RRB NTPC CBT 2 Level 6 May-2-2022 Shift 1 Exam Previous Question Paper with Solutions

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111) For which novel did Valeria Luiselli win the International Dublin Award 2021?

A) Lost Children Archive
B) The Story of My Teeth
C) Sidewalks
D) Faces in the Crowd

View Answer

A) Lost Children Archive

112) In Vaishnavism, how many avatars or incarnation of deity were recognised?

A) 5
B) 7
C) 9
D) 10

View Answer

D) 10

113) The outer radius of a spherical shell is 9 cm and the thickness of the shell is 1 cm. Find the volume of the metal used for the shell (in cubic cm).
(Use π = $\frac{22}7$ )

A) $912\frac23$
B) $915\frac13$
C) $909\frac13$
D) $909\frac35$

View Answer

C) $909\frac13$

Explanation:Given:
– Outer radius of the spherical shell, ( R = 9 \, \text{cm} )
– Thickness of the shell, ( t = 1 cm )
– Inner radius of the spherical shell, ( r = R – t = 9 – 1 = 8 cm )
We are asked to find the volume of the metal used for the shell.
Step 1: Formula for the volume of a spherical shell
The volume of a spherical shell is given by the difference in the volumes of two spheres: one with the outer radius ( R ) and the other with the inner radius ( r ).
The formula for the volume ( V ) of a sphere with radius ( r ) is:
$V = \frac{4}{3} \pi r^3$
So, the volume of the spherical shell is:
$V_{\text{shell}} = \frac{4}{3} \pi (R^3 - r^3)$
Step 2: Substitute the values
We know that ( R = 9 cm ) and ( r = 8 cm ). Also, we are given $( \pi = \frac{22}{7} )$ .
Substitute the values into the formula for the volume of the shell:
$V_{\text{shell}} = \frac{4}{3} × \frac{22}{7} × \left(9^3 - 8^3\right)$
Step 3: Calculate the cubes of the radii
First, calculate the cubes of ( R ) and ( r ):
$9^3 = 729, \quad 8^3 = 512$
Now, substitute these values into the volume formula:
$V_{\text{shell}} = \frac{4}{3} × \frac{22}{7} × (729 - 512)$
$V_{\text{shell}} = \frac{4}{3} × \frac{22}{7} × 217$
Step 4: Simplify the expression
First, simplify the product:
$\frac{4}{3} × \frac{22}{7} = \frac{88}{21}$
Now, calculate the volume:
$V_{\text{shell}} = \frac{88}{21} × 217 = \frac{88 × 217}{21} = \frac{19096}{21} = 909 \frac{1}{3} \, \text{cubic cm}$
Final Answer:
The volume of the metal used for the shell is 909 $( \frac{1}{3} )$ cubic cm.

114) Eight colleagues K, L, M, N, O, P, Q and R are seated in a circle facing the centre. N is an immediate neighbour of both L and Q. P is an immediate neighbour of both K and R. O is second to the right of K. Which of the following is definitely true about M’s position?

A) Second to the left of L
B) Exactly between K and O
C) Exactly between Q and O
D) Exactly between O and P

View Answer

B) Exactly between K and O

115) The average of 30 numbers is 55. The average of the first 15 numbers is 51 and the average of the last 12 numbers is 56. What is the average of 68, 74 and the remaining 3 numbers?

A) 71
B) 70
C) 72.2
D) 7

View Answer

A) 71

Explanation:We are given the following information:
– The average of 30 numbers is 55.
– The average of the first 15 numbers is 51.
– The average of the last 12 numbers is 56.
We need to find the average of the numbers 68, 74, and the remaining 3 numbers.
Step-by-Step Solution:
Let the 30 numbers be denoted as $( x_1, x_2, \dots, x_{30} )$ .
»1. Total sum of 30 numbers:
The average of the 30 numbers is 55. Therefore, the total sum of the 30 numbers is:
Total sum = 30 × 55 = 1650
»2. Sum of the first 15 numbers:
The average of the first 15 numbers is 51. Therefore, the sum of the first 15 numbers is:
Sum of the first 15 numbers} = 15 × 51 = 765
»3. Sum of the last 12 numbers:
The average of the last 12 numbers is 56. Therefore, the sum of the last 12 numbers is:
Sum of the last 12 numbers = 12 × 56 = 672
»4. Sum of the remaining 3 numbers:
The sum of the remaining 3 numbers can be found by subtracting the sum of the first 15 numbers and the sum of the last 12 numbers from the total sum of the 30 numbers:
Sum of the remaining 3 numbers} = 1650 – 765 – 672 = 213
»5. Average of 68, 74, and the remaining 3 numbers:
The numbers we are interested in are 68, 74, and the remaining 3 numbers. The sum of these 5 numbers is:
68 + 74 + 213 = 355
The average of these 5 numbers is:
Average = $\frac{355}{5}$ = 71

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