This is a TCS model placement paper for aptitude preparation. This placement paper will cover aptitude questions that are asked in TCS recruitment drives and also strictly follows the pattern of questions asked in TCS interviews. It is recommended to solve each one of the following questions to increase your chances of clearing the TCS interview.

- Identify the missing number in the series: 2, 5, __, 19, 37, 75?

a) 16

b) 12

c) 10

d) 9**Answer:**d) 9**Solution:**

2 * 2 + 1 = 5

5 * 2 – 1 = 9

9 * 2 + 1 = 19

19 * 2 – 1 = 37 and so on - A rectangle is divided into four rectangles with area 70, 36, 20, and x. What is the value of ‘x’?

a) 350/7

b) 350/11

c) 350/9

d) 350/13**Answer:**c) 350/9**Solution:**

Since the areas of the rectangles are in proportion we can say,

=> 70/x = 36/20

=> x = 350/9

- If VXUPLVH is written as SURMISE, what is SHDVD written as?

a) PEASA

b) PBASA

c) PEBSB

d) None of the above**Answer:**a) PEASA**Solution:**

It is a question of coding-decoding where,

V is written as S (V – 3 = S)

X is written as U (X – 3 = U)

and so on.

Similarly, SHDVD will be written as PEASA - Aman owes Bipul Rs 50. He agrees to pay Bipul over a number of the consecutive day starting on Monday, paying a single note of Rs 10 or Rs 20 on each day. In how many different ways can Aman repay Bipul. (Two ways are said to be different if at least one day, a note of a different denomination is given)

a) 5

b) 6

c) 7

d) 8**Answer:**d) 8**Solution:**

Aman can pay Bipul in all 10 rupees note in 5 days = 5 * 10 = 50 rupees = 1 way

Aman can pay Bipul in 3 ten rupee note and 1 twenty rupee note = 4!/(3! * 1!) = 4 ways

Aman can pay Bipul in 1 ten rupee note and 2 twenty rupee note = 3!/(1! * 2!) = 3 ways

So in all Aman can pay Bipul in 8 ways. - Salim bought a certain number of oranges at a rate of 27 oranges for rupees 2 times M, where M is an integer. He divided these oranges into two equal halves, one part of which he sold at the rate of 13 oranges for Rs M and the other at the rate of 14 oranges for Rs M. He spent and received an integral no of rupees, but bought the least number of oranges. How many did he buy?

a) 980

b) 9828

c) 1880

d) 102660**Answer:**b) 9828**Solution:**

Let Salim buy 2x number of oranges.

So he buys 27 oranges at a price of 2M.

He buys 1 orange at a price of 2M/27

or, x oranges cost him Rs. 2Mx/27

Now he sells x oranges at the rate of 13 oranges for Rs. M

So he sells 1 orange at Rs. M/13

and x oranges at Rs Mx/13

Same goes for 14 oranges which are Mx/14,

According to the question, 2Mx/27, Mx/13, Mx/14 are integers

So, x oranges must be divisible by 27, 13 and 14

The lcm of 27, 13 and 14 = 4914 or 2x = 9828 - In a football match, 16 teams participate and are divided into 4 groups. Every team from each group will play with each other once. The top 2 winning teams will move to the next round and so on the top two teams will play the final match. So how many minimum matches will be played in that tournament?

a) 40

b) 14

c) 43

d) 50**Answer:**c) 43**Solution:**

Total matches to be played = 4C2 = 6 matches.

So total number of matches played in the first round = 6 * 4 = 24 matches

Now top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.

Total matches played in the second round = 6 × 2 = 12 matches

Now 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matches

From this round, 2 teams progress to the next round. And final will be played between them.

Total matches = 24 + 12 + 6 + 1 = 43 - There are 12 letters and exactly 12 envelopes. There is one letter to be inserted randomly into each envelope. What is the probability that exactly 1 letter is inserted in an improper envelope?

a) 1

b) 0

c) 10!

d) None of these**Answer:**b) 0**Solution:**

This is a question of very common sense in which,

12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0. - A hollow space on the earth surface is to be filled. The total cost of filling is Rs. 20000. The cost of filling per cubic-meter is Rs 225. How many times is a size of 3 cubic-meter soil required to fill the hollow space?

a) 29.62

b) 30.32

c) 88.88

d) 43.64**Answer:**a) 29.62**Solution:**

The total cost of filling = 20, 000

Cost of filling 1 cubic meter = Rs. 225

So cubic meters to be filled = 20, 000/225 = 88.89 meter-cube

Now we need to find the three times of 88.89 to be filled = 88.89/3 = 29.63

So the closest match is 29.62 - A 7-digit number is to be formed with all different digits. If the digits at the extreme right and extreme left are fixed to 5 and 6 respectively, find how many such numbers can be formed?

a) 120

b) 30240

c) 6720

d) None of these**Answer:**c) 6720**Solution:**

If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8

So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways

= 6720 ways - There are five tires in a sedan (four road tires and one spare) which is to be used equally in a journey to travel 40, 000 km. The number of km of use of each tyre was

a) 32000

b) 8000

c) 4000

d) 10000**Answer:**a) 32000**Solution:**

The total km travelled by the sedan = 40, 000 km

Since every tire capacity’s = 40, 000/5 = 8000 km each

So total distance covered by each tire = 8000*4 = 32000 km each will be travelled by each tire after being worn out after every 8000 km.

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