option A
Explanation :
3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3} + 3} + \frac {1}{\sqrt {3} - 3}[/latex]
= 3 + [latex]\frac {1}{\sqrt {3}} + \frac {\sqrt {3} - 3 + \sqrt {3} + 3}{(\sqrt {3} + 3) (\sqrt {3} - 3)} + \frac {1}{\sqrt {3} - 3}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac {2 \sqrt {3}}{3 - 9}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac { \sqrt {3}}{3}[/latex]
3 + [latex]\frac {1}{\sqrt {3}} + \frac {1}{\sqrt {3}}[/latex] = 3
6. The area of the triangle formed by the graphs of 3x + 4y = 12. x-axis and y-axis (in sq. units)'

A. 4 B. 12 C. 6 D. 8

Answer Option C
Explanation :
Putting y=0 in 3x + 4y = 12, we get x = 4
Co-ordinates of point of intersection on x-axis = (4, 0)
Similarly for x =0 we get y =3
Co-ordinates of point of intersection on y-axis= (0, 3)
Area of Triangle formed = [latex]\frac {1}{2}[/latex] × 4× 3 = 6 sq units
Hence option C is correct
7. The greatest among the numbers [latex]\sqrt {0.09}[/latex], [latex]\sqrt {0.064}[3][/latex], 0.5 and [latex]\frac {3}{5}[/latex] is

A. [latex]\sqrt {0.09}[/latex] B. [latex]\sqrt {0.064}[3][/latex] C. 0.5 D. [latex]\frac {3}{5}[/latex]

Answer Option D
Explanation :
[latex]\sqrt {0.09}[/latex] = 0.3
[latex]\sqrt [3] {0.064}[/latex] = 0.4
0.5
[latex]\frac {3}{5}[/latex] = 0.6 Hence, it is clear that [latex]\sqrt {0.09}[/latex] < [latex]\sqrt {0.064}[3][/latex] < 0.5 < [latex]\frac {3}{5}[/latex]
8. Simplify : [latex]\sqrt {3 \frac {33}{64}} \div \sqrt {9 \frac {1}{7}} \times 2 \sqrt {3 \frac {1}{9}}[/latex]

A. 2 B. 44 C. 49 D. 2 [latex] \frac {3}{16}[/latex]

Answer Option D
Explanation :
[latex]\sqrt {3 \frac {33}{64}} \div \sqrt {9 \frac {1}{7}} \times 2 /sqrt {3 \frac {1}{9}}[/latex]
[latex]= \sqrt {\frac {225}{64}} \div \sqrt {\frac {64}{7}} \times 2 \sqrt { \frac {28}{9}}[/latex]
[latex]= \frac {15}{8} \div \frac {8}{\sqrt {7}} \times 2 \times 2 \times \frac {/sqrt {7}}{9}[/latex]
[latex]= \frac {5 \times 7}{16} = \frac {35}{16} = 2 \frac {3}{16}[/latex]
9. If [latex]= \frac {a}{1 - a}+ \frac {b}{1 - b} + \frac {c}{1 - c}[/latex] = 1 then the value of [latex] = \frac {1}{1 - a} + \frac {1}{1 - b} + \frac {1}{1 - c}[/latex] is

A. 1 B. 2 C. 3 D. 4

Answer Option D
Explanation :
[latex]= \frac {a}{1 - a} = \frac {b}{1 - b} + \frac {c}{1 - c}[/latex] = 1
[latex]= ( \frac {a}{1 - a} + 1) = (\frac {b}{1 - b} + 1) + ( \frac {c}{1 - c} + 1)[/latex] = 3 + 1 = 4
[latex]= \frac {a + 1 - a}{1 - a} = \frac {b + 1 - b}{1 - b} + \frac {c + 1 - c}{1 - c}[/latex] = 4
[latex] = \frac {1}{1 - a} + \frac {1}{1 - b} + \frac {1}{1 - c}[/latex] = 4
10. [latex] = 3 \frac {1}{2}[/latex] - ([latex]2 \frac {1}{4}[/latex] + ([latex] 1 \frac {1}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] = 1 \frac {1}{2} - \frac {1}{3} - \frac {1}{6}[/latex]))) is

A. [latex]\frac {1}{2}[/latex] B. [latex]2 \frac {1}{2}[/latex] C. [latex]3 \frac {1}{2}[/latex] D. [latex]9 \frac {1}{2}[/latex]

