HCL Aptitude Test Questions and Answers

HCL Aptitude Test Questions and Answers Pattern For Freshers. HCL Written Test Interview Questions with Solution and Explanation.   

You can also see:   

Tech Mahindra Aptitude Test

Wipro Aptitude Test

Accenture Aptitude Test

HCL Written Test Questions and Answers:

1) If the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is?

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

Answer: D

Explanation:

2 * 22/7 * 14 * x = 1056 => x = 12


2) Rs. 6000 is lent out in two parts. One part is lent at 7% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?

A.5 : 1        B.4 : 1        C.3 : 2        D.2 : 1        E.None of these

Answer: A

Explanation:

Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 - x) = 600 - 3x/100
=> 600 - 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5 : 1

3) Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?

A.Rs.81000    B.Rs.80000    C.Rs.75000    D.Rs.64000

Answer: A

Explanation:

64000* 9/8 * 9/8 = 81000


4) 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: E


5) A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
A.20:23        B.34:43        C.32:45        D.37:45

Answer: C

Explanation:
    A             :             B
(8000*4)+(4000*8) :    (9000*6)+(6000*6)
64000             :    90000
32                :    45


6) The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?
A.5 : 3 : 1    B.5 : 3 : 2    C.10 : 7 : 4    D.10 : 8 : 5    E.None of these

Answer: B

Explanation:

Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.


7) Three 6 faced dice are thrown together. The probability that all the three show the same number on them is -.

A.1/216        B.1/36        C.5/9        D.5/12        E.7/12

Answer: B

Explanation:

It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36


8) Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case?

A.3    B.4    C.5    D.6

Answer: C

Explanation:

Calculate the differences, taking two numbers at a time as follows:
(215-47) = 168
(365-215) = 150
(365-47) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.


9) What is the remainder when 263251 is divided by 9?
A.8        B.3        C.6        D.1        E.None of these

Answer: D


10) A trader marks his articles 20% more than the cost price. If he allows 20% discount, then find his gain or loss percent?

A.No profit No loss    B.4% profit    C.2% loss    D.2% profit    E.None of these

Answer: E

Explanation:

Let CP of an article = RS. 100
MP= Rs. 120
Discount = 20%
SP = M[(100 - d%)/100] = 120(80/100) = Rs. 96
Clearly, the trader gets 4% loss.


11)  What is the least number to be subtracted from 11, 15, 21 and 30 each so that resultant numbers become proportional?
A.1        B.2    C.3    D.4    E.None of these

Answer: C

Explanation:

Let the least number to be subtracted be x, then 11 - x, 15 - x, 21 - x and 30 - x are in proportion.
<=> (11 - x) : (15 - x) = (21 - x) : (30 - x)
=> (11 - x)(30 - x) = (15 - x)(21 - x)
From the options, when x = 3
=> 8 * 27 = 12 * 18

12) Alok ordered 16 chapatis, 5 plates of rice, 7 plates of mixed vegetable and 6 ice-cream cups. The cost of each chapati is Rs.6, that of each plate of rice is Rs.45 and that of mixed vegetable is Rs.70. The amount that Alok paid the cashier was Rs.961. Find the cost of each ice-cream cup?

A.Rs.25        B.Rs.22.50    C.Rs.20        D.Rs.17.50    E.None of these

Answer: A

Explanation:

Let the cost of each ice-cream cup be Rs.x
16(6) + 5(45) + 7(70) + 6(x) = 961
96 + 225 + 490 + 6x = 961
6x = 150 => x = 25.


13) Anil invested a sum of money at a certain rate of simple interest for a period of five years. Had he invested the sum for a period of eight years for the same rate, the total intrest earned by him would have been sixty percent more than the earlier interest amount. Find the rate of interest p.a.
A.6%    B.9%    C.12.5%        D.Cannot be determined    E.None of these

Answer: D

Explanation:

Let the sum be Rs.P. Let the rate of interest be R% p.a.
(P)(8)(R)/100 = [1 + 6/100][ (P)(5)(R) /100] = 1.6{5PR/100]
8PR/100 = 8PR/100 which is anyway true.
R cannot be found.


14) (98 / 7 / 2) / ? = 14

A.1/2        B.1/3        C.1/4        D.1/5        E.None of these

Answer: A

Explanation:

? = (98 / 7 / 2) / 14 = 1/2


15) Two trains of equal lengths are running at speeds of 30 kmph and 60 kmph. The two trains crossed each other in 30 seconds when travelling in opposite direction. In what time will they cross each other when travelling in the same direction?

A.90 sec    B.75 sec    C.85 sec    D.80 sec    E.None of these

Answer: A

Explanation:
 
Let the length of each train be x m.
 (x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5) = 90 sec.


16) Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days, six men left and six women joined, then in hoe many more days will the work be completed?

A.1(1/2)    B.1        C.2        D.2(1/2)    E.None of these

Answer: D

Explanation:

Work done by a women in one day = 1/2 (work done by a man/day)
 One women's capacity = 1/2(one man's capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women

30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women - 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.



17) The diameters of two spheres are in the ratio 1:2 what is the ratio of their surface area?

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

Answer: B

Explanation:

1:4

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Tech Mahindra Aptitude Test Questions and Answers

Tech Mahindra Aptitude Test Questions and Answers Pattern For Freshers. Tech Mahindra Written Test Interview Questions with Solution and Explanation. 

