# Placement Questions & Answers :: TCS

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231.

What is the next number?

11, 23, 47, 83, 131, ....

**Answer:**** Option D **

**Explanation:**

23â€“11 = 12

47â€“23 = 24

83â€“47 = 36

131â€“83 = 48

Therefore, 131+60=191

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232. A sum of Rs.3000 is distributed among A, B, and C. A gets 2/3 of what B and C got together and c gets 1/3 of what A and B got together, C's share is?

**Answer:**** Option A **

**Explanation:**

As A+B+C=3000. And A=2/3(B+C).

Also C=1/3(A+B).

So, 3C=A+B.

Hence now we can substitute A+B by 3C.

So final equation will be 3C+C=3000.

Hence , 4C=3000 => C=750.

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233.

### On door A - It leads to freedom

On door B - It leads to Ghost house

On door C - door B leads to Ghost house

The statement written on one of the doors is wrong.

Identify which door leads to freedom.

**Answer:**** Option C **

**Explanation:**

Case 1: A, B are true. In this case, Statement C also correct. So contradiction.

Case 2: B, C are true. In this case, B leads to ghost house and C confirms it. Now A is wrong. So door A does not lead to freedom. So Door C leads to freedom.

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234. What is the probability of getting sum 3 or 4 when 2 dice are rolled

**Answer:**** Option B **

**Explanation:**

Required number of ways of sum 3 or 4 = (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5

Total ways = 6^2 = 36

Probability = 5/36

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235. a bb ccc dddd eeeee .........What is the 120th letter?

**Answer:**** Option D **

**Explanation:**

Number of letters in each term are 1, 2, 3, 4, 5,….

The above number are in series of A.P.

So n*(n+1)/2 ? 120

For n = 15 above eq. condition will be true.

15th term contains 15 O's.

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236. Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300 (Hint 44^2=1936)

**Answer:**** Option A **

**Explanation:**

The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.

Given that 44^2 = 1936.

Shortcut: To find the next perfect square, add 45th odd number to 44^2.

So 45^2 = 1936 + (2 x 45 -1) = 2025

46^2 = 2025 + (2 x 46 - 1) = 2116

47^2 = 2116 + (2 x 47 - 1) = 2209

Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.

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237. What is in the 200th position of 1234 12344 123444 1234444....?

**Answer:**** Option C **

**Explanation:**

Here is no explanation for this answer

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238.

What is the next number in the given series?

70, 54, 45, 41,.....

**Answer:**** Option D **

**Explanation:**

Consecutive squares are subtracted from the numbers.

70 - 54 = 16

54 - 45 = 9

45 - 41 = 4

So next we have to subtract 1.

So answer = 41 - 1 = 40

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239. The sticks of the same length are used to form a triangle as shown below. If 87 such sticks are used then how many triangles can be formed?

**Answer:**** Option B **

**Explanation:**

First triangle is formed by using 3 sticks, but any subsequent triangle may be formed by using 2 sticks.

Therefore, If 1st triangles uses 3 sticks,

Remaining sticks = 87 - 3 = 84.Â With these 84, we can form 42 triangles. So total = 42 + 1 = 43

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240. The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?

**Answer:**** Option A **

**Explanation:**

From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.

We have to find the HCF of the given numbers 272725, 232325.

HCF = 25.

So sum of the digits = 7.

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Here is the list of questions asked in TCS assessment test questions TCS aptitude test papers 2016. Practice TCS Written Test Papers with Solutions and take Q4Interview TCS Online Test Questions to crack TCS written round test. Overall the level of the TCS Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of TCS