**Infosys Quantitative Aptitude Placement Test Questions**

Is Infosys visiting your campus soon? Are you looking for resources to ready yourself for the placement challenge? Here are a set of Quantitative aptitude practice questions based on previous years

Infosys written test includes three sections, Quantitative Aptitude, Verbal Ability & Reasoning Ability, with 10, 15 and 40 questions respectively. Time allotted for Quantitative Aptitude Section is 35 minutes. Questions are expected to be based on Ratios & Proportions, Time & Work, Time & Distance, Number Systems, Probability, Permutations & Combinations, Mensuration, Percentages and algebra. For complete recruitment pattern, click here Infosys Campus Recruitment Process

**Infosys Aptitude Questions**

**Students of St. Peter college are lined up in row. Lata who is standing 16th from the left stands on the immediate right of Sita. When they exchange their places, Sita stands 14th from the right. Find out the total number of students in the row.**a) 29

b) 30

c) 28

d) 31

Answer/ Explanation: a) 29

Number of students in the row = 15+ 14 =29

**A man travels 40 Km by a bicycle and 20 Km by walk in 42 minutes. Instead, if he covers 25 Km by bicycle and 35 Km by walk, he takes 18 more minutes. What is his speed on bicycle in Kmph?**a) 100

b) 80

c) 40

d) 200

Answer/ Explanation: d) 200

Let the speed of the man by bicycle be b and walk be w in Kmph respectively.

According to the question, (40/b) +(20/w) = (42/60)

(25/b) + (35/w) = 1

Solving for b and w, we get the value of b as 200 Kmph and w as 40 Kmph respectively.

**A box contains 5 red and 6 green shoes. If two shoes are drawn, what is the probability of getting red shoes?**a) 5C₂/ 11C₂

b) 3C₂/ 11C₂

c) 5C₂/ 10C₂

d) 4C₂/ 10C₂

Answer/ Explanation: a) 5C₂/ 11C₂

Probability = number of probable events/ Total number of evets

Number of ways two red shoes can be picked = 5C₂

Number of ways any two shoes can be picked is 11C₂as the total number of balls is 11.

**There are few people in the village Haripuram. Each house has as many number of people in it as the number of houses. Find the total number of people if the number is in between 1000 and 1050.**a) 1034

b) 1024

c) 1004

d) 1044

Answer/ Explanation: b) 1024

Let the number of houses be h and number of people in each house be h.

According to the question, h×h is a number between 1000 and 1050. It has to be 1024 and the number of people in each house is 32 members.

**A clock loses 7 seconds in one hour and was set right at 4 a.m. What time will it show at 8 p.m. on the same day?**a) 7: 57: 08 p.m.

b) 7: 59: 08 p.m.

c) 8: 01: 52 p.m.

d) 7: 58: 08 p.m.

Answer/ Explanation: d) 7: 58: 08 p.m.

A right clock would have completed 16 hours by 8 p.m. whereas this particular clock loses 7 seconds every hour. Hence, total time lost = 7×16 = 112 seconds = 1 minute 52 seconds.

Therefore, the slower clock shows 7: 58: 08 p.m.

**A horse is tied to a place which is located at the vertex of an equilateral triangle of side 8 m with a rope of length 12 m. Given that the horse can graze in the area only around the equilateral triangle, how much area can the horse graze in?**a) 144π - 16√3

b) 16√3 - 144π

c) 144√3 - 16π

d) Π - 16√3

Answer/ Explanation: a) 144π - 16√3

The horse will be able to graze in the circular area formed by the horse with the length of the rope being 12 m.

Area of circle = π×12×12 = 144π sq.m

Area of the equilateral triangle in which the horse cannot graze = (√3/4) ×8×8 = 16√3 sq.m

Therefore, the area in which the horse cannot graze =(144π - 16√3) sq.m

**Coffee worth Rs. 183 per Kg, 173 per Kg and a third variety are mixed in the ratio of 2: 2: 3 and sold at Rs. 200 per Kg at a price of no loss/ profit. What is the approximate cost per Kg of the third variety of coffee?**a) 220

b) 229

c) 239

d) 240

Answer/ Explanation: b) 229

Let the third variety of coffee be Rs. x Per Kg.