Tech Mahindra Aptitude Test Questions:

1) The radius of a circular wheel is 1.75 m, how many revolutions will it make in traveling 1 km?
A.1000        B.2000        C.3000        D.4000

Answer: A

Explanation:

2 * 22/7 * 1.75 * x = 11000
x = 1000


2) A trader purchased two colour televisions for a total of Rs. 35000. He sold one colour television at 30% profit and the other 40% profit. Find the difference in the cost prices of the two televisions if he made an overall profit of 32%?
A.Rs. 21000    B.Rs. 17500    C.Rs. 19000    D.Rs. 24500    E.None of these

Answer: A

Explanation:

Let the cost prices of the colour television sold at 30% profit and 40% profit be Rs. x and Rs. (35000 - x) respectively.
Total selling price of televisions = x + 30/100 x + (35000 - x) + 40/100 (35000 - x)
=> 130/100 x + 140/100 (35000 - x) = 35000 + 32/100 (35000)
x = 28000
35000 - x = 7000
Difference in the cost prices of televisions = Rs. 21000


3) Ravi invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Ravi's investments if the interests from those investments are equal?
A.4 : 3        B.3 : 2        C.6 : 5        D.7 : 6        E.None of these

Answer: D

Explanation:

Let x be the investment of Ravi in 6% and y be in 7%
x(6)(n)/100 = y(7)(n)/100
=> x/y = 7/6
x : y = 7 : 6


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: B
 
Explanation:

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) A and B invests Rs.10000 each, A investing for 8 months and B investing for all the 12 months in the year. If the total profit at the end of the year is Rs.25000, find their shares?
A.Rs.8000, Rs.17000    B.Rs.9000, Rs.16000    C.Rs.18000, Rs.7000    D.Rs.10000, Rs.15000

Answer: D

Explanation:

The ratio of their profits A:B = 8:12 = 2:3
Share of A in the total profit = 2/5 * 25000 = Rs.10000 Share of A in the total profit = 3/5 * 25000 = Rs.15000


6) Ten percent of Ram's monthly salary is equal to eight percent of Shyam's monthly salary. Shyam's monthly salary is twice Abhinav's monthly salary. If Abhinav's annual salary is Rs. 1.92 lakhs, find Ram's monthly salary?

A.Rs. 18000    B.Rs. 20000    C.Rs. 25600    D.Rs. 32000    E.None of these

Answer: C

Explanation:

Let the monthly salaries of Ram and Shyam be Rs. r and Rs. s respectively.
10/100 r = 8/100 s
r = 4/5 s
Monthly salary of Abhinav = (1.92 lakhs)/12 = Rs. 0.16 lakhs
s = 2(0.16 lakhs) = 0.32 lakhs
r = 4/5(0.32 lakhs) = Rs. 25600


7) 10 books are placed at random in a shelf. The probability that a pair of books will always be together is -.
A.1/10        B.9/10        C.1/5        D.3/10        E.1/2

Answer: C

Explanation:

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
Required probability = 1/5


8) Find the greatest number which, while dividing 19, 83 and 67, gives a remainder of 3 in each case?
A.16    B.17    C.18        D.19

Answer: A

Explanation:

Subtract the remainder 3 from each of the given numbers: (19-3)=16, (67-3)=64 and (83-3)=80. Now, find the HCF of the results 16, 64 and 80, we get 16. Thus, the greatest number is 16.


9) Find the greatest number which leaves the same remainder when it divides 25, 57 and 105.

A.18        B.8        C.12        D.16        E.None of these

Answer: D

Explanation:

105 - 57 = 48
57 - 25 = 32
105 - 25 = 80
The H.C.F of 32, 48 and 80 is 16.


10) Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article?

A.Rs. 15000    B.Rs. 16200    C.Rs. 14700    D.Rs. 15900    E.None of these

Answer: A

Explanation:

Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 - 10/100(18000) = 18000 - 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000


11) If p, q and r are positive integers and satisfy x = (p + q -r)/r = (p - q + r)/q = (q + r - p)/p, then the value of x is?
A.1/2        B.1    C.-1/2        D.-1    E.None of these

Answer: B

Explanation:

When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.
Hence, x = (p + q -r)/r = (p - q + r)/q = (q + r - p)/p
=> x = (p + q - r + p - q + r + q + r - p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is non-zero.


12) Three consecutive odd integers are in increasing order such that the sum of the last two integers is 13 more than the first integer. Find the three integers?
A.9, 11, 13    B.11, 13, 15    C.13, 15, 17    D.7, 9, 11    E.None of these

Answer: D

Explanation:

Let the three consecutive odd integers be x, x + 2 and x + 4 respectively.
x + 4 + x + 2 = x + 13 => x = 7
Hence three consecutive odd integers are 7, 9 and 11.


13) Vijay lent out an amount Rs. 10000 into two parts, one at 8% p.a. and the remaining at 10% p.a. both on simple interest. At the end of the year he received Rs. 890 as total interest. What was the amount he lent out at 8% pa.a?
A.Rs. 6000    B.Rs. 5500    C.Rs. 4500    D.Rs. 5000    E.None of these

Answer: B

Explanation:

Let the amount lent out at 8% p.a. be Rs. A
=> (A * 8)/100 + [(10000 - A) * 10]/100 = 890
=> A = Rs. 5500.


14) (24 * 5 * 7 * 9) / ? = 216

A.15        B.20        C.25        D.35        E.None of these

Answer: D

Explanation:

? = (24 * 5 * 7 * 9) / 216 = 35


15) A train leaves Mumabai at 9 am at a speed of 40 kmph. After one hour, another train leaves Mumbai in the same direction as that of the first train at a speed of 50 kmph. When and at what distance from Mumbai do the two trains meet?

A.1:00pm, 220 km    B.1:00pm, 200km        C.2:00pm, 200 km    D.2:00pm, 220 km    E.None of these

Answer: C

Explanation:

When the second train leaves Mumbai the first train covers 40 * 1 = 40 km
So, the distance between first train and second train is 40 km at 10.00am
Time taken by the trains to meet = Distance / relative speed = 40 / (50 -40) = 4 hours
So, the two trains meet at 2 p.m. The two trains meet 4 * 50 = 200 km away from Mumbai.


16) Two men can complete a piece of work in four days. Two women can complete the same work in eight days. Four boys can complete the same work in five days. If four men, eight women and 20 boys work together in how many days can the work be completed?

A.1/2 day    B.1(1/2) days        C.1 day        D.2 days    E.None of these

Answer: A

Explanation:

Two men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.
Similarly four women would take two days to complete it and 20 children would take one day to complete it.
All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day
= 2 times the unit work in a day.
They will take 1/2 a day to complete it working together.


