1.Simplify the expression using BODMAS rule:: ([latex]\frac {3}{2}[/latex]) of ([latex]\frac {4}{7}[/latex]) {(10 × 3) – (8 × 2)}
Answer - Option B
Explanation - Applying BODMAS rule = ([latex]\frac {3}{2}[/latex]) of ([latex]\frac {4}{7}[/latex]) {30 – 16} = ([latex]\frac {12}{14}[/latex]) × 14 = 12
Therefore, the correct answer is option 2.
2. The six digit number 54321A is divisible by 9 where A is a single digit whole number. Find A.
Answer - Option D
Explanation -A number is divisible by 9, when the sum of its digits is divisible by 9. Here, 5 + 4 + 3 + 2 + 1 + A = 15 + A should be divisible by 9. Therefore, A = 3 gives 15+ 3 = 18 as the sum of digits, which is divisible by 9. So, answer is option 4.
3. Find the fractional equivalent of 6789.
A. [latex]\frac {5111}{9000}[/latex]
B. [latex]\frac {6117}{9000}[/latex]
C. [latex]\frac {6119}{9000}[/latex]
D. None of these
Answer - Option D
Explanation - 6789 = [latex]\frac {(6789-678)}{9000}[/latex] = [latex]\frac {6111}{9000}[/latex]
4. In a library 2/3 of books are in English and the remaining 300 are of Hindi. Find the total books in library
A. 400
B. 900
C. 275
D. None of these
Answer - Option B
Explanation -[latex]\frac {2}{3}[/latex] of books are in English means remaining 1 – [latex]\frac {2}{3}[/latex] = [latex]\frac {1}{3}[/latex] are in Hindi which is equal to 300. So 1/3rd of total = 300 ⇒ Total books = 300 x 3 = 900.
5. Which of the following fraction is the largest?
A. [latex]\frac {8}{9}[/latex]
B. [latex]\frac {18}{23}[/latex]
C. [latex]\frac {13}{21}[/latex]
D. [latex]\frac {14}{17}[/latex]
Answer - Option A
Explanation -Converting the given fractions into decimal form, we have,
[latex]\frac {8}{9}[/latex] = 0.88
[latex]\frac {18}{23}[/latex] = 0.78
[latex]\frac {13}{21}[/latex] = 0.76
[latex]\frac {14}{17}[/latex] = 0.82
From the above, 0.88 is the largest.
The corresponding fraction is.
= [latex]\frac {8}{9}[/latex]
6. Bala and Aruna can complete a work in 30 days. Bala can do it alone in 50 days. How many days would Aruna take to complete the work alone?
A. 100 days
B. 100 days
C. 90 days
D. 75 days
Answer - Option D
Explanation -By using the alternate method as mentioned in Question 1, we are going to solve further problems.
LCM of 30 and 50 is 150.
Bala and Aruna can complete the whole work in 30 days. In 1 day they can complete 5(150/30) of the work.
Bala alone can complete the whole work in 50 days. In 1 day he can complete 3(150/30) of the work.
Aruna alone can complete in 1 day (5 - 3)of the work.
Therefore, Aruna will complete the whole work in [latex]\frac {150}{2}[/latex] = 75 days.
7. Prakash can complete a piece of work in 24 days and Praveen can complete the same piece of work in 36 days. If Prakash and Praveen work together, how long would they take to complete the work?
A. 14.4 days
B. 15.5 days
C. 12 days
D. 20 days
Answer - Option A
Explanation - Prakash can complete in one day = [latex]\frac {1}{24}[/latex]
of work.
Praveen can complete in one day = [latex]\frac {1}{36}[/latex]
of work.
Prakash and Praveen can complete in one day = [latex]\frac {1}{24}[/latex] + [latex]\frac {1}{36}[/latex]
of work.
= 3 + [latex]\frac {2}{72}[/latex]
= [latex]\frac {5}{72}[/latex]
Both can complete the work in 72/5 days. i.e., 14.4 days
Usually, we will solve this type of problem in the above method. There is another alternative method to solve these types of problems.
i.e., we can solve by using the LCM method instead of the fraction method.
