SarkariPYQ

SSC CHSL 2019 PYQ — Question 72

SSC CHSL - 3rd July 2019 - Shift 3 • Quantitative Aptitude • Q72

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If 9a2 + 16b2 + c2 + 25 = 24(a + b), then the value of (3a + 4b + 5c) is:

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Correct answer: A

Explanation

Correct Answer is option 1 i.e. 7

Application

9a2 + 16b2 + c2 + 25 = 24(a + b)

⇒ 9a2 – 24a + 16 + 16b2 – 24b + 9 + c2 = 0

⇒ (3a – 4)2 + (4b – 3)2 + c2 = 0

Hence, 3a = 4, 4b = 3 and c = 0

Now,

(3a + 4b + 5c) = 4 + 3 + 0 = 7