TCS Aptitude Questions and Answers with Explanations

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TCS Aptitude Questions

TCS Aptitude Questions Are Will Help you to prepare TCS Interview, most of the TCS exams will have similar questions to the previous one. We made a list of few Aptitude Questions and Answers with Answer Explanations.

TCS Aptitude Questions and Answers:

1) A, B, and C can together do some work in 72 days. A and B can together do
two times as much work as C alone, and A and C together can do four times as
much work as B alone. Find the time taken by C alone to do the whole work.

a. 144 days

b. 360 days

c. 216 days

d. 180 days

Answer: 216 days

The work done by A, B and C together = A + B + C = 72 days
A + B = 2C

A + C = 4B

On solving, we get 3C = 72 days and hence C = 72*3 = 216 days

2) A and B completed certain work together in 5 days. Had A worked at twice his
own speed and B half his own speed, it would have taken them 4 days to
complete the job. How much time would it take for A alone to do the job?

a. 10 days

b. 20 days

c. 25 days

d. 15 days

Answer: 10 days

A and B can together do a work in 5 days = A + B = 1/5 days
2A + B/2 = 1/4

On solving these equations, we get A = 1/10 and hence A will take 10 days to
complete
the work all alone.

3) A sum of Rs 2387 is divided into three parts in such a way that one-fifth of the first part, one half of the second part, and the fourth one and the third part are equal. Find the sum of five times the first part, three times the second part and four times the third part (in rupees).

a. 9982

b. 7812

c. 9114

d. 10199

Answer: 10199

Let the amount be divided into three parts X, Y, and Z.
X + Y + Z = 2387

X/5 = Y/2 = Z/4 = K
X = 5K

Y = 2K
Z = 4K

Hence, 5K + 2K + 4K = 2387
11K = 2387

K = 217

5 times of 1st part + 3 times of 2nd part + 4 times of 3rd part = 5X + 3Y + 4Z

= 5(5K) + 3(2K) + 4(4K) = 5(5*217) + 3(2*217) + 4(4*217)

= 5425 + 1302 + 3472 = 10199

4) What is the greatest possible positive integer n if 16^n divides (44)^44
without leaving a remainder.

a. 14

b. 15

c. 28

d. 29

Answer: 29

5) In a test with 26 questions, five points were deducted for each wrong answer, and eight points were added for every correct answer. How many were answered correctly if the score was zero?

a. 11

b. 10

c. 13

d. 12

Answer: 10

Let the number of correct answers be y and number of wrong answers be x.
(-5)x + 8(y) = 0

x + y = 26

On solving these, we get x = 16 and y = 10

6) The air-conditioned bus service from Siruseri industry park runs at regular
intervals throughout the day. It is now 3:12 pm and it has arrived 1 minute ago but it was 2 minutes late. The next bus is due at 3:18 pm. When is the next bus due?

a. 3:27 pm

b. 3:29 pm

c. 3:24 pm

d. 3:25 pm

Answer: 3:27 pm

Time right now = 3:12 pm

Time at which the bus should have arrived = 3:09 pm
The next bus timing = 3:18 pm

The interval between 1st bus and 2nd bus = 0.09 min
so the next bus will be at = 3:18 +0.09= 3:27 pm

7) How many numbers of plates can be made if the number of plates has two letters of the English alphabets (A-Z) followed by two digits (0-9) if the repetition of digits or alphabets is not allowed?

a. 56800

b. 56500

c. 52500

d. 58500

Answer: 58500

The number of English alphabets (a-z) = 26
The number of digits (0-9) = 10

Number of ways to arrange two alphabets without repetition = 26*25
Number of ways to arrange two digits without repetition = 10*9
Number of number plates that can be made = 26*25*10*9 = 58500

8) In a cricket tournament, 16 school teams participated. A sum of Rs. 8000 is to be awarded among them as prize money. If the team placed last is awarded Rs.
275 as prize money and the award increases by the same amount for successive
finishing teams, how much will the team placed first receive?

a. 1000

b. 500

c. 1250

d. 725

Answer: 725

Let the team which got placed first receive an amount a.

Since the award money increases by the same amount for successive finishing
teams,
the series will be in AP. Let the constant amount be d.

Now, l = 275 , n = 16 and S16 = 8000

l = a + (n – 1) d and hence 275 = a + 15d

S16 = 16/2 [2a + (16 -1)(d)] and hence 8000 = 8 (2a + 15d)

On solving these equations,
275 = a + 15d

1000 = 2a + 15d

(2a + 15d) – (a + 15d) = 1000 – 275
a = 725

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9) Eesha’s father was 34 years of age when she was born. Her younger brother,
Shashank, now that he is 13, is very proud of the fact that he is as tall as her, even though he is three years younger than her. Eesha’s mother, who is shorter than Eesha, was only 29 when Shashank was born. What is the sum of the ages of Eesha’s parents now?

a. 92

b. 76

c. 66

d. 89

Answer: 92

Let Eesha’s present age be x.
Eesha’s father’s present age = x + 34
Shashank’s age = 13

Eesha’s present age = 13 + 3 = 16

Eesha’s mother’s present age = 29 + 13 = 42

Sum of the ages of Eesha’s parents now = 42 + 16 + 34 = 92

10) Fishing is a serious environmental issue. It has been determined by the
scientists that if the net of a trawler has a mesh size x cm by x (square mesh)
then the percentage of fish entering the net that is caught in the net is

(100-0.02x^2-0.05x). For example, if the mesh size is zero 100% of the fish that
enters the net will be caught.

