Partnership
1. A, B and C started a business by investing Rs. 1,20,000, Rs. 1,35,000 and ,Rs.1,50,000 respectively. Find the share of each, out of an annual profit of Rs. 56,700.
A. 1500,1300,1200
B. 12345,12354,18967
C. 16800,18900,21000
D. 12300,11500,17300
Answer: Option (C)
Explanaton:
Ratio of shares of A, Band C = Ratio of their investments
= 120000 : 135000 : 150000 = 8 : 9 : 10.
A’s share = Rs. (56700 x (8/27))= Rs. 16800.
B's share = Rs. ( 56700 x (9/27)) = Rs. 18900.
C's share = Rs. ( 56700 x (10/27))=Rs. 21000.
2. Alfred started a business investing Rs. 45,000. After 3 months, Peter joined him with a capital of Rs. 60,000. After another 6 months, Ronald joined them with a capital of Rs. 90,000. At the end of the year, they made a profit of Rs. 16,500. Find the lire of each.
A. 6600,6600,3300
B. 6660,6660,3330
C. 6620,6670,3340
D. 3600,3600,6600
Answer: Option (A)
Explanaton:
Clearly, Alfred invested his capital for 12 months, Peter for 9 months and Ronald for 3 months.
So, ratio of their capitals = (45000 x 12) : (60000 x 9) : (90000 x 3)
= 540000 : 540000 : 270000 = 2 : 2 : 1.
Alfred's share = Rs. (16500 x (2/5)) = Rs. 6600
Peter's share = Rs. (16500 x (2/5)) = Rs. 6600
Ronald's share = Rs. (16500 x (1/5)) = Rs. 3300.
3. A, B, and C start a business each investing Rs. 20,000. After 5 months A withdrew Rs.6000 B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.
A. 20400,20500,18700
B. 21500,22600,21700
C. 20500,21200,28200
D. 21900,22700,27300
Answer: Option (C)
Explanaton:
Ratio of the capitals of A, Band C
= 20000 x 5 + 15000 x 7 : 20000 x 5 + 16000 x 7 : 20000 x 5 + 26000 x 7
= 205000:212000 : 282000 = 205 : 212 : 282.
A’s share = Rs. 69900 x (205/699) = Rs. 20500
B's share = Rs. 69900 x (212/699) = Rs. 21200;
C's share = Rs. 69900 x (282/699) = Rs. 28200.
4. A, B, and C enter into a partnership. A invests 3 times as much as Band B invests two-third of what C invests. At the end of the year, the profit earned is Rs. 6600. What is the hare of B?
A. 1200
B. 1400
C. 1600
D. 1800
Answer: Option (A)
Explanaton:
Let C's capital = Rs. x. Then, B's capital = Rs. (2/3)x
A’s capital = Rs. (3 x (2/3).x) = Rs. 2x.
Ratio of their capitals = 2x : (2/3)x : x = 6 : 2 : 3.
Hence, B's share = Rs. ( 6600 x (2/11))= Rs. 1200.
5. Four milkmen rented a pasture. A grazed 24 cows for 3 months; B 10 for 5 months; C 35 cows for 4 months and D 21 cows for 3 months. If A's share of rent is Rs. 720, find the total rent of the field.
A. 3000
B. 3200
C. 3250
D. 3300
Answer: Option (C)
Explanaton:
Ratio of shares of A, B, C, D = (24 x 3) : (10 x 5) : (35 x 4) : (21 x 3)
= 72 : 50 : 140 : 63.
Let total rent be Rs. x. Then, A’s share = Rs. (72x)/325
(72x)/325=720 => x=(720 x 325)/72 = 3250.
Simple Interest
1. 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)
Explanaton:
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.
2. 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
E. None of these
Answer: Option (D)
Explanaton:
Principal
= Rs. [latex]\frac{100 \times 4016.25}{9 \times 5}[/latex]
= Rs. [latex]\frac{401625}{45}[/latex]
= Rs. 8925.
3. 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
E. None of these
Answer: Option (B)
Explanaton:
Let rate = R% and time = R years.
Then, [latex]\frac{1200 \times R \times R}{100}[/latex] = 432
12R[latex]^{2}[/latex] = 432
R[latex]^{2}[/latex] = 36
R = 6.
4. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
A. 3%
B. 4%
C. 5%
D. 6%
E. None of these
Answer: Option (D)
Explanaton:
S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate = [latex]\frac{100 \times 3000}{12500 \times 4}[/latex]% = 6%
5. A man took a loan from a bank at a rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
A. Rs. 2000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 20,000
Answer: Option (C)
Explanaton:
Principal = Rs. [latex]\frac{100 \times 5400}{12 \times 3}[/latex]= Rs. 15000.
