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.

Topic | Number of questions expected | Difficulty Level |
---|---|---|

Percentages | 1 – 2 | Easy – Medium |

Profit and Loss | 1 – 2 | Easy – Medium |

Permutations and Combinations | 1 – 2 | Easy – Medium |

Probability | 1 – 2 | Easy – Medium |

Speed Time and Distance | 1 – 2 | Easy – Medium |

Boats and Streams | 1 – 2 | Easy – Medium |

Time and Work | 1 – 2 | Easy – Medium |

Data Interpretation | 1 – 2 | Easy – Medium |

Mixtures and Allegation | 1 – 2 | Easy – 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 ^{10}c_{3} 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 ^{10}c_{3}

ie, =(^{10}c_{3})-(^{5}c_{3})-(^{3}c_{3})

=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.