SSC CHSL 2021 PYQ — Question 75

SSC CHSL - 6 Aug 2021 - Shift 3 • Quantitative Aptitude • Q75

The average of the squares of four consecutive odd natural numbers is 201. The average of 7 times of the largest number and 3 times of the smallest number is:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 76

Explanation:

Let x be an odd natural number.

Then 4 consecutive odd numbers = (x - 2), (x + 2), (x + 4)

Acording to the given question,

(x - 2)2 + x2 + (x + 2)2 + (x + 4)2 = (201 × 4)

⇒ x2 - 4x - 4 + x2 + x2 + 4x + 4 + x2 + 8x + 16 = 804

⇒ 4x2 + 8x + 24 = 804

⇒ x2 + 2x + 6 = 201

⇒ x2 + 2x - 195 = 0

⇒ x2 + 15x - 13x - 195 = 0

⇒ x(x + 15) - 13(x + 15) = 0

⇒ (x - 13)(x + 15) = 0

⇒ x = 13; x = -15(not possible)

Now, Largest number = (x + 4) = 13 + 4 = 17

Smallest Number = (x -2 ) = 13 - 2 = 11

So, required value = (17×17+3×11)2(17 \times 17 + 3 \times 11) \over 2 =(119+33)2(119 + 33) \over 2 = (152)2(152) \over 2 = 76

Hence the required answer is 76.