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SSC CHSL 2019 PYQ — Question 69

SSC CHSL - 2nd July 2019 - Shift 1 • Quantitative Aptitude • Q69

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If a2 + 4b2 + 49c2 + 18 = 2(2b + 28c – a), then the value of (3a + 2b + 7c) is:

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Correct answer: C

Explanation

Correct Answer is option 3 i.e. 2

Concept

Application

Perfect square expression

a2 + 4b2 + 49c2 + 18 = 2(2b + 28c – a)

a2 + 4b2 + 49c2 + 18 = 4b + 56c – 2a

a2 + 2a + 4b2 – 4b + 49c2 – 56c + 18 = 0

(a2 + 2a + 1) + (4b2 – 4b + 1) + (49c2 – 56c + 16) = 0

(a + 1)2 + (2b – 1)2 + (7c – 4)2 = 0

Hence, a = -1, b = ½, c = 4/7

Now

(3a + 2b + 7c) = 3 × -1 + 2 × ½ + 7 × 4/7

= -3 + 1 + 4 = 2