Mindtree Company Campus Recruitment - Quantitative Questions

   

Mindtree Quantitative Aptitude Placement Test Questions

Is Mindtree visiting your campus soon? Are you looking for resources to prepare yourself for the placement challenge? Here are a set of Quant related practice questions based on previous years’ Mindtree test pattern. The test will cover Logical Reasoning, Quantitative aptitude, Verbal Ability and Coding questions. In the Quantitative Ability Section, there will be 25 questions. This blog covers aptitude questions related to Probability, time and distance, profit-loss, percentage, and log questions which have appeared in previous year's papers. You can go through the following sample questions to make yourself ready for Mindtree Test Pattern.

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Directions for Questions 1 to 8: Choose the correct alternative

1. A bag contains 7 red balls, 3 white balls and 8 black balls. 3 balls are drawn at random one by one. What is the probability that all the balls are red balls?

a) 3/18

b) 7C3/18C3

c) (7/18) x (6/17) x (5/16)

d) 3/(18 x 17 x 16)

Explanation:

Total number of balls = 7+3+8 = 18

The probability of drawing first ball as red = total red balls/total balls = 7/18

The probability of drawing the second ball as red = (total red balls – 1) /(total balls – 1) = 6/17 (Since we have already drawn one ball in the previous turn, so 1 will be deducted from both numerator and denominator)

Probability of drawing third ball as red = (total red balls – 2) / (total balls – 2) = 5/16

Now, probability of picking up three red balls one by one = (7/18) x (6/17) x (5/16)

Therefore, the correct answer is Option C

2. Arun invited his 11 friends to his birthday party. Every one shook hand with each other. Find the total number of handshakes at the party?

a) 24

b) 66

c) 55

d) 48

Explanation:

The formulae for number of handshakes = {n x (n-1)} / 2

12 x (12-1) / 2 = 66

Therefore, the correct answer is Option B

3.A man ran one kilometre in the direction of the wind in three minutes and came back the same way (opposite to the direction of wind) in four minutes. If we assume that the man always puts constant force on the road, how much time he will take to run one kilometre without wind?

a) 24/7

b) 43/12

c) 13/3

d) 17/9

Explanation:

Suppose the speed of the man is x km/h and speed of wind is y km/h

Therefore, 1/(x+y)=3/60 =>x+y == 20 – 1st equation

And 1/(x-y)=4/60 => x-y== 15 – 2nd equation 

Solving both the equations, we will get x = 35/2 and y = 5/2

Now, time taken by man in absence of wind =Distance / Speed = 1 / (35/2)*60=24/7 min

Therefore, the correct answer is Option A

4. Arun invested Rs. 1,20,000 partially in two schemes. The first scheme has a rate of interest of 10 % and the second scheme has a rate of interest of 20 %. At the end of one year his income is 20,000. What is the amount that he invested on scheme having rate of interest as 10 %?

a) 60,000

b) 30,000

c) 80,000

d) 40,000

Explanation: 

Total Income = Simple Interest earned in First Scheme + Simple Interest earned in Second Scheme

Let money invested in Scheme First is “X”. Therefore, money invested in Scheme Second is “120000-X”

20000 = (X*10*1) / 100 + {(120000-X) * 20 * 1} / 100

On solving this equation. X = 40000

Therefore, Amount invested in Scheme first is 40,000. The correct answer is Option D

5. Arun bought a Scooter at 30% discount on its original price. After that, he sold the scooter at a 20% increase on the price he bought it. Find out the profit/loss percentage that he made on the original price.

a) 16 percent profit

b) 16 percent loss

c) 12 percent loss

d) 15 percent loss

Explanation:

Let Original Price = 100P

Cost price for Arun = 100P – 0.30 x 100P = 70P

Selling price for Arun = 70P + 0.20 X 70P = 84P

Now. Profit/loss on original Price = {(84P-100P) / 100P} x 100 = -16%

Therefore, he made a loss of 16 percent on the original price of the scooter. The Correct answer is Option B. 

6. The HCF and LCM of two numbers are 90 and 270 respectively, and the sum of two numbers is 180. Find the sum of reciprocal of these two numbers

a) 1/90

b) 1/180

c) 1/135

d) 1/270

Explanation:

The formulae that we know is LCM * HCF = Product of two Numbers

Let suppose that two numbers are X and Y. Thus X + Y = 180 and XY = 90 x 270 = 600

So, 1/X + 1/Y = (X + Y)/(XY) = (180/90x270) = 1/135

So, the correct answer is Option C

7. Convert 0.01187 to fraction?

a) 19/1623

b) 38/3200

c) 24/1937

d) 23/1937

Explanation:

We have to calculate for every option. The Correct answer is Option D

8. Find out the value of log (x^2/yz) + log (y^2/xz) + log (z^2/xy)?

a) 0

b) 1

c) 2

d) 3

Explanation:

Formulae: log a + log b + log c = log abc

Using this formulae, log (x^2/yz) + log (y^2/xz) + log (z^2/xy) = log (x^2 * y^2 * z^2 / x^2 * y^2 * z^2)

= log 1 => 0

Therefore, The Correct answer is Option A

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Read 112 times Last modified on Saturday, 22 September 2018 17:59
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