Discount
1. The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is:
A. Rs. 1386
B. Rs. 1764
C. Rs. 1575
D. Rs. 2268
Answer: Option (B)
2. A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:
A. gains Rs. 55
B. gains Rs. 50
C. loses Rs. 30
D. gains Rs. 30
Answer: Option (B)
3. A man purchased a bicycle for Rs.4800 and sold it immediately at Rs.5995, allowing the buyer a credit of one and a half years. If the rate of interest is 6%, then the man gains:
A. Rs.700
B. Rs.700
C. Rs.495
D. Rs.900
Answer: Option (A)
4. Gopal has to pay Rs.440 to Ajay after 1 year. Ajay asks Gopal to pay Rs.220 in cash and defer the payment of Rs.220 for 2 years. If the rate of interest is 10% per annum, in this mode of payment:
A. Gopal gains Rs.3.33
B. Ajay gains Rs.3.33
C. Ajay loses Rs.16.67
D. Gopal loses Rs.16.67
Answer: Option (B)
5. The present worth of Rs. 1404 due in two equal half-yearly instalments at 8% per annum simple interest is:
A. Rs. 1325
B. Rs. 1300
C. Rs. 1350
D. Rs. 1500
Answer: Option (A)
Average
1. A library has an average of 510 visitors on Sundays and 240 on other day. The average number of visitors in a month of 30 days starting with sunday is
A. 280
B. 285
C. 290
D. 295
Answer: Option (B)
Explanaton:
As the month begins with Sunday, so there will be five Sundays in the month. So the result will be:
= [latex]\frac{510 \times 5 + 240 \times 25}{30}[/latex] = [latex]\frac{8550}{30}[/latex] = 285.
2. A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17[latex]^{th}[/latex] match
Answer: Option (D)
Explanaton:
Let the average after 17[latex]^{th}[/latex] match is x
then the average before 17[latex]^{th}[/latex] match is x-3
so 16(x-3) + 87 = 17x
=> x = 87 - 48 = 39.
3. The average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then the weight of the new man is
Answer: Option (C)
Explanaton:
Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60.
4. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is
Answer: Option (A)
Explanaton:
Number is (5*27) - (4*25) = 135-100 = 35.
5. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then average for the last four matches is
A. 33.25
B. 32.25
C. 34.25
D. 34.50
Answer: Option (C)
Explanaton:
=[latex]\frac {38.9 \times 10 − 42 \times 6}{4}[/latex]
= [latex]\frac{1216−750}{4}[/latex]
= 34.25
Time & Work
1. A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?
A. 11 days
B. 12 days
C. 13 days
D. 14 days
Answer: Option (A)
Explanaton:
One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A, B and C in the initial number of days.
The number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
2. A works twice as fast as B.If B can complete a work in 18 days independently, the number of days in which A and B can together finish the work is:
A. 4 days
B. 6 days
C. 8 days
D. 10 days
Answer: Option (B)
Explanaton:
The ratio of rates of working of A and B =2:1. So, the ratio of times taken =1:2
Therefore, A's 1 day's work=1/9
B's 1 day's work=1/18
(A+B)'s 1 day's work= 1/9 + 1/18 = 1/6
so, A and B together can finish the work in 6 days.
3. A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 20 days and C alone in 60 days, then B alone could do it in:
A. 20days
B. 40 days
C. 50 days
D. 60 days
Answer: Option (D)
Explanaton:
(A+B)'s 1 day's work=1/20
C's 1 day work=1/60
(A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15
Also, A's 1 day's work =(B+C)'s 1 day's work
Therefore, we get 2 x (A's 1 day 's work)=1/15
=>A's 1 day's work=1/30
Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60
So, B alone could do the work in 60 days.
4. 4 men can repair a road in 7 hours. How many men are required to repair the road in 2 hours?
A. 17 men
B. 14 men
C. 13 men
D. 16 men
Answer: Option (B)
Explanaton:
M x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal workload.
Clearly, substituting in the above equation we get, M2 = 14 men.
5. A, B, C together can do a piece of work in 10 days. All the three started working at it together and after 4 days, A left. Then, B and C together completed the work in 10 more days. In how many days can complete work alone?
Answer: Option (A)
Explanaton:
(A+B+C) do 1 work in 10 days.
So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so work is done by them in 4 days=4/10=2/5
Remaining work=1-2/5=3/5
(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50
Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25
A does 1/25 work in 1 day
Therefore 1 works in 25 days.
Permutation and Combination
1. In how many ways can the letters of the word 'LEADER' be arranged?
A. 72
B. 144
C. 360
D. 720
E. None of these
Answer: Option (C)
Explanaton:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways = [latex]\frac{6!}{(1!)(2!)(1!)(1!)(1!)}[/latex] = 360.
2. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer: Option (D)
Explanaton:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
3. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
A. 266
B. 5040
C. 11760
D. 86400
E. None of these
Answer: Option (C)
Explanaton:
Required number of ways = [latex]^8C_{5}[/latex] x [latex]^{10}C_{6}[/latex]
= [latex]^8C_{3}[/latex] x [latex]^{10}C_{4}[/latex]
= [latex]\frac{8 \times 7 \times 6}{3 \times 2 \times 1}[/latex] x [latex]\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}[/latex]
= 11760.
4. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A. 63
B. 90
C. 126
D. 45
E. 135
Answer: Option (A)
Explanaton:
Required number of ways = [latex]^7C_{5}[/latex] x [latex]^3C_{2}[/latex] = [latex]^7C_{2}[/latex] x [latex]^3C_{1}[/latex] = ([latex]\frac{7 \times 6}{2 \times 1}[/latex] x 3) = 63.
5. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS' if repetition of letters is not allowed?
A. 40
B. 400
C. 5040
D. 2520
Answer: Option (C)
Explanaton:
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= [latex]^{10}P_{4}[/latex]
= (10 x 9 x 8 x 7)
= 5040.