8. if the numerator and denominator of a fraction is diminished and increased by 70% respectively then the resultant fraction is [latex]\frac {39}{34}[/latex]
. find the original fraction.
A. [latex]\frac {17}{11}[/latex]
B. [latex]\frac {13}{11}[/latex]
C. [latex]\frac {13}{2}[/latex]
D. [latex]\frac {17}{2}[/latex]
Answer - Option C
Explanation - let the original fraction be [latex]\frac {X}{Y}[/latex]
then (30% of x)/(170% of y) = [latex]\frac {39}{34}[/latex]
x/y = ([latex]\frac {39}{34}[/latex])*([latex]\frac {170}{30}[/latex])
= [latex]\frac {13}{2}[/latex]
9. The percentage of numbers from 11 to 75 which are divided by 5 or 7 is:
Answer - Option A
Explanation - Clearly, the numbers which divided by 5 or 7 from 11 to 75 are 14, 15, 20, 21, 25, 28, 30, 35, 40, 42, 45, 49, 50, 55, 56, 60, 63, 65, 70, 75.
The number of such numbers = 20
Total numbers from 11 to 75 = 65
Then the required percentage = (20/65) x 100 = 30.76% = 31 %
Hence the answer is 31%
10. Ramu buys 3 dozen tomatoes for Rs. 360. He buys carry bags to take them to his shop. Only one dozen can be filled into a carry bag and cost of each carry bag is Rs.10. He then sells the tomatoes along with carry bags for a cost of 200 per dozen. What is his profit percentage.
A. 53.84%
B. 60%
C. 56.67%
D. 63.84%
Answer - Option A
Explanation - Cost of 3 dozens of tomatoes is Rs. 360
Cost of 1 carry bag = Rs. 10. Therefore, three of them cost Rs. 30
Overall CP of tomatoes = 360 + 30 = Rs. 390
SP of each dozen of tomatoes = Rs. 200
Therefore SP of three dozens = 600
Profit percentage = (SP - CP) / CP x 100% = (600 - 390)/390 x 100% = 210/390x 100% = 2100/39 % = 53.84%
11. Ravi buys a second hand car for Rs. 2,00,000. He then repaints it for Rs. 2000. He attaches new threading to all 4 tyres. Cost of threading per tyre is Rs. 200. At what price he should resell the car so that he gains 10%.
A. Rs. 223080
B. Rs. 223090
C. Rs. 232080
D. Rs. 223880
Answer - Option A
Explanation - Cost of second hand car = Rs. 2,00,000
Repainting cost = Rs. 2000
Cost of threading per tyre = Rs. 200
Cost of threading for all tyres = Rs. 200 x 4 = Rs. 800
Total costs involved = Actual cost + Repainting costs + Threading costs = 2,00,000 + 2000 + 800 = 202800
Profit Percentage = 10% = (SP - CP)/CP x 100%
Or 10% = (SP - 202800)/202800 x 100%
Or 10 = (SP - 202800)/2028
Or SP = 20280 + 202800 = 223080
12. A bike at a speed of 60km/hr, it covers certain distance in 3 hours and the train can cover the same distance in 1 and half hours then the speed of train is
A. 80km/hr
B. 120km/hr
C. 160km/hr
D. 50km/hr
Answer - Option B
Explanation - The diatance covered by the bike with speed 60km/hr in 3 hours = (60 x 3) = 180 km
Now, Speed = Distance/Time
since the train covers the same 180 km in 3/2 hours (we can write 1 and half hours as 3/2 hours)
Then the speed of the train = 180/ (3/2) = 120 km/hr.
Hence the train traveled at the speed of 120km/hr to cross 180km in 1 and half hours.
13. A car starts at A and reaches a destination in 8 hours. If it travels first and second half at the speed of 80 km/hr and 100 km/hr respectively then the distance between A and destination is:
A. 818km
B. 979km
C. 711km
D. 850km
Answer - Option C
Explanation - Let the distance between A and destination be X.
The total time taken by the car to cover X = 8 hours
Since X/2 by 80km/hr and remaining X/2 by 100km/hr
Then by Time = distance/speed, we have
(X/2)/80 + (X/2)/100 = 8
X/80 + X/100 = 16
5X + 4X = 6400
9X = 6400
X = 6400/9 = 711
Hence 711km is the required answer.
