Tech Mahindra Aptitude Questions | Quantitative Aptitude Questions – FACE Prep

This article will discuss some Tech Mahindra Aptitude Questions that might come in the Quantitative Aptitude Section of the first round in Tech Mahindra Recruitment Process. This written Test assesses skills like analytical, problem solving and general English. Following are the parts of the Tech Mahindra Aptitude Test:

  1. Tech Mahindra Aptitude Question
  2. Tech Mahindra English Verbal Ability 
  3. Tech Mahindra Verbal Reasoning
  4. Tech Mahindra Non Verbal Reasoning

To see the marking scheme, the number of questions in each section, check the complete Tech Mahindra Test Pattern here.

You can check the complete Tech Mahindra section-wise syllabus here.

In this article, we will be focussing on some types of Tech Mahindra Aptitude Questions. Note that it is important to practice these questions before the real test. This will make sure you are comfortable and familiar with the questions. Gain more practice for Tech Mahindra Aptitude Questions – Quantitative Ability here. 

Tech Mahindra Aptitude Questions – Syllabus

Quantitative Aptitude is the kind of test that assesses a candidate’s Mathematical, Calculation, and Accuracy Skills. A good knowledge of High School Mathematics basics is enough to get through with this section. However, it is advised to gain practice in Quantitative Aptitude Questions.

The following Topics describe the complete syllabus of Tech Mahindra Aptitude Question – Quantitative Aptitude Section:

  • Time and Work
  • Time and Distance
  • L.C.M and H.C.F
  • Simple Interest
  • Areas
  • Problems on Trains
  • Boats and Streams
  • Partnership
  • Ratio and Proportion
  • Averages
  • Simple Equations
  • Quadratic Equations
  • Mixtures and Allegations
  • Percentages
  • Problems on Numbers
  • Compound Interest
  • Profit and Loss
  • Age-related Questions
  • Mensuration
  • Permutations and Combinations
  • Probability

Tech Mahindra Aptitude Questions

1) A is thrice as good a workman as B and therefore A takes 6 days less than B to finish the work individually. If A and B working together complete the work in 2 1/4 days, then how many days are required by B to complete the work alone?

  1. 8 B. 10 C. 12 D. 9

 

Answer – Option D

Since efficiency and days are inversely related, if B takes Q days to complete the work and A takes Q-6 days to complete the work, 3 x (Q-6) = Q

Solving that, we get Q = 9.



2) In a village the average age of n people is 39 years. But after the verification it was found that one person was 50 years older than his age considered in the average. So the new average, after correction increased by 1. The value of n is?

  1. 40 B. 45 C. 50 D. 55

 

Answer – Option C

It is the same as a person with 50 years more age replacing an existing person in the village.

Since the total age of the village having n persons, is being increased by 50 years, and the average age is being increased by 1 year, there are 50 people in the village.



3) Two numbers are in the ratio 5:8. If 5 is added to each, the ratio between the numbers becomes 2:3. Find the numbers.

  1. 5,8 B. 10,16 C. 20,32 D. 25, 40

 

Answer – Option D

Let the numbers be A and B

A/B = 5/8 —> A = 5B/8

(A+5) / (B+5) = 2/3 —> 3 (5B/8 + 5) = 2 (B+5)

Solving, we get B/8 = 5. —> B = 40

Thus, A = 25.



4) A  sum of money under compound interest doubles itself in 4 years. In how many years will it become 32 times itself?

  1. 16 B. 20 C. 24 D. 30

 

Answer – Option B

Let the money be P. We need 32P

P becomes 2P in 4 years

2P becomes 4P in 4 years

4P becomes 8P in 4 years

8P becomes 16P in 4 years

16P becomes 32P in 4 years

So, P becomes 32P in 20 years.



5) A took a loan at 8% per Annum simple interest for a period of 4 years. At the end of 4 years, he paid 62700 to clear his loan. How much loan did he take?

  1. 45000 B. 47500 C. 50000 D. 48750

 

Answer – Option B

Let the loan amount be = 100

Therefore amount to be repaid = 100 + (100x8x4)/100 = 132

So, if 62700 was repaid, the initial amount was 62700 x 100/132 = 47500



6) A starts from Lucknow and B starts from Kanpur to meet each other. Speed of A is 30 km/h, while the speed of B is 50 km/h. If the distance between Lucknow and Kanpur is 200 km and both  A and B start their journey at the same time, at what distance from Lucknow do they meet?

  1. 70 km B. 75 km C. 80 km D. 85 km

 

Answer – Option B

Time taken to meet each other = 200 / 50+30 = 2.5 hours.

Distance covered by A in 2.5 hours = 2.5 x 30 = 75 km from Lucknow.