Linear Equation
1. Seven subtracted from twice a number gives 17.
A. 2x - 7 = 17
B. 2(x - 7) = 17
C. 7 - 2x = 17
D. 7(2 - x) = 17
Answer: Option (A)
2. A number added to -19 is equal to -21.
A. x - 21 = -19
B. -19 + x = -21
C. x = -19 + 21
D. -19 - 21 = x
Answer: Option (B)
3. Ten subtracted from a number yields 15.
A. 10 + x = 15
B. 10 - x = 15
C. x - 10 = 15
D. 15 - 10 = x
Answer: Option (C)
4. The quotient of 24 and a number is 4.
A. [latex]\frac{x}{24}[/latex] = 4
B. 24 - x = 4
C. 24x = 4
D. [latex]\frac{24}{x}[/latex] = 4
Answer: Option (D)
5. The sum of 2, 6, and a number amounts to 15. Find the number.
Answer: Option (A)
Ratio & Proportion
1. Find the fourth proportion to 2,3,6
Answer: Option (C)
Explanaton:
2: 3:: 6:x
=> 2/3 = 6/x
=> x = 18/2
=> x = 9
2. If 2: 9:: x: 18, then find the value of x
Answer: Option (C)
Explanaton:
Treat 2:9 as 2/9 and x:18 as x/18, treat :: as =
So we get 2/9 = x/18
=> 9x = 36
=> x = 4
3. Rs. 120 are divided among A, B, C such that A's share is Rs. 20 more than B's and Rs. 20 less than C's. What is B's share
A. Rs 10
B. Rs 20
C. Rs 24
D. Rs 28
Answer: Option (B)
Explanaton:
Let C = x. Then A = (x — 20) and B = (x — 40).
x + x - 20 + x - 40 = 120 Or x=60.
A:B:C = 40:20:60 = 2:1 :3.
B's share = Rs. 120*(1/6) = Rs. 20
4. In a college, the ratio of the number of boys to girls is 8: 5. If there are 200 girls, the total number of students in the college is
A. 420
B. 520
C. 620
D. 720
Answer: Option (B)
Explanaton:
Let the boy are 8x and Girls are 5x
=> 5x = 200
=> x = 40
Total students = 8x+5x = 13x = 13(40) = 520
5. In a mixture of 60 litres, the ratio of milk and water 2: 1. If this ratio is to be 1: 2, then the quantity of water to be further added is
A. 20 litres
B. 30 litres
C. 50 litres
D. 60 litres
Answer: Option (D)
Explanaton:
Quantity of Milk = 60*(2/3) = 40 liters
Quantity of water = 60-40 = 20 liters
As per the question, we need to add water to get quantity 2:1
=> 40/(20+x) = 1/2
=> 20 + x = 80
=> x = 60 liters
Number Systems
1. When writing numbers from 1 to 10,000, how many times is the digit 9 written?
A. 3200
B. 3600
C. 4000
D. 4200
Answer: Option (C)
Explanaton:
The digits 9 occurs in the thousands place in 1000 numbers.
It occurs in the hundreds place in 1000 numbers and so on
The digit occurs 4000 times.
2. Which digits should come in place of @ and # if the number 62684@# is divisible by both 8 and 5?
A. 4, 0
B. 0, 4
C. 4, 4
D. 1, 1
Answer: Option (A)
Explanaton:
Since the given number is divisible by 5, so 0 or 5 must come in place of #.
But, a number ending with 5 is never divisible by 8.
So, 0 will replace #.
Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4.
Hence, digits in place of @ and # are 4 and 0 respectively.
3. How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?
A. 3001
B. 2893
C. 2704
D. 2890
Answer: Option (B)
Explanaton:
While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 keystrokes.
There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers.
There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.
Then 1000 is a four-digit number which requires 4 keystrokes.
Totally, therefore, one requires 9+180+2700+4= 2893 keystrokes.
4. How many natural numbers below 660 are divisible by 5 and 11 but not by 3?
Answer: Option (A)
Explanaton:
If the number is divisible by 5 and 11 it must be divisible by 55.
The numbers are less than 660.
Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).
The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.
The numbers of such, multiples are the quotient of 11 by 3.
Quotient of [latex]\frac{11}{3}[/latex] =3.
Out of 11 multiples of 55, 3 are multiples of 3.
Hence, numbers less than 660 and divisible by 5 and 11 but not by 3 = 11 − 3 = 8.
5. A number, when divided by 342, gives a remainder 47. When the same number is divided by 19, what would be the remainder?
Answer: Option (B)
Explanaton:
On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then,
number − 342k + 47 = (19×18k+19×2+9) = 19(18k+2)+9
⇒ The given number when divided by 19, gives 18k+2 as quotient and 9 as remainder.
Decimal and Fractions
1. 34.95 + 240.016 + 23.98 = ?