17) The sides of a cube are in the ratio 1:2 the ratio of their volume is?

A.1:2        B.1:4        C.1:8        D.2:1

Answer: C

Explanation:

1:8

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Wipro Aptitude Test Questions and Answers

Wipro Aptitude Test Questions and Answers Pattern For Freshers. Wipro Written Test Interview Questions with Solution and Explanation.

You can also Practice:

• TCS Aptitude Test
• Infosys Aptitude Test
• Accenture Aptitude Test

Wipro Aptitude Test Questions with Explanation:

1) Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?

A.Rs.150    B.Rs.132    C.Rs.100    D.Rs.125

Answer: B

Explanation:

2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132


2) The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?

A.21    B.22    C.23    D.24    E.None of these

Answer: C

Explanation:

Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n - 48n = 69 => n = 23
Therefore there were 23 members in the group initially.


3) The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing 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: D

Explanation:

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 compound and the simple interests on a certain sum at the same rate of interest for two years are Rs.11730 and Rs.10200 respectively. Find the sum.

A.Rs.18000    B.Rs.17000    C.Rs.18500    D.Rs.17500    E.None of these.

Answer: B

Explanation:

The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year
So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a.
So P = (100 * 10200)/ (30 * 2) = Rs.17000



1:85) A and B starts a business with Rs.8000 each, and after 4 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months?

A.18:11        B.22:13        C.23:12        D.11:9

Answer: A

: Explanation
6) Two tests had the same maximum mark. The pass percentages in the first and the second test were 40% and 45% respectively. A candidate scored 216 marks in the second test and failed by 36 marks in that test. Find the pass mark in the first test?

A.136        B.128        C.164        D.214        E.None of these

Answer: E

Explanation:

Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.


7) What is the probability that a leap year has 53 Sundays and 52 Mondays?

A.0        B.1/7        C.2/7        D.5/7        E.6/7
   
Answer: B

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7


8) What is the probability that a leap year has 53 Sundays and 52 Mondays?
A.0        B.1/7        C.2/7        D.5/7        E.6/7
   
Answer: B

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7


9) Find the greatest number that exactly divides 35, 91 and 840?

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

Explanation:

The greatest number that exactly divides 35, 91 and 840 is the HCF of the three numbers. So, calculating HCF we get the answer 7.


10) Find the greatest four digit number which leaves respective remainders of 2 and 5 when divided by 15 and 24.
A.9974        B.9125        C.9565        D.9997        E.None of these

Answer: E

Explanation:

Since the difference between the divisors and the respective remainders is not constant, back substitution is the convenient method. None of the given numbers is satisfying the condition.


11) Ravi purchased a refrigerator and a mobile phone for Rs. 15000 and Rs. 8000 respectively. He sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 10 percent. Overall he make a.

A.loss of Rs. 200    B.loss of Rs. 100    C.profit of Rs. 100    D.profit of Rs. 200    E.None of these

Answer: D

Explanation:

Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 - 4/100) = 15000 - 600
m = 8000(1 + 10/100) = 8000 + 800
Total SP - Total CP = r + m - (15000 + 8000) = -600 + 800 = Rs. 200
As this is positive, an overall profit of Rs. 200 was made.


12) The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?

A.32 kg        B.36 kg        C.40 kg        D.44 kg        E.None of these

Answer: B

Explanation:

Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.


13) There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits?

A.30        B.46        C.40        D.44        E.None of these

Answer: E

Explanation:

Let the number of rabbits and peacocks be 'r' and 'p' respectively. As each animal has only one head, so r + p = 60 --- (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 --- (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.

14) A certain sum becomes four times itself at simple interest in eight years. In how many years does it become ten times itself?

A.21    B.25    C.23    D.27    E.None of these

Answer: E

Explanation:

Let the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.
R = (100 * 3x)/(x * 8) = 300/8 %
If the sum becomes ten times itself, then interest is 9x.
The required time period = (100 * 9x)/(x * 300/8) = (100 * 9x * 8)/(x * 300) = 24 years.


15) Roja and Pooja start moving in the opposite directions from a pole. They are moving at the speeds of 2 km/hr and 3 km/hr respectively. After 4 hours what will be the distance between them?

A.12 km        B.20 km        C.24 km        D.4 km        E.None of these

Answer: B

Explanation:

Distance = Relative Speed * Time
= (2 + 3) * 4 = 20 km
[ They are travelling in the opposite direction, relative speed = sum of the speeds].


16) A and B can do a work in 12 days and 36 days respectively. If they work on alternate days beginning with B, in how many days will the work be completed?
A.20 2/3    B.9        C.24        D.26 5/9    E.None of these

Answer: E

Explanation:

The work done in the first two days = 1/12 + 1/36 = 1/9
so, 9 such two days are required to finish the work.
i.e., 18 days are required to finish the work.


17) A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?

A.12 cm        B.10 cm        C.18 cm        D.24 cm

Answer: B

Explanation:

30 * 20 * x = (8 * 5.5 * 1.5)/2
x = 10

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Accenture Aptitude Test Questions and Answers

Accenture Aptitude Test Questions and Answers Pattern For Freshers. Accenture Written Test Interview Questions with Solution and Explanation.

Also see: TCS Aptitude Test & Infosys Aptitude Test 

1) The length of rectangle is thrice its breadth and its perimeter is 96 m, find the area of the rectangle?

A.432 sq m    B.356 sq m    C.452 sq m    D.428 sq m

Answer: A

Explanation:

2(3x + x) = 96
l = 36   b = 12
lb = 36 * 12 = 432


2) The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 40 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.

A.20    B.15    C.25    D.35    E.None of these.

Answer: C

Explanation:

Let the number of students who wrote the exam be x.
Total marks of  students = 80 x.
Total marks of (x - 5) students = 90(x - 5)
80x - (5 * 40) = 90(x - 5)
250 = 10x  => x = 25


3) Find the amount on Rs.5000 in 2 years, the rate of interest being 4% per first year and 5% for the second year?

A.Rs.460    B.Rs.5640    C.Rs.5460    D.Rs.5604

Answer: C

Explanation:

5000 * 104/100 * 105/100 => 5460


4) The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.

