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# MindTree Aptitude Questions for Mindtree Aptitude Test 2019

Published on 06 Nov 2019

MindTree has started using the AMCAT Test for their recruitment process, and you can use our free MindTree Aptitude Questions to prepare for the exam. To know more about the career opportunities at MindTree, click here. Aspirants should be aware of the latest test pattern before attending the Mindtree recruitment process.

## MindTree Aptitude Questions Syllabus

Mindtree online test consists of the following sections:

MindTree Aptitude Questions syllabus is given here. You can expect questions from these topics.

TopicsDifficultyExpected No. of Questions
LCM & HCFMedium2-3
Numbers, decimal fractions and power Medium2
DivisibilityMedium2-3
InverseMedium1-2
Simple and Compound InterestMedium2
Time, Speed and DistanceMedium3
Profit and LossMedium3
LogarithmsMedium2
ProbabilityHard2-3
Permutation and CombinationsHard3-4

## Mindtree Aptitude Questions with Answers

Attempt the free MindTree Aptitude questions given below. These are taken from previous year Mindtree placement papers. Hence they would give you a good idea of the question types.

### Mindtree Aptitude questions set -1

1. A cricketer played 80 innings and scored an average of 99 runs. His score in the last inning was zero run. To have an average of 100 at the end, his score in the last innings should have been?

a) 60

b) 80

c) 10

d) 1

Ans: Option b

Let S be the score in the last inning to make the average of 100.

Sum = (Average*N). When average is 99, total score of 80 innings = 80 × 99 = 7920,

If the average is 100, the total score of 80 innings will be 80 × 100 = 8000, in order to get the average 100, the sum should be 8000. So the difference between the sum will be the run he has to score in the last innings. Therefore, S = 8000 – 7920 = 80

2. What is the largest number which divides 150, 180 and 144 leaving the same remainder in each case?

a) 12

b) 6

c) 3

d) 8

Ans: Option b

The largest number which divides x, y, z and leaves the same remainder in each case is

HCF ((x-y), (y-z), (x-z)) → HCF ((180-150), (180-144), (150-144))

HCF (30,36,6) = 6

3. Which of the following will not be the unit digit of a perfect square of a number?

a) 1

b) 6

c) 0

d) 2

Ans: Option d

No perfect square has 2,3,7,8 as the unit digit (Last digit). From the given options, the answer will be 2.

4. What is the remainder if 2^81 is divided by 5?

a) 1

b) 2

c) 3

d) 4

Ans: Option b

Raise the power of 2 so that the answer should be near to multiple of 5.

24 = 16, (16/5) ➔ R (1)

R (281/5) = R ((280 * 2)/ 5)

R ([(24)20] * 2/5) = R ((24)20/5) * R (2/5)

= 120 * 2 = 1 * 2 = 2

5. A man sold a plot at 6% profit. If he had sold it at 10% profit he would receive Rs. 200 more. What is the selling price of the plot in Rs?

a) 5000

b) 5300

c) 5700

d) 6000

Ans: Option b

If he sold it at 10% profit, the extra amount gained is Rs. 200.

That extra 4% profit is Rs.200, 4% of CP = Rs.200

Multiplying 25 on both sides, 100% of CP = Rs. 5000. Finally, the CP of Plot = Rs.5000.

Profit is 6% = 6% of CP = Rs.300, SP of the Plot = CP of the plot + Profit earned

SP = Rs.5000 + Rs.300 = Rs.5300

6. A and B are two integers differ by 4. The sum of their reciprocals is 1/2. Find the integers.

a) 2, 8

b) 3, 2

c) 8, 2

d) None

Ans: Option d

Given that difference between two integers A and B is 4.i.e., A – B = 4

Sum of their reciprocals = (1/A + 1/B) = (1/2)

Since A – B = 4, A = 4 + B. We have,

(1/A) + (1/B) = (1/2) → [1/ (4+B)] + (1/B) = (1/2)

(B + 4 + B)/B (4 + B) = (1/2) → (4 + 2B)/ (4B + B2) = (1/2)

2(4 + 2B) = 4B + B2 → 8 + 4B = 4B + B2 → 4B+B2= 8+4B → B2= 8 → B = sqrt (8)

Since we are getting one value (B) is sqrt (8) which is not an integer. Answer is none of these.