14. If A and B are connected by the relation [latex]{A}^{2}[/latex]
+ 4[latex]{B}^{2}[/latex]
= 4AB then A:B is:
A. 2:1
B. 1:2
C. 3:1
D. 2:1
Answer - Option A
Explanation - Given that the relation of A and B is [latex] {A}^{2}[/latex] + 4[latex]{B}^{2}[/latex] = 4AB
[latex]{A}^{2}[/latex] + 4[latex]{B}^{2}[/latex]- 4AB = 0
[latex]{(A - 2B)}^{2}[/latex] = 0 (by [latex]{(X - Y)}^{2}[/latex] = [latex]{X}^{2}[/latex] + [latex]{Y}^{2}[/latex] - 2xy )
(A - 2B) = 0
A = 2B
[latex]\frac {A}{B}[/latex] = [latex]\frac {2}{1}[/latex]
Hence, the answer is A:B = 2:1
15. An amount of money was lent for 3 years. What will be the difference between the simple and the compound interest earned on it at the same rate?
I. The rate of interest was 8 p.c.p.a.
II. The total amount of simple interest was Rs. 1200.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are necessary to answer
Answer - Option D
Explanation - Given: T = 3 years.
I. gives: R = 8% p.a.
II. gives: S.I. = Rs. 1200.
Thus, P = Rs. 5000, R = 8% p.a. and T = 3 years.
Difference between C.I. and S.I. may be obtained.
So, the correct answer is (E).
16. What will be the least number which when doubled will be exactly divisible by 12,18,21 and 30?
A. 630
B. 360
C. 603
D. 306
Answer - Option A
Explanation - LCM of 12,18,21 and 30 = 1260
Doubled (divide by 2) = [latex]\frac {1260}{2}[/latex] = 630
17. HCF and LCM of two numbers are 11 and 385 .If one number lies between 75 and 125 then that number is
A. 123
B. 73
C. 77
D. 154
Answer - Option C
Explanation - Product of the two number = 11a×11b = 4235
11a×11ab = [latex]\frac {4235}{12}[/latex] = 35
35 = 7 × 5 (co prime)
77 × 11 = 77
18. The HCF of 2511 and 3402 is
Answer - Option B
Explanation - 2511 = 81×31
3402 = 81×42
Hence HCF is 81
19. In a rectangle the ratio of the length and breadth is 3:2. If each of the length and breadth is increased by 4m their ratio becomes 10:7. The area of the original rectangle in m² is?
A. 846
B. 864
C. 840
D. 876
Answer - Option B
Explanation - [latex]\frac {3x + 4}{2x + 4}[/latex] = [latex]\frac {10}{7}[/latex]
x = 12
Area of the original rectangle = 3x x 2x = 6x²
Area of the original rectangle = 6 x 144 = 864m²
20. The area of a rectangle gets reduced by 9 square units,if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, then the area is increased by 67 square units. Find the area of the rectangle?
A. 159m
B. 179m
C. 147m
D. 153m
Answer - Option D
Explanation - Length = x; Breadth =y
xy – (x-5)(y+3) = 9
3x – 5y – 6 = 0 —(i)
(x + 3)(y + 2) – xy = 67
2x + 3y - 61 = 0 —(ii)
solving (i) and (ii)
x = 17m ; y = 9m
Area of the Rectangle = 153 m
21. The width of a rectangular piece of land is 1/4 th of its length. If the perimeter of the piece of land is 320m its length is?
A. 140M
B. 128M
C. 120M
D. 156M
Answer - Option B
Explanation - length = l ; breadth = [latex]\frac {1}{4}[/latex]
2(l + b) = 320
2(l + [latex]\frac {1}{4}[/latex]) = 320
l = 320 x [latex]\frac {4}{10}[/latex] = 128m
22. The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 500 cm. If the area of the rectangle is less than that of a Square then find the area of the rectangle?