A. 288.946
B. 298.946
C. 198.946
D. 188.946
Answer: Option (B)
2. Find the value of X => 3889 + 12.952 - X = 3854.002
A. 47.95
B. 44.95
C. 43.95
D. 40.95
Answer: Option (A)
3. If 144/0.144 = 14.4/x, then x = ?
A. 0.144
B. 0.0144
C. .00144
D. 1.44
Answer: Option (B)
4. Evaluate [latex]\frac{3.6 \times 0.48 \times 2.50}{0.12 \times 0.09 \times 0.5}[/latex]
A. 80
B. 800
C. 8000
D. 80000
Answer: Option (B)
[latex]\frac{3.6 \times 0.48 \times 2.50}{0.12 \times 0.09 \times 0.5}[/latex]
[latex]\frac{36 \times 48 \times 250}{12 \times 9 \times 5}[/latex] = 800.
5. What is increasing the order of the fractions 6/7, 8/9, 7/8, 9/10?
A. 6/7, 8/9, 7/8, 9/10
B. 9/10, 7/8, 8/9, 6/7
C. 6/7, 7/8, 8/9, 9/10
D. 9/10, 8/9, 7/8, 6/7
Answer: Option (C)
Percentages
1. What will be the fraction of 20%
A. 1/4
B. 1/5
C. 1/10
D. None of above
Answer: Option (B)
Explanaton:
It will 20*1/100 = 1/5
2. What will be the fraction of 4%
A. 1/20
B. 1/50
C. 1/75
D. 1/25
Answer: Option (D)
Explanaton:
4*1/100 = 1/25.
Friends I know it is quite simple, but trust me while solving percentage questions in Hurry we use to do these types of mistake only. So I recommend you to have a bit practise of this.
3. The ratio 5:20 expressed as percent equals to
A. 50 %
B. 125 %
C. 25 %
D. None of the above
Answer: Option (C)
Explanaton:
Actually, it means 5 is what percent of 20, which can be calculated as,
(5/20)*100 = 5 * 5 = 25
4. 2.09 can be expressed in terms of percentage as
A. 2.09%
B. 20.9%
C. 209%
D. 0.209%
Answer: Option (C)
Explanaton:
While calculation in terms of the percentage we need to multiply by 100, so
2.09 * 100 = 209.
5. Half of 1 percent written as a decimal is
A. 5
B. 0.5
C. 0.05
D. 0.005
Answer: Option (D)
Explanaton:
It will be 1/2(1%) = 1/2(1/100) = 1/200 = 0.005.
Fundamental Arithmetical Operations
1. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours
Answer: Option (A)
2. A man completes a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and the second half at the rate of 24 km/hr. Find the total journey in km.
A. 220 km
B. 224 km
C. 230 km
D. 234 km
Answer: Option (B)
3. A car travelling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
A. 17 6/7 km/hr
B. 25 km/hr
C. 30 km/hr
D. 35 km/hr
Answer: Option (D)
4. In covering a distance of 30 km, Celine takes 2 hours more than Sam. If Celine doubles his speed, then he would take 1 hour less than Sam. Celine's speed is:
A. 5 kmph
B. 6 kmph
C. 6.25 kmph
D. 7.5 kmph
Answer: Option (A)
Explanaton:
Let Celine's speed be x km/hr.
Then, 30/x - 30/2x = 3
6x = 30
x = 5 km/hr.
5. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
A. 14 km
B. 15 km
C. 16 km
D. 17 km
Answer: Option (C)
Explanaton:
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So, x/4 + (61 -x)/9 = 9
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
Mensuration
1. The parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. (Round off your answer to two decimal places)
A. 77.14 cm
B. 47.14 cm
C. 84.92 cm
D. 94.94 cm
E. 23.57 cm
Answer: Option (E)
Explanaton:
Let the side of the square be a cm.
Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= 1/2(∏)(15)
= 1/2(22/7)(15) = 330/14 = 23.57 cm to two decimal places
2. A cube of side one-meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?
A. 10
B. 100
C. 1000
D. 10000
E. None of these
Answer: Option (C)
Explanaton:
Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge, 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000
3. The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of whitewashing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each?
A. Rs. 4800
B. Rs. 3600
C. Rs. 3560
D. Rs. 4530
E. None of these
Answer: Option (D)
Explanaton:
Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530.
4. The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 500 resolutions?
A. 252 m
B. 704 m
C. 352 m
D. 808 m
E. None of these
Answer: Option (B)
Explanaton:
In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.
= 500 * 2 * 22/7 * 22.4 = 70400 cm = 704 m
5. The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?
A. 140
B. 156
C. 175
D. 214
E. None of these
Answer: Option (A)
Explanaton:
Given that the area of the square = 1225 sq.units
=> Side of square = [latex]\sqrt {1225}[/latex] = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units.
Time & Distance
1. A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
A. 100 m
B. 150 m
C. 190 m
D. 200 m
Answer: Option (A)
Explanaton:
Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr
Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m
Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.
2. How long will a boy take to run around a square field of side 35 meters, If he runs at the rate of 9 km/hr?
A. 50 sec
B. 52 sec
C. 54 sec
D. 56 sec
Answer: Option (D)
Explanaton:
Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec
Distance = (35 x 4) m = 140 m.
Time taken = 140 x (2/5) sec= 56 sec
3. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is
A. 35.55 km/hr
B. 36 km/hr
C. 71.11 km/hr
D. 71 km/hr
Answer: Option (C)
Explanaton:
Total time taken = (160/64 + 160/8)hrs
= 9/2 hrs.
Average speed = (320 x 2/9) km.hr
= 71.11 km/hr.