A.200 sq cm    B.72 sq cm    C.162 sq cm    D.Cannot be determined    E.None of these

Answer: D


5) A started a business with an investment of Rs. 70000 and after 6 months B joined him investing Rs. 120000. If the profit at the end of a year is Rs. 52000, then the share of B is?

A.Rs. 28000    B.Rs. 24000    C.Rs. 30000    D.Rs. 26000    E.None of these

Answer: B

Explanation:

Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 52000
Share of B = 6/13 (52000) = Rs. 24000


6) There are two numbers. If 40% of the first number is added to the second number, then the second number increases to its five-fourth. Find the ratio of the first number to the second number?

A.8 : 25    B.25 : 8    C.8 : 5        D.5 : 8        E.None of these

Answer: D

Explanation:

Let the two numbers be x and y.
40/100 * x + y = 5/4y
=> 2/5 x = 1/4 y => x/y = 5/8

7) A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good.

A.6/63        B.2/63        C.125/126    D.1/126        E.1/63

Answer: C

Explanation:

Required probability = 1 - 1/126 = 125/126

8) The wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?

A.1 min        B.2 min        C.3 min        D.None

Answer: A

Explanation:

Time for one revolution = 60/15 = 4
60/ 20 = 3
60/48 = 5/4
LCM of 4, 3, 5/4
LCM of Numerators/HCF of Denominators = 60/1 = 60


9) A gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same. If there are 4 rows or 5 rows or 6 rows, then no tree will be left. Find the least number of trees required.

A.30        B.60        C.120        D.240        E.None of these

Answer: B

Explanation:

The least number of trees that are required = LCM(4, 5, 6) = 60.


10) A fruit vendor purchased 20 dozens of bananas at Rs. 15 per dozen. But one-fourth of the bananas were rotten and had to be thrown away. He sold two-third of the remaining bananas at Rs. 22.50 per dozen. At what price per dozen should he sell the remaining bananas to make neither a profit nor a loss?

A.Rs. 20    B.Rs. 15    C.Rs. 22.50    D.Rs. 7.50    E.None of these

Answer: B

Explanation:

CP of 20 dozen of bananas = 15 * 20 = Rs. 300
Number of bananas which are rotten = 1/4 * 20 = 5 dozen.
SP of two-third of remaining bananas = (2/3 * 15) * 22.5 = Rs. 225
SP of remaining 5 dozens of bananas to make no profit and no loss =(300 - 225) = Rs. 75.
SP of 1 dozen bananas = 75/5 = Rs. 15.


11) In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case.

A.420        B.528        C.494        D.464        E.None of these

Answer: B

Explanation:

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k
As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)
x = [(1200)(30)(3)] / (25)(2.5) - 1200 => x = 528.

12) A man traveled a total distance of 1800 km. He traveled one-third of the whole trip by plane and the distance traveled by train is three-fifth of the distance traveled by bus. If he traveled by train, plane and bus, then find the distance traveled by bus?

A.450 km    B.850 km    C.1200 km    D.750 km    E.None of these

Answer: D

Explanation:

Total distance traveled = 1800 km.
Distance traveled by plane = 600 km.
Distance traveled by bus = x
Distance traveled by train = 3x/5
=> x + 3x/5 + 600 = 1800
=> 8x/5 = 1200 => x = 750 km.


13) Rs.4500 amounts to Rs.5544 in two years at compound interest, compounded annually. If the rate of the interest for the first year is 12%, find the rate of interest for the second year?

A.10%    B.12%    C.15%    D.20%    E.None of these

Answer: A

Explanation:

Let the rate of interest during the second year be R%. Given,
4500 * {(100 + 12)/100} * {(100 + R)/100} = 5544
R = 10%


14) 2003 * 2004 - 2001 * 2002 = ?

A.8000        B.8010        C.8020        D.8040        E.8030

Answer: B

Explanation:

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 * 2000) - (2000 * 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> ? = 14012 - 6002 = 8010


15) In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?

A.145        B.150        C.155        D.160        E.None of these

Answer: A

Explanation:

By the time A covers 1000 m, B covers (1000 - 50) = 950 m.
By the time B covers 1000 m, C covers (1000 - 100) = 900 m.
So, the ratio of speeds of A and C =  1000/950 * 1000/900 = 1000/855  So, by the time A covers 1000 m, C covers 855 m.
So in 1000 m race A beats C by 1000 - 855 = 145 m.


16) X men can do a work in 120 days. If there were 20 men less, the work would have taken 60 days more. What is the value of X?

A.60        B.40        C.50        D.70        E.None of these

Answer: A

Explanation:

We have M1 D1 = M2 D2
120X = (X - 20)180
=> 2X = (X - 20) 3 => 2X = 3X - 60
=> X = 60


17) The diameters of two spheres are in the ratio 1:2 what is the ratio of their volumes?

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

Answer: C

Explanation:

1:8

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Infosys Aptitude Test Questions and Answers

Infosys Aptitude Test Questions and Answers For Freshers. Infosys Written Test Interview Questions with Solution and Explanation.

You can also see: TCS Aptitude Test Questions and Answers
 
1) Twelve men can complete a piece of work in 32 days. The same work can be completed by 16 women in 36 days and by 48 boys in 16 days. Find the time taken by one man, one woman and one boy working together to complete the work?

A.(64 * 36)/13 days    B.(32 * 36)/13 days    C.(96 * 36)/13 days    D.(128 * 36)/13 days    E.None of these

Answer: A

Explanation:

12 men take 32 days to complete the work. One man will take (12 * 32) days to complete it. Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.

One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.


2) By travelling at 40 kmph, a person reaches his destination on time. He covered two-third the total distance in one-third of the total time. What speed should he maintain for the remaining distance to reach his destination on time?

A.20 kmph    B.30 kmph    C.25 kmph    D.15 kmph    E.None of these.

