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# TCS Placement Paper | MCQ 6

• Difficulty Level : Medium
• Last Updated : 11 Jan, 2023

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. 1. Crusoe hatched from a mysterious egg discovered by Angus was growing at a fast pace that Angus had to move it from home to the lake. Given the weights of Crusoe in its first weeks of birth as 5, 15, 30, 135, 405, 1215, 3645. Find the odd weight out.
a) 3645
b) 135
c) 15
d) 30

`Answer: d) 30`

Solution:
Looking at the series closely we find that the 3rd number is oddly placed.
The series is in the form:
5 * 3 = 15
15 * 3 = 45
45 * 3 = 135
135 * 3 = 405 and so on

2. Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?
a) 5436
b) 4831
c) 5508
d) 4832

`Answer: d) 4832`

Solution:
We need to use f(1) to calculate the value of f(17)
f(17) can be written as f(1+16)
f(16) can be written as f(8+8)
f(8) can be written as f(4+4)
f(4) can be written as f(2+2)
f(2) can be written as f(1+1)
f(1) = 0, so f(2) = f(1+1) = f(1)+f(1)+4(9*1*1-1) = 32.
or, f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204.
or, f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
or, f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
or, f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832

3. A sum of Rs.3000 is distributed among P, Q, and R. P gets 2/3 of what Q and R got together and R gets 1/3 of what P and Q got together, R’s share is?
a) 750
b) 850
c) 800
d) 700

`Answer: a) 750`

Solution:
According to the question,
case 1: P = 2(Q + R)/3
or, (Q+R)/P = 3/2
case 2: Also, R = (P+Q)/3
or, (P+Q)/R = 3/1
Simply using componendo-dividendo, we get,
for case 1, (P+Q+R)/P = 3+2/2 = 5/2 = 20/8
for case 2, (P+Q+R)/R = 3+1/1 = 4/1 = 20/5
On solving we get, P = 8, Q = 7, R = 5
or R’s share = 5/(8+7+5) * 3000 = 750

4. In the given series 11, 23, 47, 83, 131, … What is the next number?
a) 145
b) 178
c) 191
d) 176

`Answer: c) 191`

Solution:

The given series follows the order of multiple of 12
23 – 11 = 12
47 – 23 = 24
83 – 47 = 36
131 – 83 = 48
x – 131 = 60
or x = 191

5. If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?
a) 9
b) 3
c) 7
d) 5

`Answer: d) 5 `

Solution:
Let the number be N when divided by 357 leaves remainder 5 and quotient q.
So, N = 357k + 5 = 17 * 21 * k + 5
So, 357 is exactly divisible by 17 so remainder is 5

6. A pole of height 36m is on one edge of a road broke at a certain height. It fell in such a way that the top of the pole touches the other edge of the road. If the breadth of the road is 12m, then what is the height at which the pole broke?
a) 12
b) 16
c) 24
d) 18

`Answer: b) 16`

Solution:
Let the point at which the pole broke be ‘x’ from the ground, so the length of the broken piece be (36-x).
So applying Pythagoras theorem we get, => => 72x = 1296 – 144
=> x = 16

7. There is a hall consisting of 23 people. They are shaking hands together. So how many hands shakes possible if they are in a pair of cyclic sequence?
a) 23
b) 22
c) 253
d) 250

`Answer: c) 253`

Solution:
Since there are 23 people, number of handshakes possible = 23C2 = 253 handshakes.

8. In a basement, there are some bicycles and cars. On Tuesday there are 182 wheels in the basement. How many bicycles are there?
a) 20
b) 19
c) 18
d) 16

`Answer: b) 19`

Solution:
This is a very ambiguous question and must be calculated using the options.
If there are 20 bicycles, there must be 20*2 = 40 wheels
Remaining wheels = 182-40 = 142 wheels = 142/4 is not an integer so there cannot be 20 bicycles.
Similarly checking for 19 bicycles = 19*2 = 38 wheels
Remaining wheels = 182 – 38 = 144 = 144/4 = 36 cars hence this is the answer.

9. There is a rectangular ground 17 × 8 m surrounded by a 1.5 m width path. The depth of the path is 12 cm. Sand is filled and find the quantity of sand required.
a) 5.5
b) 10.08
c) 6.05
d) 7.05

`Answer: b) 10.08`

Solution:
Area of the inner rectangle = 17 * 8 = 136 meter-square
Area of the outer rectangle = (17 + 2*1.5) * (8 * 2*1.5) = 220 meter-square
So area of the remaining path = 220 – 136 = 84 meter-square
So sand required to fill the path = 84 * (12/100) = 10.08 meter-square

10. 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?
a) 5
b) 4
c) 7
d) 8

`Answer: c) 7`

Solution:
So according to the question, (272738 – 13) and (232342 – 17) are exactly divisible by n.
So if we find the HCF of these two numbers, we get n,
The HCF of 272725 and 232325 is 25
So the sum of the digits = 7.

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