Answer Option A
Explanation :
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] \frac {3}{2} - \frac {1}{3} - \frac {1}{6}[/latex])))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4}[/latex] - [latex] \frac {1}{2}[/latex]([latex] \frac {9 - 2 - 1}{6}[/latex])))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5}{4} - \frac {1}{2}[/latex]))
[latex] = \frac {7}{2}[/latex] - ([latex]\frac {9}{4}[/latex] + ([latex] \frac {5 - 2}{4}[/latex]))
[latex] = \frac {7}{2}[/latex] - ([latex] \frac {9}{4} + \frac {3}{2}[/latex])
[latex] = \frac {7}{2} - 3[/latex] = [latex] \frac {7 - 6}{2} = \frac {1}{2}[/latex]

1. If sin[latex]\theta[/latex] - cos [latex]\theta[/latex] = [latex]\frac {7}{13}[/latex] and 0 < [latex]\theta[/latex] < 90, then the value of sin[latex]\theta[/latex] + cos [latex]\theta[/latex] is

A. [latex] \frac {17}{13}[/latex] B. [latex] \frac {13}{17}[/latex] C. [latex] \frac {1}{13}[/latex] D. [latex] \frac {1}{17}[/latex]

Answer Option A
Explanation : sin[latex]\theta[/latex] - cos [latex]\theta[/latex] = [latex] \frac {7}{13}[/latex] ....(i)
sin[latex]\theta[/latex] + cos [latex]\theta[/latex] = x ....(ii)
On squaring both equations and adding,
2( [latex]{sin}^{2} \theta + {cos}^{2}\theta[/latex]) = [latex] \frac {49}{169} + {x}^{2}[/latex]
[latex] {x}^{2} = 2 \frac{338 - 49} {169} = \frac{289} {169}[/latex]
x = [latex] \sqrt {\frac{289} {169}} = \frac{17} {13}[/latex]
2. Successive discounts of 50% and 50% is equivalent to

A. 100% B. 75% C. 50% D. 25%

Answer Option B
Explanation :
Let X be the price
Price after first 50% discount = x - [latex] \frac {50 x}{100}[/latex] = - [latex] \frac {x}{2}[/latex] = [latex] \frac {x}{2}[/latex]
Price after second 50% discount
[latex] \frac {x}{2}[/latex] - [latex] \frac {50 \times\frac{x}{2}}{100}[/latex] = [latex] \frac {x}{2} - \frac {x}{4}[/latex] = [latex] \frac {x}{4}[/latex]
Hence final price = [latex] \frac {x}{4}[/latex]
Total discount = x - [latex] \frac { x}{4}[/latex] = [latex] \frac {3 x}{4}[/latex]
% total discount = ([latex] \frac {3 x}{4}[/latex]) × [latex] \frac {100}{x}[/latex] = 75%
3. The numerical value of ([latex] \frac {1}{cos \theta} + \frac {1}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {1}{cos \theta}[/latex]) is?

A. 0 B. -1 C. 1 D. 2

Answer Option C
Explanation : ([latex] \frac {1}{cos \theta} + \frac {1}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {1}{cos \theta}[/latex])
([latex] \frac {1}{cos \theta} + \frac {sin \theta}{cos \theta}[/latex])([latex] \frac {1}{cos \theta} - \frac {sin \theta}{cos \theta}[/latex])
[latex] \frac {1 + sin \theta}{cos \theta} + \frac {1 - sin \theta}{cos \theta}[/latex]
[latex] \frac {1 - {sin}^{2} \theta}{cos \theta} = \frac {{cos}^{2} \theta}{{cos}^{2} \theta}[/latex] = 1
4. The value of tan4°× tan43°× tan47° × tan86° is

A. 0 B. 1 C. [latex]\sqrt {3} [/latex] D. [latex]\frac {1}{\sqrt {3}} [/latex]

Answer Option B
Explanation : = tan 4°× tan 43°× tan 47° × tan 86°
= (tan 4°× tan 86°)× (tan 47° × tan 43°)
= (tan 4°× cot 4°)× (tan 43° × cot43°)
= 1
5. There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of this pyramid is of length [latex]\frac {5 a}{2}[/latex] cm, then the volume of this pyramid is

A. [latex]3{a}^{2} {cm}^{3}[/latex] B. [latex]3 \sqrt {2}{a}^{2} {cm}^{3}[/latex] C. [latex]3 \sqrt {3}{a}^{3} {cm}^{3}[/latex] D. [latex]6{a}^{2} {cm}^{3}[/latex]

Answer Option C
6. [latex]\frac {sin A}{(1 + cosA)} + \frac {(1 + cosA)}{sin A}[/latex] is equal to