4. A man takes 6 hours 15 minutes in walking a distance and riding back to the starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride both ways is
A. 4 hours
B. 4 hours 30 minutes
C. 4 hours 45 minutes
D. 5 hours
Answer: Option (C)
Explanaton:
Time is taken in walking both ways = 7 hours 45 minutes --------(i)
Time is taken in walking one way and riding back= 6 hours 15 minutes-------(ii)
By equation (ii)*2 -(i), we have
Time taken to man ride both ways, = 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes.
5. A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge (in metres) is
A. 600
B. 750
C. 1000
D. 1250
Answer: Option (D)
Explanaton:
speed = (5x5/18)m/sec
= 25/18 m/sec.
Distance covered in 15 minutes = (25/18 x 15 x 60)m
= 1250 m.
Use of Tables & Graphs
Direction (1 - 5):The following table shows the number of new employees added to different categories of employees in a company and also the number of employees from these categories who left the company every year since the foundation of the Company in 1995.
1. What is the difference between the total number of Technicians added to the Company and the total number of Accountants added to the Company during the years 1996 to 2000?
A. 128
B. 112
C. 96
D. 88
Answer: Option (D)
Explanaton:
Required difference
= (272 + 240 + 236 + 256 + 288) - (200 + 224 + 248 + 272 + 260)
= 88.
2. What was the total number of Peons working in the Company in the year 1999?
A. 1312
B. 1192
C. 1088
D. 968
Answer: Option (B)
Explanaton:
Total number of Peons working in the Company in 1999
= (820 + 184 + 152 + 196 + 224) - (96 + 88 + 80 + 120)
= 1192.
3. For which of the following categories the percentage increase in the number of employees working in the Company from 1995 to 2000 was the maximum?
A. Managers
B. Technicians
C. Operators
D. Accountants
Answer: Option (A)
Explanaton:
Number of Managers working in the Company:
In 1995 = 760.
In 2000 = (760 + 280 + 179 + 148 + 160 + 193) - (120 + 92 + 88 + 72 + 96) = 1252.
Therefore Percentage increase in the number of Managers
= ([latex]\frac{(1252 - 760)}{760}[/latex] x 100) % = 64.74%.
Number of Technicians working in the Company:
In 1995 = 1200.
In 2000 = (1200 + 272 + 240 + 236 + 256 + 288) - (120 + 128 + 96 + 100 + 112) = 1936.
Therefore Percentage increase in the number of Technicians
= [latex]\frac{(1936 - 1200)}{1200}[/latex] x 100 ) % = 61.33%.
Number of Operators working in the Company:
In 1995 = 880.
In 2000 = (880 + 256 + 240 + 208 + 192 + 248) - (104 + 120 + 100 + 112 + 144) = 1444.
Therefore Percentage increase in the number of Operators
= ([latex]\frac{(1444 - 880)}{880}[/latex] x 100 ) % = 64.09%.
Number of Accountants working in the Company:
In 1995 = 1160.
In 2000 = (1160 + 200 + 224 + 248 + 272 + 260) - (100 + 104 + 96 + 88 + 92) = 1884.
Therefore Percentage increase in the number of Accountants
= ([latex]\frac{(1884 - 1160)}{1160}[/latex] x 100 ) % = 62.41%.
Number of Peons working in the Company:
In 1995 = 820.
In 2000 = (820 + 184 + 152 + 196 + 224 + 200) - (96 + 88 + 80 + 120 + 104) = 1288.
Therefore Percentage increase in the number of Peons
= [latex]\frac{(1288 - 820)}{820}[/latex] x 100 ) % = 57.07%.
Clearly, the percentage increase is maximum in case of Managers.
4. Total number of employees of various categories working in the Company in 1997 are:
A. 1325
B. 1195
C. 1265
D. 1235
Answer: Option (B)
Explanaton:
Managers = (760 + 280 + 179) - (120 + 92) = 1007.
Technicians = (1200 + 272 + 240) - (120 + 128) = 1464.
Operators = (880 + 256 + 240) - (104 + 120) = 1152.
Accountants = (1160 + 200 + 224) - (100 + 104) = 1380.
Peons = (820 + 184 + 152) - (96 + 88) = 972.
Therefore Pooled average of all the five categories of employees working in the Company in 1997
= [latex]\frac{1}{5}[/latex]x (1007 + 1464 + 1152 + 1380 + 972)
=[latex]\frac{1}{5}[/latex] x (5975)
= 1195.
5. During the period between 1995 and 2000, the total number of Operators who left the Company is what percent of the total number of Operators who joined the Company?
A. 19%
B. 21%
C. 27%
D. 29%
Answer: Option (D)
Explanaton:
Total number of Operators who left the Company during 1995 - 2000
= (104 + 120 + 100 + 112 + 144) = 580.
Total number of Operators who joined the Company during 1995 - 2000
= (880 + 256 + 240 + 208 + 192 + 248)
= 2024.
Therefore Required Percentage = ([latex]\frac {580}{2024}[/latex] x 100 ) % = 28.66% = 29%.
