# Infosys Aptitude Questions with solutions 2019 for Infosys Aptitude Test

Infosys most repeated aptitude questions with solutions are given here. Infosys aptitude questions will give you an idea of expected questions in Infosys online test. The overall Quantitative Aptitude section of the Infosys drive is of the medium difficulty level. According to various test-takers, the following can be interpreted:

• Five Infosys questions asked in the Quants section are easy and the remaining 5 are tricky and difficult.
• The easy Infosys questions were from Percentages, Time and Distance, Profit and Loss, and Mensuration.

The syllabus for Infosys aptitude questions section is given below. This Analysis was based on the number of questions asked in previous Infosys placement papers.

TopicNumber of questions expectedDifficulty Level
Percentages1 – 2Easy – Medium
Profit and Loss1 – 2Easy – Medium
Permutations and Combinations1 – 2Easy – Medium
Probability1 – 2Easy – Medium
Speed Time and Distance1 – 2Easy – Medium
Boats and Streams1 – 2Easy – Medium
Time and Work1 – 2Easy – Medium
Data Interpretation1 – 2Easy – Medium
Mixtures and Allegation1 – 2Easy – Medium

## Infosys Aptitude Questions and Answers

1) There was a cycle race going on. 1/5th of those in front of a person and 5/6 of those behind him gives the total number of participants. How many people took part in the race?

Solution:
let us assume number of participants = x
Based on the condition  x.1/5+x.6/5=x+1/5+1/6
(x-1)1/5+(x-1)6/5=x
x=31

2) X^1/3 -X^1/9  = 60.then find X?

Solution:
let us take X=a^9 . Then, ( a^9)^1/3-(a^9)^1/9 =60
a^9/3-a^9/9=60
a^3-a=60
a(a^2-1)=60
by solving this , a=4

3) There are 10 points in three parallel planes, the first plane contain 5 points, the second plane contains 3 points and the third plane contain rest of them, then how many triangles to be formed?
a) 104
b) 109
c) 105
d) 106
e) None of these

Solution:
In general, 3 points are needed for forming a triangle. Therefore from 10 points, we can form 10c3 triangles but
out of 10 points 5 (1st line), 3 (2nd line), 2 (3rd line) points are collinear.  So we have to eliminate those points and hence we subtract those colinear points from 10c3
ie, =(10c3)-(5c3)-(3c3)
=120-10-1
=109

4) 200meter train travel with the speed of 120kmph and other train travel in opposite direction with speed of 80kmph in 9 seconds. Then find the length of the other train?
a) 240
b) 300
c) 260
d) 270

Solution:
Time = Distance / Speed
Distance = Length1+Length2 = 200+Length2
Speed = S1+S2 = 120+80 = 200
Time = 9sec = 9*(5/18) = 5/2 = 2.5
So now L2 = 300m

5) Class P has 30 students of which 20 like Music. In class Q, 10 students like Music. Find the number of students in Class Q if the average number of students who like Music in a class is 16.

Solution:
let the number of students in class Q be x, so here the total number of students in P is 30 and 20 people like music and in Q 10 like music, so the combined mean of students who like music in both the classes is 16 (the average number of students who like music in a class)
hence the applying formula of combined mean , ((30*20)+(x*10))/(30+x) = 16, solving we get x=20.

6) Ram & Shyam started from a point X and Y respectively and started moving towards each other. After they met Ram took 4 hours to reach Y and Shyam took 16 hours to reach X. Ram’s speed is 48 kmph. What is the speed of Shyam?

1) 24kmph
2) 56kmph
3) 32kmph
4) 12kmph
5) 24 kmph

Solution:
Let the speed of Shyam be ‘x’ kmph, then the ratio of speed of Ram and Shyam = Square root of (Time taken by Shyam to Reach X after they meet / Time taken by Ram to Reach Y after they meet)
= 48/x = Sqrt(16/4)
= 48/x = 2
x = 24

7) A train leaves Meerut at 5 a.m. and reaches Delhi at 9 a.m. Another train leaves Delhi at 7 a.m. and reaches Meerut at 10.30 a.m. At what time do the two trains travel in order to cross each other?

Solution:
The first train takes 4 hours and the second train takes 3.5 hours.
Time ratio is 8:7. Therefore, the speed ratio will be 7:8.
Let the speeds be 7x and 8x, and distance is 28x ( 4×7 or 3.5×8).
At 7 AM, the first train must have covered a distance of 14x. Therefore, at 7 A.M. the distance between the two trains is 28x-14x=14x.
Time taken to meet = 14x/(7x+8x)=14/15 hour or 56 minutes.
Hence, the two trains meet at 7.56 AM

8) A man reaches his office 20 min late if he walks from his home at 3 km per hour and reaches 30 min early if he walks 4 km per hour. How far is his office from his house?

1) 20 km
2) 16 km
3) 14 km
4) 10 km

Solution:
Let distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
time taken at 4 kmph : x/4 = 30 min earlier
difference between time taken : 30-(-20) = 50 mins = 50/60 hours.
x/3- x/4 = 50/60
x/12 = 5/6
x = 10 km

9) Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:

1) 30%
2) 33.33%
3) 35%
4) 44%

Solution:
Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.
C.P. of 30 articles = Rs. 5/6 x 30 = Rs. 25.
S.P. of 30 articles = Rs. 6/5 x 30 = Rs. 36.
Gain % = 11/25 x 100 % = 44%.

10) How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?

1) 36
2) 42
3) 54
4) 63

Solution:
S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.
C.P. of 1 kg of mixture = Rs. (100/110) x 9.24 = Rs. 8.40
By the rule of alligation, we have:
Ratio of quantities of 1st and 2nd kind = 14:6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
x = (7 x 27)/3 = 63 kg.