Infosys Aptitude Test Questions and Answers

Infosys Aptitude Test

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 

Infosys Aptitude Test Questions with Explanation:

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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