The trawler with the net with a square mesh that was suspected of using an illegal size net dropped its net to the ocean near the damans and coast guard officials arrested the crew.
The scientists later looked at the size of the fish caught and estimated that the  net
used by the trawler at least 97.93% of the fish entering the net would be caught.

What is the maximum value of x for the net by the trawler?

a. 8.5

b.9

c. 11

d. None of the answers

Answer: 9

11) In this question, x^y stands for x raised to the power y. For example, 2^3=8
and 4^1.5=8. If a,b are real numbers such that a+b=3, a^2+b^2=7, the value of
a^4+b^4 is?

a. 49

b. 45

c. 51

d. 47

Answer: 47

12) The set A (0) is (1,2,3,4). For n > 0, A(n+1) contains all possible sums that can be obtained by adding two different numbers from what is the number of integers in  A(10). (This is an advanced question)

Answer: 67

13) Considering a hash table with 100 slots. Collisions are resolved using chaining. Assuming simple uniform hashing, what is the probability that the first 3 slots are unfilled after the first 3 insertions? (NOTE:100 ^ 3 means 100 raised to the power 3)
(This is an advanced question)

a.(97*96*95)/100^3

b.(97*96*95)/(6*100^3)

c.(97*97*97)/100^3

d. (99*98*97)/100^3

Answer: (97*97*97)/100^3

Step-by-step explanation:

A simple Uniform hashing function is a hypothetical hashing function that evenly distributes items into the slots of a hash table. Moreover, each item to be hashed has an equal probability of being placed into a slot, regardless of the other elements already placed.

Probability that the first 3 slots are unfilled after the first 3 insertions =

  • (the probability that the first item doesn’t go in any of the first 3 slots)*
  • (the probability that the second item doesn’t go in any of the first 3 slots)*
  • (the probability that the third item doesn’t go in any of the first 3 slots) = (97/100) * (97/100) * (97/100)

14) Advanced In this question x^y stands for x raised to the power y. For example 2^3=8 and 4^1.5=8. Find the number of positive integers n>2000 which can be expressed as n=2^m+2^n where m and n are integers (for example, 33=2^0+2^5)  (This is an advanced question)

Answer: 65

Step-by-step explanation:

We know that 2^10 = 1024 < 2000 < 2048 = 2^11

Since 2^10+2^9 = 1536 < 2000, any combination of 2^m + 2^n < 2000 (as long as both m or n are not 10)

So, we would have the following possibilities;

2^10 + 2^n, (n = 1,2….9)

2^9 + 2^n, (n = 1,2….8)

And so on;

Therefore, the answer is (10*11)/2 = 65.

15) A road network covers some cities. City C can be reached only from the city
A or city B. The distance from A to C is 65 km and that from B to C is 30 km. The shortest distance from A to B is 58 km. The shortest distance from city P to A is 420 km and the shortest distance from city P to B is 345 km. The shortest distance from city P  to city C in km is:

a. 153

b. 478

c. 403

d. 375

Answer: 375

Step-by-step explanation:

As we can see there are only 4 routes from city P to C

a) from P to A and then A to C

So, distance in this route will be

= 420 + 65

= 485 kms

b) From P to B and then B to C

So, distance in this route will be

= 345 + 30

= 375 kms

c) from P to A then A to B and then B to C

So, distance in this route will be

= 420 + 58 + 30

= 508 kms

d) From P to B then B to A and then A to C

So, distance in this route will be

= 345 + 58 + 65

= 468 kms

Hence we can see that the shortest distance from P to C is 375 kms via route b) i.e from P to B and then B to C.

16) In a 200 meters race, A beats B by 20 meters, while in a 100 meters race, B beats C by 5 meters. A beats C in a kilometer race by

    1. 105 meters
    2. 225 meters
    3. 205 meters
    4. 145 meters

Answer: Option D

Step-by-step explanation:

When A = 200, B = 180. When B = 100, C = 95. Make B = 900. A will be 1000. C will be 855.

TCS Aptitude Questions – FAQs

Q: The above-mentioned Questions are previously Asked in TCS Interview?

Ans: Yes, These Questions are Asked by TCS Aptitude.

Q: What is the Cut Off For TCS Aptitude Section?

Ans: TCS never reveled about cut off.

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