7. If 30 Oxen can plough 1/7th of a field in 2 days, then how many days 18 oxen will take to do the remaining?

a) 10

b) 20

c) 30

d) 40

Ans: Option b

Completed work = (1/7) and the remaining work = (6/7)

Let x be the number of days required to complete the remaining work.

[30 * 2] / (1/5) = [18 * x] / (6/7)

X = 20

8. A ball is dropped from a height of 10 feet. Every time it re-bounces to half of the height when it reaches the ground. How many feet would the ball have traveled before it comes to rest?

a) 25

b) 35

c) 30

d) 50

Ans: 3

The ball is dropped from a height of 10 feet. When it reaches the ground it bounces back to half of the distance (5 feet) and it will drop to ground from a height of 5 feet.

Total distance travelled is 10 + (5+5) + (2.5+2.5) + (1.25+1.25) +…   = 10 + 10 + 5 + 2.5 + 1.25 + .625…

Except the first term 10, the remaining terms are forming Geometric Progression (G.P). The total distance travelled by the ball is nothing but 10 + sum of the terms in G.P.

Sum of GP infinite series (Since we don’t know the number of terms in series) is:

Sn = a / (1 – r)

Common ration, r → Second term/ First term = Third term/ Second term

(5/ 10) = (2.5/ 5) = 0.5

Sn = 10 / 1-0.5 = 10 / 0.5

Sn = 20 feet

Total distance travelled = 20 + 10 = 30 feet.

(10 feet – initial distance)

9. When n is divided by 6, the remainder is 4. When 2n is divided by 6, then what will be the remainder?

a) 2

b) 0

c) 1

d) 4

Ans: Option a

It was given that when n is divided by 6, the remainder is 4. Hence, the number can be written as n = 6k +4, where ‘k’ is the quotient. Hence, 2n = 12k + 8 When 2n is divided by 6, the remainder will be

R{ (12k + 8)/6} = 0 + 2 = 2 [Because when divided by 6, 12k gives remainder 0 and 8 gives remainder 2 So, the overall remainder will be 0 + 2 = 2]

10. A reduction of 20% in the price of rice enables a customer to purchase 12.5 kg more for Rs. 800. The original price of rice per kg is:

a) Rs. 16

b) Rs. 12

c) Rs. 14

d) Rs. 15

Ans: Option a

Let the price of the rice is X, the quantity of the rice is Y.

So, the total amount is XY = 800 => Y = (800/X), because of the reduction of 20% in the price of rice a customer was able to purchase 12.5 kg more for Rs. 800=> (0.8X)(Y+12.5) = 800 = (Y+12.5) = (800/0.8X).

The difference between them is, (Y+12.5) – Y = (800/0.8X) – (800/X), On solving X = 16. So, the original price of the rice, X is Rs 16.

### Mindtree Aptitude questions set - 2

11. In two alloys A and B, the ratio of zinc to tin is 5: 2 and 3: 4 respectively. Seven kg of the alloy A and 21 kg of the alloy B are mixed together to form a new alloy. What will be the ratio of zinc and tin in the new alloy?

a) 2:1

b) 1:2

c) 1:1

d) 2:3

Ans: Option c

Alloy A:

Alloy A has zinc and tin in the ratio of 5:2.

So in 7 kg of the alloy, the amount of zinc and tin will be 5 kg and 2 kg respectively. {Splitting 7 kg in the ratio of 5:2}

Alloy B:

Alloy B has zinc and tin in the ratio of 3:4.

So in 21 kg of the alloy, amount of zinc and tin will be 9 kg and 12 kg respectively. {Splitting 21 kg in the ratio of 3:4}

Now in the new alloy, Amount of zinc = (5+9) = 14 kg

Amount of tin = (2+12) = 14 kg, Ratio of zinc and tin in the new alloy is 14 : 14 = 1:1.