According to the question

((183×2) + (173×2) + (x×3))/ (2+2+3) = 200

Solving the above equation, we get the value of x as Rs.229.30

**If 40 chocolates are distributed among some men and women in a party such that a woman gets 4 chocolates and a man gets 2 chocolates,****Statement 1: The number of men in the group is not more than 3****Statement 2: If each man gets 4 chocolates and each woman gets 2 chocolates, the total number of chocolates needed would be 14 less****What is the number of men and women in the party?**a) Statement 1 alone is enough to solve the question but statement 2 alone is not sufficient to answer the question.

b) Statement 2 alone is sufficient to answer the question but statement 1 alone is not enough to answer the question

c) Both statements taken together are sufficient to answer the question but neither statement alone is not sufficient

d) Either of the statements alone is sufficient to answer the question.

e) Statements 1 and 2 together are not sufficient to answer the question and additional data is needed to answer the question.

Answer/ Explanation: d) Either of the statements alone is sufficient to answer the question.

Let the number of men be M and women be W. According to the question, the equation is

4W + 2M = 40

The following are the solutions to this equation.

M = 2; W = 9

M = 4; W = 8

Looking at the statement 1, second solution isn’t possible as the number of men cannot exceed 10.

Hence, the question can be answered using statement 1 alone.

Let’s look at statement 2.

The equation is 4M+ 2W = 26

Solving the two simultaneous equations, we have the number of men as 2 and women as 9.

Hence, the right answer is d.

**A shopkeeper marks an article at 40 % higher price than the cost price and offers a discount of 25%. What is the actual profit percentage of the shopkeeper?**a) 105

b) 5

c) 10.5

d) 10

Answer/ Explanation: b) 5

Let the cost price of article be x

Marked Price = 1.4x

Selling Price = 1.4x ×0.75 = 1.05x

Hence, the profit percentage = 5%

**40 Quintals are what percent of 4 metric tonnes?**a) 15

b) 150

c) 5

d) 1500

Answer/ Explanation: b) 150 Kg

1 Quintal = 100 Kg

1 Metric Ton = 1000 Kg

Hence, the equation is

40 ×100 = (x/100) ×2×1000

Solving for x, we get its value as 150%

**Vineeth starts working on a task and works on it for 10 days and completes 40 % of the work. To help him complete the job, he employs Rajath and together they work for another 14 days and the work gets completed. How much more efficient is Vineeth than Rajath?**a) 14

b) 25

c) 28

d) 0

Answer/ Explanation: d) 0

Let the times taken by Vineeth and Rajath to complete the tasks alone be V and R respectively.

(1/V) ×10 = (40/100)

From the equation, time taken by Vineeth to complete the task alone is 25 days

((1/v) + (1/R)) ×14 = (60/100)

Substituting the value of V in the second equation, we get

((1/25) +(1/R))×14 = (3/5)

R = 25 days.

They are equally efficient.

**Lavanya has Rs. 480 in the denominations of one-rupee notes, five rupee notes and ten rupee notes. The number of notes of each denomination is equal. What is the total number of notes?**a) 30

b) 90

c) 120

d) 480

Answer/ Explanation: a) 30

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

**The average salary to be paid to 45 workers is Rs. 10000 and 15 workers is x. The overall average salary is 15000. Find the average monthly salary of these 15 workers.**a) 15000

b) 30000

c) 45000

d) 60000

Answer/ Explanation: c) 30000

According to the question,

((45×10000) + (15×x))/60 = 15000

Hence, the value of x is 30000

**What is the cost of fencing round a circular field of radius 14 m at the rate of Rs. 2 per m?**a) 167

b) 176

c) 180

d) 190

Answer/ Explanation: b) 176

Given, the radius of the circle is 14 m.

Perimeter of the circle = 2×π×r = 88 m

Cost of fencing = 88×2 = 176

**If I divide the number of biscuits with me into two unequal numbers, 78 times the difference between the numbers equals difference between squares of the numbers. How many biscuits do I have?**a) 87

b) 78

c) 70

d) 65

Answer/ Explanation: b) 78

Let the numbers into which 78 biscuits are divided be x and y.

78(x-y) = x² - y² = (x+y)(x-y)

x+y = 78

Therefore, the total number of biscuits with me is 78.

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