Answer: A

Explanation:

Let the time taken to reach the destination be 3x hours. Total distance = 40 * 3x = 120x km
He covered 2/3 * 120x = 80x km in 1/3 * 3x = x hours So, the remaining 40x km, he has to cover in 2x hours. Required speed = 40x/2x = 20 kmph.


3) Calculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?

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

Answer: C

Explanation:

10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000


4) 7 2/5 of 110 / ? = 844

A.22        B.24        C.30        D.28        E.None of these

Answer: C

Explanation:

? = 844 - 7 2/5 of 110 = 844 - 37 * 22 = 30


5) An amount of Rs. 3000 becomes Rs. 3600 in four years at simple interest. If the rate of interest was 1% more, then what was be the total amount?

A.Rs. 3800    B.Rs. 3780    C.Rs. 3720    D.Cannot be determined        E.None of these

Answer: C

Explanation:

A = P(1 + TR/100)
=> 3600 = 3000[1 + (4 * R)/100] => R = 5%
Now R = 6%
=> A = 3000[1 + (4 * 6)/100] = Rs. 3720.


6) On the independence day, bananas were be equally distributed among the children in a school so that each child would get two bananas. On the particular day 360 children were absent and as a result each child got two extra bananas. Find the actual number of children in the school?

A.600        B.620        C.500        D.520        E.None of these

Answer: E

Explanation:

Let the number of children in the school be x. Since each child gets 2 bananas, total number of bananas = 2x.
2x/(x - 360) = 2 + 2(extra)
=> 2x - 720 = x => x = 720.


7) Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves one-fourth of his income, find the ratio of their monthly savings?

A.3 : 5        B.3 : 10    C.3 : 8        D.1 : 2        E.None of these

Answer: B

Explanation:

Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) - (His expenditure) = 5x - 2y.
=> 5x = 20x - 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x - 3y : 1/4(5x) = 6x - 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.


8) The successive discounts 20% and 15% are equal to a single discount of?

A.35%        B.38%        C.32%        D.29%        E.None of these
   
Answer: C

Explanation:

Let the CP of an article be Rs. 100
Given that successive discounts are 20% and 15%.
SP = 85% of 80% of 100 = (85/100)(80/100)(100)
=> SP = Rs. 68
Clearly, single discount is 32%.


9) Two numbers have a H.C.F of 16 and a product of two numbers is 2560. Find the L.C.M of the two numbers?

A.140    B.150    C.160    D.170    E.None of these

Answer: C

Explanation:

L.C.M of two numbers is given by
(Product of the two numbers) / (H.C.F of the two numbers) = 2560/16 = 160.


10) Find the lowest 4-digit number which when divided by 3, 4 or 5 leaves a remainder of 2 in each case?

A.1020        B.1026        C.1030        D.1022

Answer: D

Explanation:

Lowest 4-digit number is 1000.
LCM of 3, 4 and 5 is 60.
Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020.
Now, add the remainder 2 that's required. Thus, the answer is 1022.


11) Out of 15 consecutive numbers, 2 are chosen at random. The probability that they are both odds or both primes is -.

A.10/17        B.10/19        C.46/105    D.11/15        E.Cannot be determined

Answer: E

Explanation:

There is no definite formula for finding prime numbers among 15 consecutive numbers. Hence the probability cannot be determined.

12) In an election between two candidates A and B, the number of valid votes received by A exceeds those received by B by 15% of the total number of votes polled. If 20% of the votes polled were invalid and a total of 8720 votes were polled, then how many valid votes did B get?

A.2160        B.2420        C.2834        D.3150        E.None of these

Answer: C

Explanation:

Let the total number of votes polled in the election be 100k.
Number of valid votes = 100k - 20% (100k) = 80k
Let the number of votes polled in favour of A and B be a and b respectively.
a - b = 15% (100k) => a = b + 15k
=> a + b = b + 15k + b
Now, 2b + 15k = 80k and hence b = 32.5k
It is given that 100k = 8720
32.5k = 32.5k/100k * 8720 = 2834
The number of valid votes polled in favour of B is 2834.

13) A and B start a business with Rs.6000 and Rs.8000 respectively. Hoe should they share their profits at the end of one year?

A.1:2        B.3:4        C.2:5        D.3:7

Answer: B

Explanation:

They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
6000 : 8000 => 3:4


14) The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?

A.91.5 cm    B.93.5 cm    C.94.5 cm    D.92.5 cm    E.None of these

Answer: A

Explanation:

Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm


15) A sum of Rs.4800 is invested at a compound interest for three years, the rate of interest being 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years.

A.Rs.2520    B.Rs.3120    C.Rs.3320    D.Rs.2760    E.None of these

Answer: B

Explanation:

Let A be the amount received at the end of the three years.
A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4800 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.7920
So the interest = 7920 - 4800 = Rs.3120


16) The total marks obtained by a student in Mathematics and Physics is 60 and his score in Chemistry is 20 marks more than that in Physics. Find the average marks scored in Mathamatics and Chemistry together.

A.40    B.30    C.25    D.Data inadequate    E.None of these.

Answer: A

Explanation:

Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 60 and C - P = 20 M + C / 2 = [(M + P) + (C - P)] / 2 = (60 + 20) / 2 = 40.


17) Find the area of circle whose radius is 7m?

A.124 sq m    B.154 sq m    C.145 sq m    D.167 sq m

Answer: B

Explanation:

22/7 * 7 * 7 = 154

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TCS Aptitude Test Questions and Answers

TCS Aptitude Test Questions and Answers For Freshers. Tata Consultancy Services written test paper interview questions with solutions.
 
1) The cross-section of a cannel is a trapezium in shape. If the cannel is 10 m wide at the top and 6 m wide at the bottom and the area of cross-section is 640 sq m, the depth of cannel is?

A.20 m        B.60 m        C.40 m        D.80 m

Answer: D

Explanation:

1/2 * d (10 + 6) = 640
d = 80

2) Five years ago the average of the ages of A and B was 40 years and now the average of the ages of B and C is 48 years. What will be the age of the B ten years hence?