A. 1800 cm²
B. 1500 cm²
C. 1100 cm²
D. 1600 cm²
Answer - Option C
Explanation - Perimeter of rectangle = Perimeter of Square = 160
4a = 160 => a = 40
Area of square = 1600
1600 – lb = 500
lb = 1100 cm²
23. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A. [latex]\frac {1}{2}[/latex]
B. [latex]\frac {2}{5}[/latex]
C. [latex]\frac {8}{15}[/latex]
D. [latex]\frac {9}{20}[/latex]
Answer - Option D
Explanation - Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) =[latex]\frac {n(E)}{n(S)}[/latex] =[latex]\frac {9}{20}[/latex]
24. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
A. [latex]\frac {1}{2}[/latex]
B. [latex]\frac {3}{4}[/latex]
C. [latex]\frac {3}{8}[/latex]
D. [latex]\frac {5}{16}[/latex]
Answer - Option B
Explanation - In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
=> n(E) = 27.
P(E) = [latex]\frac {n(E)}{n(S)}[/latex][latex]\frac {27}{36}[/latex][latex]\frac {3}{4}[/latex]
25. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
A. [latex]\frac {1}{10}[/latex]
B. [latex]\frac {2}{5}[/latex]
C. [latex]\frac {2}{7}[/latex]
D. [latex]\frac {5}{7}[/latex]
Answer - Option C
Explanation - P (getting a prize) = [latex]\frac {10}{10+25}[/latex]=[latex]\frac {10}{35}[/latex]=[latex]\frac {2}{7}[/latex]
26. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
A. [latex]\frac {1}{13}[/latex]
B. [latex]\frac {3}{13}[/latex]
C. [latex]\frac {1}{4}[/latex]
D. [latex]\frac {9}{52}[/latex]
Answer - Option B
Explanation - Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = [latex]\frac {12}{52}[/latex]=[latex]\frac {3}{13}[/latex]
27. The value of sin (45° + θ) – cos (45° – θ) is
A. 1
B. 0
C. 2 cos θ
D. 2 sin θ
Answer - Option B
Explanation - sin(45° + θ) – cos(45° – θ)
= sin{90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ)
{∵sin(90 – A) = cos A }
= 0
28. If sin (A + B) = 1, where 0 < B < 45°, then what is cos (A – B) equal to?
A. sin 2B
B. sin B
C. cos 2B
D. cos B
Answer - Option A
Explanation - ∵ sin (A + B) = 1
⇒ A + B = [latex]{sin}^{-1}[/latex] 1
⇒ (A + B) = 90°
⇒ B = 90° – A
⇒ A = 90° – B
Now, cos (A – B) = cos A cos B + sin A + sin B
= cos (90 – B) cos B + sin (90 – B) sin B
= sin B cos B + cos B sin B
= 2 sin B cos B = sin 2B.
29. What is sin 25° sin 35° sec 65° sec 55° equal to?
Answer - Option A
Explanation - sin 25° sin 35° sec 65° sec 55°
= sin25°.sin35°[latex]\frac {1}{cos65°}[/latex][latex]\frac {1}{cos55°}[/latex]
= sin25°.sin35°[latex]\frac {1}{sin25°}[/latex][latex]\frac {1}{sin35°}[/latex]
[ ∵ cos(90 – Θ) = sinΘ ]
= 1.
30. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57%
B. 60%
C. 65%
D. 90%
Answer - Option A
Explanation - Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = [latex]\frac {11628}{20400}[/latex] x 100% = 57%.
31. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
A. Rs. 650
B. Rs. 690
C. Rs. 698
D. Rs. 700
Answer - Option C
Explanation - S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.
32. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
Answer - Option D
Explanation - Principal
= Rs [latex]\frac {100 x 4016.25 }{9 x 5}[/latex]
= Rs [latex]\frac {401625}{45}[/latex]
= Rs. 8925.
33. Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
A. 3.6
B. 6
C. 18
D. Cannot be determined
Answer - Option B
Explanation - Let rate = R% and time = R years.
Then,[latex]\frac {1200 x R x R }{100}[/latex]= 432
=> 12[latex]{R}^{2}[/latex] = 432
=[latex]{R}^{2}[/latex]= 36
=R = 6.
34. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
Answer - Option B
Explanation - N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
35. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A. 9600
B. 9800
C. 9500
D.
Answer - Option A
Explanation - Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.