Profit & Loss
1. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
A. No profit, no loss
B. 5%
C. 8%
D. 10%
Answer: Option (B)
Explanaton:
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
Gain =(80/1600*100) % = 5%
2. If the selling price is doubled, the profit triples. Find the profit percent?
A. 100%
B. 200%
C. 300%
D. 400%
Answer: Option (A)
Explanaton:
Let the C.P be Rs.100 and S.P be Rs.x, Then
The profit is (x-100)
Now the S.P is doubled, then the new S.P is 2x
New profit is (2x-100)
Now as per the given condition;
=> 3(x-100) = 2x-100
By solving, we get
x = 200
Then the Profit percent = (200-100)/100 = 100
Hence the profit percentage is 100%
3. If books bought at prices ranging from Rs. 200 to Rs. 350 are sold at prices ranging from Rs. 300 to Rs. 425, what is the greatest possible profit that might be made in selling eight books?
A. 600
B. 1200
C. 1800
D. none of these
Answer: Option (C)
Explanaton:
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
4. If the cost price is 25% of the selling price. Then what is the profit percent?
A. 150%
B. 200%
C. 300%
D. 350%
Answer: Option (C)
Explanaton:
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = (75/25) * 100 = 300%
5. A man buys oranges at Rs 5 a dozen and an equal number at Rs 4 a dozen. He sells them at Rs 5.50 a dozen and makes a profit of Rs 50. How many oranges does he buy?
A. 30 dozens
B. 40 dozens
C. 50 dozens
D. 60 dozens
Answer: Option (C)
Explanaton:
Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9.
Sell price of 2 dozen oranges = Rs. 11.
If profit is Rs 2, oranges bought = 2 dozen.
If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.
Data Interpretation
Direction (1 - 5): A cosmetic company produces five different products. The sales of these five products(in lakh number of packs) during 1995 and 2000 are shown in the following bar graph.
1. The questions given below are based on this graph. The sales have increased by nearly 55% from 1995 to 2000 in the case of :
A. Lipsticks
B. Nail Enamels
C. Talcum Powders
D. Shampoos
E. Conditioners
Answer: Option (D)
Explanaton:
Lipsticks = [(48.17 - 20.15)/20.15 * 100] = 139.06%
Nail Enamels = [(37.76 - 5.93)/5.93 * 100] = 536.76% Talcum Powders = [(29.14 - 14.97)/14.97 * 100] = 94.66%
Shampoos = [(12.21 - 7.88)/7.88 * 100] = 54.95% = 55%
Conditioners = [(10.19 - 5.01)/5.01 * 100] = 103.39%
2. During the period 1995-2000, the minimum rate of increase in sales is in the case of :
A. Lipsticks
B. Nail Enamels
C. Talcum powders
D. Shampoos
E. Conditioners
Answer: Option (D)
Explanaton:
Lipsticks = [(48.17 - 20.15)/20.15 * 100] = 139.06%
Nail Enamels = [(37.76 - 5.93)/5.93 * 100] = 536.76% Talcum Powders = [(29.14 - 14.97)/14.97 * 100] = 94.66%
Shampoos = [(12.21 - 7.88)/7.88 * 100] = 54.95% = 55%
Conditioners = [(10.19 - 5.01)/5.01 * 100] = 103.39%.
The minimum rate of increase in sales from 1995 to 2000 is in the case of Shampoos.
3. The sales of lipsticks in 2000 was by what percent more than the sales of nail enamels in 2000? (rounded off to the nearest integer)
A. 33%
B. 31%
C. 28%
D. 22%
E. 21%
Answer: Option (C)
Explanaton:
Required percentage = [(48.17 - 37.76)/37.76 * 100]% = 27.57% = 28%
4. The sales of conditioners in 1995 was by what percent less than the sales of shampoos in 1995? (rounded off to the nearest integer)
A. 57%
B. 36%
C. 29%
D. 25%
E. 21%
Answer: Option (B)
Explanaton:
Required percentage =[(7.88 - 5.01)/7.88 * 100]% = 36.42% ≈ 36%.
5. What is the approximate ratio of the sales of nail enamels in 2000 to the sales of Talcum powders in 1995?
A. 7:2
B. 5:2
C. 4:3
D. 2:1
E. 5:3
Answer: Option (B)
Explanaton:
Required ratio = 37.76/14.97 ≈ 2.5 = 5/2
Number Series Simplification
Directtion (1 - 5): Look carefully for the pattern, and then choose which pair of numbers comes next.
1. 42 40 38 35 33 31 28
A. 25 22
B. 26 23
C. 26 24
D. 25 23
E. 26 22
Answer: Option (C)
Explanaton:
This is an alternating subtraction series in which 2 is subtracted twice, then 3 is subtracted once, then 2 is subtracted twice, and so on.
2. 36 31 29 24 22 17 15
A. 13 11
B. 10 5
C. 13 8
D. 12 7
E. 10 8
Answer: Option (E)
Explanaton:
This is an alternating subtraction series, which subtracts 5, then 2, then 5, and so on.
3. 3 5 35 10 12 35 17
A. 22 35
B. 35 19
C. 19 35
D. 19 24
E. 22 24
Answer: Option (C)
Explanaton:
This is an alternating addition series, with a random number, 35, interpolated as every third number. The pattern of addition is to add 2, add 5, add 2, and so on. The number 35 comes after each "adds 2" step.