A.55 years    B.56 years    C.58 years    D.Data inadequate    E.None of these

Answer: D

Explanation:

Let the present ages of A, B and C be a, b and c respectively.
Given, [(a - 5) + (b - 5)] / 2 = 40 => a + b = 90 --- (1)
(b + c)/2 = 48 => b + c = 96 --- (2)
From (1) and (2), we cannot find b.


3) Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest?

A.18% p.a.    B.22% p.a.    C.20% p.a.    D.24% p.a.    E.None of these.

Answer: C

Explanation:

Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year.
Rate of interest = (100 * 240) / (100 * 1) = 20% p.a


4) 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: A

Explanation:

Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 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

5) A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A's share in the profit is?

A.Rs.1000    B.Rs.600    C.Rs.800    D.Rs.400

Answer: D

Explanation:

2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400


6) In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are at-least 50 years old. Find the number of males aged below 50 years?

A.1040        B.2080        C.3120        D.4160        E.None of these

Answer: C

Explanation:

Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 - 25)%
= 4160 * 75/100 = 3120.


7) The probability that A speaks truth is 3/5 and that of B speaking truth is 4/7. What is the probability that they agree in stating the same fact?

A.18/35        B.12/35        C.17/35        D.19/35        E.None of these

Answer: A

Explanation:

If both agree stating the same fact, either both of them speak truth of both speak false.

Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35


8) Find the greatest 4-digit number exactly divisible by 3, 4 and 5?

A.9985        B.9960        C.9957        D.9975

Answer: B

Explanation:

Greatest 4-digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is

9999 - 39 = 9960.


9) Find the greatest number which divides 83, 125 and 209 leaving the same remainder in each case.

A.19    B.17    C.42    D.23    E.None of these

Answer: C

Explanation:

The greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q - p), (r - p) and (r - q)).
For the given problem, p = 83, q = 125 and r = 209.
Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 - 83, 209 - 83, 209 - 125) = 42.


10) Ramesh purchased a refrigerator for Rs. 12500 after getting a discount of 20% on the labelled price. He spent Rs. 125 on transport and Rs. 250 on installation. At what price should it be sold so that the profit earned would be 10% if no discount was offered?

A.Rs. 16500    B.Rs. 15525    C.Rs. 17000    D.Rs. 17600    E.None of these

Answer: D

Explanation:

Price at which the TV set is bought = Rs. 12,500
Discount offered = 20%
Marked Price = 12500 * 100/80 = Rs. 15625
The total amount spent on transport and installation = 125 + 250 = Rs. 375\Total price of TV set = 15625 + 375 = Rs. 16000

The price at which the TV should be sold to get a profit of 10% if no discount was offered = 16000 * 110/100 = Rs. 17600


11) Three persons A, B and C divide a certain amount of money such that A's share is Rs. 4 less than half of the total amount, B's share is Rs. 8 more than half of what is left and finally C takes the rest which is Rs. 14. Find the total amount they initially had with them?

A.Rs. 61    B.Rs. 85    C.Rs. 80    D.Rs. 70    E.None of these

Answer: C

Explanation:

Let the total amount be Rs. p.
Let shares of A and B be Rs. x and Rs. y respectively.
C's share was Rs. 14
we have, x + y + 14 = p ----- (1)
From the given data, x = (p/2) - 4 ----- (2)
Remaining amount = p - (p/2 - 4) => p/2 + 4.
y = 1/2(p/2 + 4) + 8 => p/4 + 10 ----- (3)
From (1), (2) and (3)
p/2 - 4 + p/4 + 10 + 14 = p
3p/4 + 20 = p
p/4 = 20 => p = Rs. 80.

12) Eight years ago, Ajay's age was 4/3 times that of Vijay. Eight years hence, Ajay's age will be 6/5 times that of Vijay. What is the present age of Ajay?

A.30 years    B.40 years    C.32 years    D.48 years    E.None of these

Answer: B

Explanation:

Let the present ages of Ajay and Vijay be 'A' and 'V' years respectively.
A - 8 = 4/3 (V - 8) and A + 8 = 6/5 (V + 8)
3/4(A - 8) = V - 8 and 5/6(A + 8) = V + 8
V = 3/4 (A - 8) + 8 = 5/6 (A + 8) - 8
=> 3/4 A - 6 + 8 = 5/6 A + 20/3 - 8
=> 10 - 20/3 = 10/12 A - 9/12 A
=> 10/3 = A/12 => A = 40.


13) A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?

A.Rs. 7000    B.Rs. 8500    C.Rs. 8000    D.Rs. 7500    E.None of these

Answer: A

Explanation:

Let the sum be Rs. x.
(x * 18 * 2)/100 - (x * 12 * 2)/100 = 840 => 36x/100 - 24x/100 =840
=> 12x/100 = 840 => x = 7000.


14) 8 2/3 + 6 4/5 = ?

A.14        B.14 1/2    C.14 3/4    D.15 1/2    E.None of these
   
Answer: E

Explanation:

8 2/3 + 6 4/5 = (8 + 6) + (10/15 + 12/15) = 14 + 1 7/15 = 15 7/15


15) A car started running at a speed of 30 km/hr and the speed of the car was increased by 2 km/hr at the end of every hour. Find the total distance covered by the car in the first 10 hours of the journey.

A.380 km    B.390 km     C.400 km    D.410 km     E.None of these

Answer: B

Explanation:

The total distance covered by the car in the first 10 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 = sum of 10 terms in AP whose first term is 30 and last term is 48 = 10/2 [30 + 48] = 390 km.


16) Sreedhar and Sravan together can do a work in 25 days. With the help of Pavan, they completed the work in 8 days and earned Rs. 225. What is the share of Sravan, if Sreedhar alone can do the work in 75 days?

A.Rs. 64    B.Rs. 52    C.Rs. 48    D.Rs. 58    E.None of these

Answer: C

Explanation:

Sravan's one day's work = 1/25 - 1/75 = 2/75
Sravan worked for 8 days. So, his 8 days work = 8 * 2/75 = 16/75
Sravan completed 16/75th of total work.
So, his share is 16/75 * 225 = Rs. 48.