A. 2sec A B. 2 cosec A C. 2 tan A D. 2 cot A

Answer Option B
Explanation : [latex]\frac {sin A}{(1 + cosA)} + \frac {(1 + cosA)}{sin A} [/latex]
[latex]\frac {sin A}{(1 + cosA)} \times \frac {1 - cosA}{1 - cosA} [/latex] + [latex]\frac {1 + cosA}{sinA)} \times \frac {sin A}{sin A} [/latex]
[latex]\frac {sin A - sin AcosA + sin A + sin AcosA}{{sin}^{2}A}[/latex]
= 2cosecA
7. If the ratio of volumes of two cones is 2 : 3 and the ratio of the radius of their bases is 1 : 2 then the ratio of their heights will be

A. 3 : 4 B. 4 : 3 C. 3 : 8 D. 8 : 3

Answer Option D
Explanation : Volume of cone = [latex]\frac {1}{3 \pi {r}^{2} h} [/latex]
Required ratio = [latex]\frac {1 \pi {{r}_{1}}^{2} {h}_{1}}{3} [/latex] : [latex]\frac {1 \pi {{r}_{2}}^{2} {h}_{2}}{3} [/latex] = 2 : 3
[latex]\frac {1}{4} \times \frac {{h}_{1}}{{h}_{2}} = \frac {2}{3}[/latex]
[latex]\frac {{h}_{1}}{{h}_{2}} = \frac {2}{3} \times = \frac {8}{3}[/latex]
Hence option D is correct
8. The ratio of radii of two cones is 3 : 4 and the ratio of their heights is 4 : 3. Then the ratio of their volumes will be

A. 3 : 4 B. 4 : 3 C. 9 : 16 D. 16 : 9

Answer Option A
Explanation : [latex]\frac {{V}_{1}}{{V}_{2}} [/latex] = [latex]\frac {\pi {{r}_{1}}^{2} {h}_{1}}{\pi {{r}_{2}}^{2} {h}_{2}}\frac {{V}_{1}}{{V}_{2}} [/latex] = [latex]\frac {\pi {{r}_{1}}^{2} {h}_{1}}{\pi {{r}_{2}}^{2} {h}_{2}}[/latex]
= [latex]{\frac {3}{4}}^{2} \times \frac {4}{3}[/latex]
= [latex] \frac {3}{4} \times \frac {3}{4} \times \frac {4}{3} = \frac {3}{4}[/latex]
9. Two numbers are respectively 25% and 40% less than a third number. What per cent of the first is the second?

A. 60 B. 70 C. 40 D. 60

Answer Option D
Explanation : Let third no = 100
[latex]{1}^{st}[/latex] and [latex]{2}^{nd}[/latex] numbers will be then 75 and 60
Required Percentage = 60 [latex]\times \frac {100}{75}[/latex] = 80 %
10. From each of the two given unequal numbers, half the smaller number is subtracted. Then of the resulting numbers, the larger one is five times than the smaller one. Then the ratio of the larger to smaller one is

A. 2 : 1 B. 3 : 2 C. 3 : 1 D. 1 : 4

Answer Option C
Explanation : Let the numbers be x and y where x > y
i.e, x - [latex] \frac {y}{2} = 5 (y - \frac {y}{2})[/latex]
x - [latex] \frac {y}{2} = 5 \frac {y}{2}[/latex]
x = [latex] \frac {5y}{2} + \frac {y}{2} = \frac {6y}{2}[/latex]
x = 3y
[latex]\frac {x}{y} = \frac {3}{1}[/latex]
i.e, x : y = 3 : 1

1. Marks of two candidates P and Q are in the ratio of 2 : 5. If the marks of P are 120, marks of Q are

A. 120 B. 240 C. 300 D. 360

Answer Option C
Explanation :
Let marks of P and Q be 2x and 5x
2x = 120 x = 60
5x = 300
Marks of Q = 300
2. If (a + b) : (b + c) : (c + a) = 6 : 7 : 8 and (a + b + c) = 14, then the value of c is

A. 6 B. 7 C. 8 D. 14

Answer Option A
Explanation :
[latex]\frac {a + b}{6} = \frac {b + c}{7} = \frac {c + a}{8}[/latex] = k
[latex]\Rightarrow a + b = 6k;b + c = 7k;[/latex]
c + a = 8k
i.e, a + b + b + c + c + a
[latex]\Rightarrow 6k + 7k + 8k [/latex]
[latex]\Rightarrow 2 \times 14 = 21 k \Rightarrow k = \frac {4}{3} [/latex]
i.e, c = (a + b+ c)- (a + b)
= 14 - 6 [latex]\times \frac {4}{3} = 14 - 8 = 6[/latex]
3. If the price of rice be raised by 25%, the percent by which a house-holder must reduce his consumption of rice so as not to increase his expenditure on rice is