4. 13 29 15 26 17 23 19
A. 21 23
B. 20 21
C. 20 17
D. 25 27
E. 22 20
Answer: Option (B)
Explanaton:
Here, there are two alternating patterns, with every other number following a different pattern. The first pattern begins with 13 and adds 2 to each number to arrive at the next; the alternating pattern begins with 29 and subtracts 3 each time.
5. 14 14 26 26 38 38 50
A. 60 72
B. 50 62
C. 50 72
D. 62 62
E. 62 80
Answer: Option (B)
Explanaton:
In this simple addition with repetition series, each number in the series repeats itself, and then increases by 12 to arrive at the next number.
Problems on Ages, Profit, Loss, and Discount
1. Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin
A. 24.5
B. 25.5
C. 26.5
D. 27.5
Answer: Option (A)
Explanaton:
If Rahul age is x, then Sachin age is x - 7,
so, (x-7)/x =7/9
9x - 63 = 7x
2x = 63
x = 31.5
So Sachin age is 31.5 - 7 = 24.5
2. The ratio of the ages of Maala and Kala is 4 : 3. The total of their ages is 2.8 decades. The proportion of their ages after 0.8 decades will be [1 Decade = 10 years]
A. 4:3
B. 12:11
C. 7:4
D. 6:5
Answer: Option (D)
Explanaton:
Let, Maala’s age = 4A and Kala’s age = 3A
Then 4A + 3A = 28
A = 4
Maala’s age = 16 years
and Kala’s age = 12 years
Proportion of their ages after 8 is = (16 + 8) : (12 + 8)
= 24 : 20
= 6 : 5
3. Six years ago Anita was P times as old as Ben was. If Anita is now 17 years old, how old is Ben now in terms of P?
A. 11/P + 6
B. P/11 +6
C. 17 - P/6
D. 17/P
Answer: Option (A)
Explanaton:
Let Ben’s age now be B
Anita’s age now is A.
(A - 6) = P(B - 6)
But A is 17 and therefore 11 = P(B - 6)
11/P = B-6
(11/P) + 6 = B
4. When I have married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me. The average age of my family during my marriage is the same as today. What is the age of Father when he died?
A. 50 yrs
B. 60 yrs
C. 70 yrs
D. 65 yrs
Answer: Option (B)
Explanaton:
Let the Father be x years when he died
Average Age 10 years ago be A
Total Age 10 years ago = 6*A
Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60
Father Died and Baby was born => the Total number of people in the family is Same (6)
Baby born today so age of baby = 0
(6A +60 - x)/6 = 6A/6
=> A + 10 -(x/6) = A
=> x/6 = 10
=> x = 60
Therefore we can conclude that the father was 60 years old when he died.
5. Sivagami is 2 years elder than Meena. After 6 years the total of their ages will be 7 times of their current age. Then the age of Sivagami is:
A. 19 years
B. 17 years
C. 15 years
D. data inadequate
Answer: Option (D)
Explanaton:
Let Meena’s age = A.
Then Sivagami’s age = A + 2
After 6 years the total of their ages will be 7 times what
Not clear.
So the given data are inadequate.
Mixture and Allegations
1. Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q, the ratio of alcohol and water in the resulting mixture is?
A. 16: 5
B. 14: 5
C. 16: 7
D. 19: 5
E. None of these
Answer: Option (D)
Explanaton:
Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.
2. A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?
A. 72
B. 72.9
C. 73.8
D. 74.7
E. None of these
Answer: Option (B)
Explanaton:
Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(90 - 9)/90][latex]^{2}[/latex] (90) = 72.9 litres.
3. In what ratio must a grocer mix teas worth Rs.60 a kg and Rs.65 a kg. So that by selling the mixture at Rs. 68.20 a kg, He may gain 10%?
A. 3:2
B. 3:4
C. 3:5
D. 4:5
Answer: Option (A)
Explanaton:
S.P of 1kg mix = RS.68.20, Gain = 10% C.P of 1kg mix =Rs.(100/110 x 68.20) =Rs.62 (Cheaper Tea) : (Dearer Tea)
= 3 : 2
4. Milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively in what ratio the liquids in both the vessels should be mixed to obtain a new mixture in vessel C containing half milk and half water?
A. 1:1
B. 1:3
C. 1:2
D. 7:5
Answer: Option (D)
Explanaton:
Milk in A = 4/7 of whole milk in B = 2/5 of whole milk in mixture A and B = 1/2 of the whole.
Let the C.P of unit quantity be Re.1
Required ratio = 1/10 = 1/14 = 14 : 10 = 7: 5
5. How much water must be added to a bucket which contains 40 litres of milk at the cost price of Rs.3.50 per litre so that the cost of milk reduces to Rs.2 per litre?
A. 25 litres
B. 28 litres
C. 30 litres
D. 35 litres
Answer: Option (C)
Explanaton:
Total cost price =Rs(40 x 7/2) = Rs.140
Cost per litre =Rs.2,
Total quantity = 140/2 = 70 Litres.
Water to be added =(70-40) =30 Litres.
Pipes and Cisterns
1. A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in?
A. 20 hrs
B. 28 hrs
C. 36 hrs
D. 40 hrs
Answer: Option (A)
Explanaton:
1/8 - 1/x = 1/10
=> x = 40 hrs
2. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 min, then the slower alone will be able to fill the tank in:
A. 81 min
B. 108 min
C. 144 min
D. 192 min
Answer: Option (C)
Explanaton:
Let the slower pipe alone fill the tank in x minutes.
