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SSC CHSL 2023 PYQ — Question 72

SSC CHSL - 07 Aug 2023 - Shift 3 • Quantitative Aptitude • Q72

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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.

If x+1x=−1x + \frac{1}{x}=-1 , then the value of x15+1x15x^{15}+\frac{1}{x^{15}} will be:

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Correct answer: B

Explanation

The correct answer is option 2 i.e. 2.

Given, x+1x=−1x + \frac{1}{x}=-1

Squarying both sides,

(x+1x)2=1(x + \frac{1}{x})^2= 1

⇒ x2+1x2+2x^2 + {1 \over x^2} + 2 = 1

⇒ x2+1x2x^2 + {1 \over x^2} = -1    ...(i)

Now, taking cube of the given expression,

(x+1x)3=−1(x + \frac{1}{x})^3=-1

⇒ x3+1x3+3(x+1x)x^3 + {1 \over x^3} +3(x + {1\over x}) = -1

⇒ x3+1x3x^3 + {1 \over x^3}  = 2   ...(ii)

Multiplying equation (i) and equation (ii),

⇒ (x2+1x2x^2 + {1 \over x^2})(x3+1x3x^3 + {1 \over x^3} ) = -2

⇒ x5+1x5+x+1xx^5 + {1 \over x^5} + x + {1 \over x} = -2

⇒ x5+1x5x^5 + {1 \over x^5}  = -1   ...(iii)

Now, taking cube of equation (iii),

(x5+1x5)3(x^5 + {1 \over x^5} )^3 = -1

⇒ x15+1x15+3(x5+1x5)x^{15} + {1 \over x^{15}} + 3 (x^5 + {1 \over x^5}) = -1

 ⇒ x15+1x15x^{15} + {1 \over x^{15}}  = -1 + 3

 ⇒ x15+1x15x^{15} + {1 \over x^{15}}  = 2