7) Two runners start running around a circular track at the same time. One is at a speed of 6 km/h, and the other at 4 km/h. The former beats the latter by 100 minutes. What is the distance of the track?

  1. 18 B. 20 C. 22 D. 24

 

Answer – Option B

100 minutes = 5/3 hours

Let D be the distance.

D/4 – D/6 = 5/3

D = 20



8) The Indian cricket team played 20 one-day matches in a particular season of an year and won 25% of the matches they played. If they wanted a minimum success rate of 75%, what is the minimum number of matches they would have to play more?

  1. 30 B. 35 C. 40 D. 45

 

Answer – Option C

20 x 25/100 = 5

So far, the Indian has won 5 cricket matches. The minimum number of matches would be in the situation where India wins all their remaining matches. Let M be the number of matches required to achieve the target.

Therefore (5+M) / (20 + M) = 75/100

(5+M) / (20 + M) = 3/4

20 + 4M = 60 + 3M

M = 40



9) he forgetful merchant A gives a discount of 15% on his marked price. Forgetting that, he took his marked price itself as the selling price, and calculated his profit as 25% on the selling price. What is his actual profit percentage?

  1. 10% B. 13 1/3% C. 13 2/3% D. 13%

 

Answer – Option B

Let the marked price be 100.

At 25%, profit = 25, cost price = 75.

At 15% discount, selling price = 100x.85 = 85

Actual profit percentage = ( 85-75 / 75 ) x 100% = 13 1/3%

 

10) If 43322K is exactly divisible by 72, what is K?

  1. 4 B. 2 C. 6 D. 8

 

Answer – Option D

If 43322K is divisible by 72, then it will be divisible by both 8 and 9

To check divisibility of 8, we have to check the last three digits (as all 000s are divisible by 8)

224 is the only one divisible by 8.



11) Find the smallest number 4 digit number, when divided by 5, 8 and 12 leaves a remainder of 3 in each case.

  1. 1203 B. 1083 C. 1233 D. 1003

 

Answer – Option B

The smallest such number will be a multiple of the LCM of 5, 8 and 12, +3 to give the remainder.

LCM of 5, 8 and 12;

5 = 5

8 = 2^3

12 = 2^2 x 3

 

LCM = 2^3 x 3 x 5 = 120.

 

Smallest 4 digit multiple of 120 = 1080

Number = 1083



12) Find the last 2 digits of 91^75

  1. 11 B. 31 C. 51 D. 71

 

Answer – Option C

For 1, the last digit if any power will always be 1.

To find the second last digit, all we need to do is to take the last digit of 5 (last digit of the power) x 9 (second last digit of the base) = 5

Therefore, last 2 digits = 51.



13) In how many ways can the letters ‘A’, ‘B’, ‘C’, ‘E’, ‘I’, ‘Z’ can be arranged so that the consonants always occupy the odd places?

  1. 18 B. 36 C. 54 D. 7

 

Answer – Option B
The three vowels can be arranged in 3! Ways

The three consonants can be arranged in 3! ways.

Total = 3! x 3! = 36



14) There are six boxes, B1 – B6, and 6 caps, C1 – C6. Each box can hold upto 6 caps. If each box can have any number caps, in how many arrangements is C1 in B1.

  1. 7676 B. 7556 C. 7776 D. 7667

 

Answer – Option C

Since we fix C1 in B1, the other 5 caps can go in any of the 6 boxes.

= 6 x 6 x 6 x 6 x 6 = 7776.

 

15) The sum of three numbers in AP is 12, and the sum of their cubes is 288. Find the product of the three numbers.

  1. 60 B.48 C. 28 D. 64

 

Answer – Option B

Let the three consecutive numbers in A.P be p, q, r such that

p = (q-d) and r = (q+d)

p+q+r = 12,

q = 4

p^3 + q^3 + r^3 = 288

p^3 + r^3 = 224

p + r = 8

(using trial and error)

p = 2, q = 6

2 x 4 x 6 = 48

 

How can FACE help you clear Tech Mahindra Placements?

FACE Prep offers a course to prepare for Tech Mahindra recruitment. This course covers a lot of previous year aptitude questions.
In Tech Mahindra’s Online Aptitude Test, a candidate is expected to solve 75 questions in 50 minutes on Quantitative Aptitude, Verbal Ability and Logical Reasoning. With an average of only 40 seconds available for reading, understanding and solving each question, it is essential that the candidate is thoroughly prepared for this high-intensity test.
This is where this online course developed by India’s largest placement training organization, FACE, will help. Also up to 50% of the questions in the Tech Mahindra Aptitude round appear from previous year test papers. By subscribing to this course, you will get access to the well-explained solutions for questions that have appeared in yesteryear’s test papers. You can check the course here – 
 
 
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