TCS aptitude questions with solutions are given here. TCS aptitude test questions are of medium to high difficulty level. A good practice is a must to solve them faster. This article will give you an idea about what type of questions are asked ad how to answer them within no time.

Table of Contents

**TCS Aptitude Questions Syllabus**

The syllabus for TCS aptitude questions is given below. Questions in the recent TCS on-campus drive (TCS Ninja) and TCS off campus drive were asked from these topics.

Topic | Number of Questions | Difficulty Level |
---|---|---|

Number system | 2 – 3 | Easy |

HCF & LCM | 1 – 2 | Easy |

Time, Speed & Distance | 2 – 3 | Medium |

Mixtures & Allegations | 2 – 3 | Medium |

Time & Work | 2 – 3 | Medium – Difficult |

Percentages | 2 – 3 | Medium – Difficult |

Permutations & Combinations | 1 – 2 | Medium |

Profit & Loss | 1- 2 | Medium – Difficult |

Functions & Equations | 1- 2 | Medium – Difficult |

Series & Progression | 1 – 2 | Easy – medium |

Blood Relations | 1 – 2 | Easy |

Averages | 1 – 2 | Easy |

Geometry | 1 – 2 | Easy |

Clocks & Calendars | 1 – 2 | Easy – Medium |

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**TCS Aptitude Questions with Solutions**

TCS aptitude questions with solutions are given here. These are the questions asked in the recent TCS drive 2018. Practicing them will give you a better idea about the question types asked and will also help you solve more number of questions appropriately.

**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 next bus will be at = 3:18 +0.09= 3:27 pm

**7) How many number plates can be made if the number plates have 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 S_{16} = 8000

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

S_{16} = 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

**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 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

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

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 kms is:

a. 153

b. 478

c. 403

d. 375

**Answer: **375

Here are more aptitude questions for you to solve and practice.

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**TCS Aptitude Questions: Preparation Strategy**

- Brush up on all the concepts specified in the TCS aptitude questions syllabus.
- Practice at least 10 – 20 TCS aptitude questions per topic. This increases your exposure in the topic and also helps you solve faster.
- Learn some shortcuts and time-saving tips and tricks. This will help you choose the easy questions first and solve faster thereby giving more time to solve harder questions.

Duration | No of questions | Question type | |

Quantitative Aptitude | 40 mins | 20 | MCQ and FIB |

**Skills Required:** Basic problem-solving capability and good mathematical conceptual understanding.

Know more about TCS Off Campus Drive Exam Pattern and Syllabus here.

**TCS Aptitude Questions – FAQs**

**1) What type of questions come in TCS aptitude test?**

TCS aptitude test consists of different types of questions. The important topics being profit and loss, time, speed and distance, probability etc. If you can practice the above-given questions, then you will have a good idea of the type of questions asked.

**2) What is the cutoff for TCS aptitude section?**

The cutoff for TCS aptitude section is not revealed by TCS officially. But according to students who have given the exam in the past, the quants section is said to have a higher cutoff ( 12-16 questions) than the other sections of the test.

**3) How to crack TCS aptitude test?**

The best way to approach TCS aptitude test is by solving TCS previous year questions.