12. There are three sections of a class in a school. The number of students in the three sections is 38, 42 and 40 and the average age of the students in these sections separately is 15.3 years, 16.5 years and 15.9 years respectively. What is the average age of the class?

a) 15.22 years

b) 15.92 years

c) 15 years

d) 15.11 years

Ans: Option b

We know that (Sum of students / No of students) = Average age of section, so we can find the sum of ages of section = (average of section * no of students) in the respective classes

38*15.3= 581.4 → 42*16.5= 693 → 40*15.9 = 636, then to find the average of entire class:

(Total sum of ages/Total sum of students) = (58.1+ 693 + 636)/ (38 + 42 + 40) = 1910.4/120

The average age of class = 15.92 years

13. How many numbers between 100 and 300, that starts and ends with 2?

a) 13

b) 12

c) 10

d) 11

Ans: Option c

Number’s starting and ending with 2 between 100 and 300 are:

No Numbers will start with 2 in the range of 100 to 199. So we can eliminate this and the numbers starting and ending with 2 are 202, 212, 222, 232, 242, 252, 262, 272, 282, and 292, so totally 10.

14. If on an item a company gives 25% discount, they earn 25% profit. If they now give 10% discount then what is the profit percentage?

a) 40%

b) 55%

c) 35%

d) 30%

Ans: Option d

Let the Market Price be Rs. X, after giving 25% discount, it becomes 0.75x

SP = 0.75x which gives 25% profit

After giving 10% discount, it becomes 0.90x → SP = 0.90x

0.75x → 25%

0.90x → ?

= 25 * 0.90x/ 0.75x = 30%

15. If a man buys 1 liter of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain?

a) 50%

b) 25%

c) 20%

d) 28%

Ans: 2

CP=Rs. 12, SP = Rs.15, since water is of no cost, the cost price will be the same.

Now profit = SP – CP = 15 – 12 = Rs.3

Now profit% = (Profit/ CP) * 100 = 3/12 * 100 = 25%

16. If prices are increased by 25%, by how much should I reduce the consumption to keep the expenditure the same?

a) 25%

b) 30%

c) 15%

d) 20%

Ans: 4

Expenditure = Price*Consumption

Price increases imply Consumption should decrease in order to keep the expenditure same.

We know that, percentage increase and decrease concept:

[1/ n] increases = [1/ (n+1)] decreases

So, (1/4) increases → then (1/ (4+1)) = 1/5 decreases = 20%

17. A 31″ x 31″ square metal plate needs to be fixed by a carpenter on to a wooden board. The carpenter uses nails all along the edges of the square such that there are 32 nails on each side of the square. Each nail is at the same distance from the neighboring nails. How many nails does the carpenter use?

a) 120

b) 128

c) 124

d) 130

Ans: 3

Here, it is given that the side of the square is of 31. And each side has 32 nails.

For 32 nails to fit on 31 unit space, the distance between each nail should be 1 unit. So, if we apply the above concept, the two opposite sides of the square will contain 32 nails and rest two sides will contain 30 nails.

So, total number of nails= 32+32+30+30=124 nails.

18. How many factors does 6400 have?

a) 24

b) 25

c) 26

d) 27

Ans: 4

Step 1: prime factorize the given number 65 → 64 x 100 = 82 x 102 = (23)2  x (2*5)2 = 26 x 22 x 52

6400 = 28 x 52

Step 2: Add one to the powers → 2(8+1) x 5(2+1) → 29 x 53

Step 3: Multiply the powers: 9 x 3 = 27

Hence 6400 has 27 factors.

19. Which of the following never comes at the end of a perfect square number?

a) 1

b) 0

c) 2

d) 6

Ans: 3

We can figure it out using unit digit multiplication itself since only unit digit is asked.

12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100

It is evident that 2 never occur at the end and hence the answer is 2.

20. By selling an umbrella for Rs. 30, a shopkeeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% of the marked price. His gain during the sale season is ______

a) 8%

b) 9%

c) 7%

d) 7.5%

Ans: a

Given, selling price of an umbrella = Rs.30

Profit percentage = 20%. Therefore, Cost price of an umbrella = (30×100)/120 = Rs.25

During the clearance sale, selling price of an umbrella = (30×90)/100 = Rs.27

Therefore, required profit percentage = (27−25)/25×100 = 8%  