17) If in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it, then what will be the maximum number of the small boxes that can be kept in it?

A.500        B.1000        C.900        D.600

Answer: B

Explanation:

6 * 5 * 4 = 60/100 * 50/100 * 40/100 * x
1 = 1/10 * 1/10 * 1/10 * x => x = 1000

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Probability - Aptitude Questions and Answers

Probability aptitude questions and answers section with explanation. Practice online test for various interview, competitive and entrance exams.

1. The probability that A speaks truth is 3/5 and that of B speaking truth is 4/7. What is the probability that they agree in stating the same fact?

A.18/35        B.12/35        C.17/35        D.19/35        E.None of these

Answer: A

Explanation:

If both agree stating the same fact, either both of them speak truth of both speak false.
Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35

2. Out of 15 consecutive numbers, 2 are chosen at random. The probability that they are both odds or both primes is -.

A.10/17        B.10/19        C.46/105    D.11/15        E.Cannot be determined

Answer: E

Explanation:

There is no definite formula for finding prime numbers among 15 consecutive numbers. Hence the probability cannot be determined.

3. What is the probability that a leap year has 53 Sundays and 52 Mondays?

A.0        B.1/7        C.2/7        D.5/7        E.6/7
   
Answer: B

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7

4. 10 books are placed at random in a shelf. The probability that a pair of books will always be together is -.

A.1/10        B.9/10        C.1/5        D.3/10        E.1/2

Answer: C

Explanation:

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
Required probability = 1/5

5. Three 6 faced dice are thrown together. The probability that all the three show the same number on them is -.

A.1/216        B.1/36        C.5/9        D.5/12        E.7/12

Answer: B

Explanation:

It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36

6. Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is -.

A.7/12        B.5/9        C.1/36        D.5/12        E.8/9

Answer: B

Explanation:

No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.

7. Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is -.

A.5/9        B.5/12        C.1/36        D.7/12        E.4/9

Answer: B

Explanation:

Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12

8. A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good.

A.6/63        B.2/63        C.125/126    D.1/126        E.1/63

Answer: C

Explanation:

Required probability = 1 - 1/126 = 125/126

9. If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -.

A.19/52        B.17/52        C.5/13        D.4/13        E.9/26

Answer: D

Explanation:

P(SᴜK) = P(S) + P(K) - P(S∩K), where S denotes spade and K denotes king.
P(SᴜK) = 13/52 + 4/52 - 1/52 = 4/13

10. If six persons sit in a row, then the probability that three particular persons are always together is -.

A.1/20        B.3/10        C.1/5        D.4/5        E.2/5
   
Answer: C

Explanation:

Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5

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Percentages - Aptitude Questions and Answers

Percentages aptitude questions and answers section with explanation. Practice online test for various interview, competitive and entrance exams.

1. In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are at-least 50 years old. Find the number of males aged below 50 years?

A.1040        B.2080        C.3120        D.4160        E.None of these

Answer: C

Explanation:

Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 - 25)%
= 4160 * 75/100 = 3120.

2. In an election between two candidates A and B, the number of valid votes received by A exceeds those received by B by 15% of the total number of votes polled. If 20% of the votes polled were invalid and a total of 8720 votes were polled, then how many valid votes did B get?

A.2160        B.2420        C.2834        D.3150        E.None of these

Answer: C

Explanation:

Let the total number of votes polled in the election be 100k.
Number of valid votes = 100k - 20% (100k) = 80k
Let the number of votes polled in favour of A and B be a and b respectively.
a - b = 15% (100k) => a = b + 15k
=> a + b = b + 15k + b
Now, 2b + 15k = 80k and hence b = 32.5k
It is given that 100k = 8720
32.5k = 32.5k/100k * 8720 = 2834
The number of valid votes polled in favour of B is 2834.

3. Two tests had the same maximum mark. The pass percentages in the first and the second test were 40% and 45% respectively. A candidate scored 216 marks in the second test and failed by 36 marks in that test. Find the pass mark in the first test?

A.136        B.128        C.164        D.214        E.None of these

Answer: E

Explanation:

Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.

4. Ten percent of Ram's monthly salary is equal to eight percent of Shyam's monthly salary. Shyam's monthly salary is twice Abhinav's monthly salary. If Abhinav's annual salary is Rs. 1.92 lakhs, find Ram's monthly salary?

A.Rs. 18000    B.Rs. 20000    C.Rs. 25600    D.Rs. 32000    E.None of these

Answer: C

Explanation:

Let the monthly salaries of Ram and Shyam be Rs. r and Rs. s respectively.
10/100 r = 8/100 s
r = 4/5 s
Monthly salary of Abhinav = (1.92 lakhs)/12 = Rs. 0.16 lakhs
s = 2(0.16 lakhs) = 0.32 lakhs
r = 4/5(0.32 lakhs) = Rs. 25600

5. The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?

A.5 : 3 : 1    B.5 : 3 : 2    C.10 : 7 : 4    D.10 : 8 : 5    E.None of these

Answer: B

Explanation:

Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.

6. In a group of 80 children and 10 youngsters, each child got sweets that are 15% of the total number of children and each youngster got sweets that are 25% of the total number of children. How many sweets were there?

A.1160        B.1100        C.1080        D.1210        E.None of these

Answer: A

Explanation:

Number of sweets each child got = 15% of 80 = 15/100 * 80 = 12.
Number of sweets 80 children got = 80 * 12 = 960.
Number of sweets each youngster got = 25% of 80 = 25/100 * 80 = 20.
Number of sweets 10 youngsters got = 10 * 20 = 200.
Total number of sweets = 960 + 200 = 1160.

7. Anil spends 40% of his income on rent, 30% of the remaining on medicines and 20% of the remaining on education. If he saves Rs. 840 every month, then find his monthly salary?

