SSC CHSL 2023 PYQ — Question 53

SSC CHSL - 07 Aug 2023 - Shift 3 • Quantitative Aptitude • Q53

In the following passage, some words have been deleted. Read the passage carefully and select the most appropriate option to fill in each blank. When Mr. Smith went to a restaurant one day, he left his coat near the door. There was (1)________ in the pockets when he left it, so he was very surprised when he took his coat after his meal and found the (2)________ full of jewellery. There was a waiter near the door, so Mr. Smith said to him, “Somebody has made a mistake. He has put some (3)________ in my coat. Take it, and when he comes back, give it to him.” The waiter took it and went away. Suddenly another man came in with a (4)________ just like Mr. Smith’s. “I am sorry,” said the man , “I made a mistake. I took your coat and you have got mine. Please give me my coat and jewellery.” Mr. Smith answered.” I gave the jewellery to the waiter. He will give it to you.” Mr. Smith called the manager of the restaurant; but the manager said, “We have (5)________ waiters here. We have only waitresses.” “You gave the jewellery to a thief!” shouted the other man. “I shall send for the police!” Mr. Smith paid the man a lot of money for the jewellery.

Simplify: (1−m)3+(m−n)3+(n−1)312(1−m)(m−n)(n−1)\frac{(1-m)^3 +(m - n)^3+(n -1)^3}{12(1 - m)(m - n)(n - 1)}

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e.  14\frac{1}{4}.

a3 + b3 = (a + b)(a2 - ab + b2)

Using this in the given expression,

(1−m+m−n)((1−m)2−(1−m)(m−n)+(m−n)2)+(n−1)312(1−m)(m−n)(n−1)\frac{(1-m + m - n)((1 - m)^2 - (1 - m)(m - n) + (m - n)^2)+ (n -1)^3}{12(1 - m)(m - n)(n - 1)}

⇒ (1−n)(1−2m+m2−m+n+m2−mn+m2−2mn+n2)+(n−1)312(1−m)(m−n)(n−1)\frac{(1 - n)(1 - 2m + m^2 - m + n + m^2 - mn+m^2 - 2mn + n^2)+ (n -1)^3}{12(1 - m)(m - n)(n - 1)}

⇒ −(n−1)(1−3m+3m2+n−3mn+n2)+(n−1)312(1−m)(m−n)(n−1)\frac{-(n-1)(1- 3m + 3m^2+ n - 3mn + n^2)+ (n -1)^3}{12(1 - m)(m - n)(n - 1)}

⇒ −(n−1)(1−3m+3m2+n−3mn+n2−(n−1)2)12(1−m)(m−n)(n−1)\frac{-(n-1)(1- 3m + 3m^2+ n - 3mn + n^2 - (n -1)^2)}{12(1 - m)(m - n)(n - 1)}

⇒ −(n−1)(1−3m+3m2+n−3mn+n2−n2+2n−1)12(1−m)(m−n)(n−1)\frac{-(n-1)(1- 3m + 3m^2+ n - 3mn + n^2 - n^2 + 2n - 1)}{12(1 - m)(m - n)(n - 1)}

⇒ −(n−1)(−3m+3m2+3n−3mn)12(1−m)(m−n)(n−1)\frac{-(n-1)(- 3m + 3m^2+ 3n - 3mn )}{12(1 - m)(m - n)(n - 1)}

⇒ −3(n−1)(m(m−1)−n(m−1))12(1−m)(m−n)(n−1)\frac{-3(n-1)(m(m - 1)- n(m - 1))}{12(1 - m)(m - n)(n - 1)}

⇒ (n−1)(1−m)(m−n)4(1−m)(m−n)(n−1)\frac{(n-1)(1 - m)(m - n)}{4(1 - m)(m - n)(n - 1)} = 14\frac{1}{4}