Then, the faster pipe will fill it in x/3 minutes.
=> [latex]\frac{1}{x}[/latex] + [latex]\frac{3}{x}[/latex] = [latex]\frac{1}{36}[/latex] => [latex]\frac{4}{x}[/latex] = [latex]\frac{1}{36}[/latex] => x = 144mins.
3. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
Answer: Option (D)
Explanaton:
Capacity of the tank =(12 x 13.5) liters =162 liters.
The capacity of each bucket =9 litres
Number of buckets needed = 162/9 =18.
4. Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 3 minutes, the cistern was full. In what time can the waste pipe empty the cistern when fill?
A. 8.21 min
B. 8 min
C. 8.57 min
D. 8.49 min
Answer: Option (C)
Explanaton:
1/10 + 1/15 = 1/6 x 3 = 1/2
1 - 1/2 = 1/2
1/10 + 1/15 - 1/x = 1/2
x = 8.57 min.
5. Pipe A can fill a tank in 16 minutes and pipe B cam empty it in 24 minutes. If both the pipes are opened together after how many minutes should pipe B be closed so that the tank is filled in 30 minutes?
A. 21 min
B. 24 min
C. 20 min
D. 22 min
Answer: Option (A)
Explanaton:
Let the pipe B be closed after 'K' minutes.
30/16 - K/24 = 1 => K/24 = 30/16 - 1 = 14/16
=> K = 14/16 x 24 = 21 min.
Square Roots
1. If (89)[latex]^{2}[/latex] is added to the square of a number, the answer so obtained is 16202. What is the (1/26) of that number?
A. 5.65
B. 2.7
C. 3.5
D. 6.66
Answer: Option (C)
Explanaton:
Let the number is = x
(89)[latex]^{2}[/latex] + x[latex]^{2}[/latex] = 16202
x[latex]^{2}[/latex] = 8281
x = 91
=> (1/26) of 91 = 3.5
2. If x × y = x + y + [latex]\sqrt{xy}[/latex], the value of 6 x 24 is:
Answer: Option (B)
Explanaton:
Given x × y = x + y + [latex]\sqrt{xy}[/latex]
Here x = 6 , y = 24
Then,
x × y = 6 + 24 + [latex]\sqrt{6 x 24}[/latex]
6 x 24 = 30 + [latex]\sqrt{144}[/latex]
= 30 + 12
= 42.
3. Cube root of 729 then square it
Answer: Option (C)
Explanaton:
729 = 9 x 9 x 9
=> Cube root of 729 = 9
Now, required square of 9 = 9 x 9 = 81.
4. If x = 7−4[latex]\sqrt{3}[/latex] then find the value of (x + [latex]\frac {1}{x}[/latex])?
A. 3[latex]\sqrt{3}[/latex]
B. 8[latex]\sqrt{3}[/latex]
C. 14
D. 14+8[latex]\sqrt{3}[/latex]
Answer: Option (C)
5. What should come in place of x in the following equation [latex]\frac {x}{\sqrt {128}}[/latex] = [latex]\frac{\sqrt{162}}{x}[/latex]
Answer: Option (B)
Explanaton:
[latex]\frac {x}{\sqrt {128}}[/latex] = [latex]\frac{\sqrt{162}}{x}[/latex]
Then x[latex]^{2}[/latex] = [latex]\sqrt {162}[/latex] x [latex]\sqrt {128}[/latex]
Sqrt of 8[latex]^{2}[/latex] x 6[latex]^{2}[/latex] x 3[latex]^{2}[/latex]
= 8 x 6 x 3
x[latex]^{2}[/latex] = 144.
x = 12.
Probability
1. 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. 1/2
B. 3/5
C. 9/20
D. 8/15
Answer: Option (C)
Explanaton:
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) = n(E)/n(S) = 9/20.
2. Two dice are tossed. The probability that a total score is a prime number is:
A. 5/12
B. 1/6
C. 1/2
D. 7/9
Answer: Option (A)
Explanaton:
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that a sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.
3. 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. 2/7
B. 5/7
C. 1/5
D. 1/2
Answer: Option (A)
Explanaton:
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
P(E) = [latex]\frac{n(E)}{n(S)}[/latex] = [latex]\frac{10}{35}[/latex] = [latex]\frac{2}{7}[/latex]
4. Three unbiased coins are tossed. What is the probability of getting at most two heads?
A. 3/4
B. 7/8
C. 1/2
D. 1/4
Answer: Option (B)
Explanaton:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
5. If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?
A. 1/6
B. 1/2
C. 1/3
D. 1/4
Answer: Option (A)
Explanaton:
P(first letter is not vowel) = [latex]\frac{2}{4}[/latex]
P(second letter is not vowel) = [latex]\frac{1}{3}[/latex]
So, probability that none of letters would be vowels is = [latex]\frac{2}{4}[/latex] × [latex]\frac{1}{3}[/latex]=[latex]\frac{1}{6}[/latex]
Introduction 
1. What is the difference between the total number of Technicians added to the Company and the total number of Accountants added to the Company during the years 1996 to 2000?
1. The questions given below are based on this graph. The sales have increased by nearly 55% from 1995 to 2000 in the case of :