A.Rs. 1800    B.Rs. 2000    C.Rs. 2200    D.Rs. 2500    E.None of these

Answer: D

Explanation:

Let's Anil's salary be Rs. 100.
Money spent on Rent = 40% of 100 = Rs. 40.
Money spent on medical grounds = 30% of (100 - 40) = 3/10 * 60 = Rs. 18.
Money spent on education = 20% of (60 - 18) = 1/5 * 42 = Rs. 8.40
Anil saves 100 - (40 + 18 + 8.40) i.e., Rs. 33.60
for 33.6 ---> 100 ; 840 ---> ?
Required salary = 840/33.6 * 100 = Rs. 2500

8. There are two numbers. If 40% of the first number is added to the second number, then the second number increases to its five-fourth. Find the ratio of the first number to the second number?

A.8 : 25    B.25 : 8    C.8 : 5        D.5 : 8        E.None of these

Answer: D

Explanation:

Let the two numbers be x and y.
40/100 * x + y = 5/4y
=> 2/5 x = 1/4 y => x/y = 5/8

9. There are three numbers. 5/7th of the first number is equal to 48% of the second number. The second number is 1/9th of the third number. If the third number is 1125, then find 25% of the first number?

A.168        B.84        C.42        D.21        E.None of these

Answer: D

Explanation:

Let the first number and the second number be F and S respectively.
5/2 F = 48/100 S ----> (1)
S = 1/9 * 1125 = 125
(1) => 5/7 F = 48/100 * 125
=> F = 84
25% of F = 1/4 * 84 = 21.

10. The monthly incomes of A and B are in the ratio 5 : 2. B's monthly income is 12% more than C's monthly income. If C's monthly income is Rs. 15000, then find the annual income of A?

A.Rs. 420000    B.Rs. 180000    C.Rs. 201600    D.Rs. 504000    E.None of these

Answer: D

Explanation:

B's monthly income = 15000 * 112/100 = Rs. 16800
B's monthly income = 2 parts ----> Rs. 16800
A's monthly income = 5 parts = 5/2 * 16800 = Rs. 42000
A's annual income = Rs. 42000 * 12 = Rs. 504000

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Mensuration - Aptitude Questions and Answers

Mensuration aptitude questions and answers section with explanation. Practice online test for various interview, competitive and entrance exams.

1. 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: A

Explanation:

Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 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

2. The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?

A.91.5 cm     B.93.5 cm     C.94.5 cm     D.92.5 cm     E.None of these

Answer: A

Explanation:

Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm

3. The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing 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: D

Explanation:

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: B

Explanation:

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 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: E

Explanation:

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

6. 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: C

Explanation: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

7. The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?

A.600 cm    B.800 cm    C.400 cm    D.1000 cm    E.None of these

Answer: B

Explanation:

Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.

8. The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.

A.200 sq cm    B.72 sq cm    C.162 sq cm    D.Cannot be determined    E.None of these

Answer: D

Explanation:

Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.

9. The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?

A.27 m        B.24 m        C.18 m        D.21 m        E.None of these

Answer: C

Explanation:

Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l2 = 324 => l = 18.

10. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

A.Rs. 3642.40    B.Rs. 3868.80    C.Rs. 4216.20    D.Rs. 4082.40    E.None of these

Answer: D

Explanation:

Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

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Problems on Numbers - Aptitude Questions and Answers

Problems on Numbers aptitude questions and answers section with explanation. Practice online test for various interview, competitive and entrance exams.

1. Find the greatest number which divides 83, 125 and 209 leaving the same remainder in each case.

A.19                  B.17              C.42            D.23       E.None of these

Answer: C

Explanation:

The greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q - p), (r - p) and (r - q)).
For the given problem, p = 83, q = 125 and r = 209.
Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 - 83, 209 - 83, 209 - 125) = 42.

2. Two numbers have a H.C.F of 16 and a product of two numbers is 2560. Find the L.C.M of the two numbers?

A.140           B.150     C.160              D.170        E.None of these

Answer: C

Explanation:

L.C.M of two numbers is given by
(Product of the two numbers) / (H.C.F of the two numbers) = 2560/16 = 160.

3. Find the greatest four digit number which leaves respective remainders of 2 and 5 when divided by 15 and 24.

A.9974  B.9125     C.9565              D.9997 E.None of these

Answer: E

Explanation:

Since the difference between the divisors and the respective remainders is not constant, back substitution is the convenient method. None of the given numbers is satisfying the condition.

4. Find the greatest number which leaves the same remainder when it divides 25, 57 and 105.

A.18           B.8        C.12        D.16 E.None of these

Answer: D

Explanation:

105 - 57 = 48
57 - 25 = 32
105 - 25 = 80
The H.C.F of 32, 48 and 80 is 16.

5. What is the remainder when 263251 is divided by 9?

A.8           B.3         C.6          D.1 E.None of these

Answer: D

Explanation:

If the sum of the digits of a number is divisible by 9, the number will be divisible by 9.
2 + 6 + 3 + 2 + 5 + 1 = 19
The nearest multiple of 9 is 18.
19 - 18 = 1.

6.If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers.

A.50           B.30 C.125 D.25 E.None of these

Answer: D

Explanation:

H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.

7. Find the smallest four-digit number which is a multiple of 112.

A.896 B.1008 C.1120 D.1024       E.None of these

Answer: B

Explanation:

The smallest four digit number is 1000. If 1000 is divided by 112, the remainder is 104.
112 - 104 = 8, if 8 is added to 1000, it will become the smallest four digit number and a multiple of 112.

8. A gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same. If there are 4 rows or 5 rows or 6 rows, then no tree will be left. Find the least number of trees required.

A.30           B.60 C.120 D.240         E.None of these

Answer: B

Explanation:

The least number of trees that are required = LCM(4, 5, 6) = 60.

9. Find the least number which when divided by 35 and 11 leaves a remainder of 1 in each case.

A.384   B.391 C.388 D.397          E.386

Answer: E

Explanation:

The least number which when divided by different divisors leaving the same remainder in each case
= LCM(different divisors) + remainder left in each case.
Hence the required least number 
= LCM(35, 11) + 1 = 386.

10. Find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5.

A.187 B.197 C.207 D.219 E.None of these

Answer: B

Explanation:

Let 'N' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) - (common difference of divisors and remainders) = (208) - (11) = 197.

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