A. 22.5 B. 25.75 C. 25 D. 20

Answer Option D
Explanation :
Price ratio = 100 : 125
Consumption ratio = 125:100
So,reduction = [latex]25 \times \frac {100}{125}[/latex] =20 %
4. If x = a(b - c), y = b(c - a) and z = c (a - b), then [latex]{\frac {x}{a}}^{3} + {\frac {y}{b}}^{3} + {\frac {z}{c}}^{3}[/latex] =

A. [latex] \frac {xyz}{3abc}[/latex] B. 3 xyzabc C. [latex] \frac {3 xyz}{abc}[/latex] D. [latex] \frac {xyz}{abc}[/latex]

Answer Option C
Explanation :
[latex] \frac {x}{a} = b - c; \frac {y}{b} = c - a; \frac {z}{c} = a - b[/latex]
Again, b - c + c - a + a - b = 0
Using identity If a + b +c = 0, then [latex]{a}^{3} + {b}^{3} + {c}^{3} [/latex] = 3abc
i.e, [latex]{\frac {x}{a}}^{3} + {\frac {y}{b}}^{3} + {\frac {z}{c}}^{3} [/latex] = [latex]{b - c}^{3} + {c - a}^{3} + {a - b}^{3} [/latex]
= 3 (b - c)(c - a)(a - b)
= [latex] \frac {3 xyz}{abc}[/latex]
5. A team played 40 games in a season and won in 24 of them. What percent of games played did the team win?

A. 70 % B. 40 % C. 60 % D. 35 %

Answer Option C
Explanation :
Total game played= 40
Number of won games = 24
% of won games = [latex]24 \times \frac {100}{40}[/latex] = 60 %
6. If the total surface area of a cube is 96 [latex]{cm}^{2}[/latex], its volume is

A. 56 [latex]{cm}^{3}[/latex] B. 16 [latex]{cm}^{3}[/latex] C. 64 [latex]{cm}^{3}[/latex] D. 36 [latex]{cm}^{3}[/latex]

Answer Option C
Explanation :
Edge of cube = x cm
i.e, [latex]6{x}^{2} \Rightarrow 96 {x}^{2} = \frac {96}{6}[/latex] = 16
[latex]\Rightarrow x = \sqrt {16} cm = 4 [/latex]
Volume of cube = [latex]{(edge)}^{2}[/latex] (edge)3 = (4)3 = 64 cu. Cm
7. The radius of cross section of a solid cylindrical rod of iron is 50 cm. The cylinder is melted down and formed into 6 solid spherical balls of the same radius as that of the cylinder. The length of the rod (in metres) is

A. 0.8 B. 2 C. 3 D. 4

Answer Option D
Explanation :
Length of rod = h cm
i.e, Volume of cylinder = Volume of 6 spheres
[latex]\Rightarrow \pi {r}^{2} h = 6 \times \frac {4}{3} \pi {r}^{3} [/latex]
[latex]\Rightarrow h = 6 \times \frac {4}{3} \times r = 8 \times 50[/latex]
= 400 cm = 4 metre
8. In a division sum, the divisor is 3 times the quotient and 6 times the remainder. If the remainder is 2, then the dividend is

A. 50 B. 48 C. 36 D. 28

Answer Option A
Explanation :
Divisor = 6 × 2 = 12
Again divisor = 3 × quotient
Quotient = [latex] \frac {12}{3}[/latex] = 4
Dividend = 12 × 4 + 2 = 50
Hence Option A is correct
9. If [latex] \frac {a}{b} + \frac {b}{a}[/latex] = 2, then the value of a - b is :

A. 2 B. -1 C. 0 D. 1

Answer Option C
Explanation :
[latex] \frac {a}{b} + \frac {b}{a}[/latex] = 2
[latex] \frac {{a}^{2} + {b}^{2}}{ab} = 2 [/latex]
[latex]{a}^{2} + {b}^{2} - 2ab = 0[/latex]
[latex]{(a - b)}^{2} = 0[/latex]
a - b = 0
[latex]\Rightarrow 12 x - 1[/latex]
New Number = 10(2x - 1) + x = 0
10. The length of the diagonal of a rectangle is 10 cm and that of one side is 8 cm. What is the area of this rectangle?

A. 80 sq cm B. 48 sq cm C. 60 sq cm D. 32 sq cm

Answer Option B
Explanation :
In rectangle we know that ,
[latex]{D}^{2} = {L}^{2} + {B}^{2}[/latex]
100= 64 + [latex]{B}^{2}[/latex]
[latex]{B}^{2}[/latex] = 100 - 64 = 36
B = 6
Area = L × B = 6 × 8 = 48 sq cm