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SSC Selection Post 2022 PYQ — Question 69

SSC Selection Post - Higher Secondary Level 2022 - 10 Feb 2022 - Shift 2 • Quantitative Aptitude • Q69

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

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Correct answer: B

Explanation

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

a2+2a+b2−2b+49c2−56c+18=0a^2 + 2a + b^2 - 2b + 49c^2 - 56c + 18 = 0

a2+2a=(a+1)2−1b2−2b=(b−1)2−1a^2 + 2a = (a + 1)^2 - 1 \\ b^2 - 2b = (b - 1)^2 - 1 \\

49c2−56c=49(c2−87c)=49((c−47)2−1649)=49(c−47)2−1649c^2 - 56c = 49\left(c^2 - \frac{8}{7}c\right) = 49\left(\left(c - \frac{4}{7}\right)^2 - \frac{16}{49}\right) = 49\left(c - \frac{4}{7}\right)^2 - 16

(a+1)2−1+(b−1)2−1+49(c−47)2−16+18=0(a + 1)^2 - 1 + (b - 1)^2 - 1 + 49\left(c - \frac{4}{7}\right)^2 - 16 + 18 = 0

(a+1)2+(b−1)2+49(c−47)2−1−1−16+18=0⇒(a+1)2+(b−1)2+49(c−47)2=0(a + 1)^2 + (b - 1)^2 + 49\left(c - \frac{4}{7}\right)^2 - 1 - 1 - 16 + 18 = 0 \\ \Rightarrow (a + 1)^2 + (b - 1)^2 + 49\left(c - \frac{4}{7}\right)^2 = 0

Since the sum of squares is zero, each term must be zero:

(a+1)2=0⇒a=−1(b−1)2=0⇒b=1(c−47)2=0⇒c=47(a + 1)^2 = 0 \Rightarrow a = -1 \\ (b - 1)^2 = 0 \Rightarrow b = 1 \\ \left(c - \frac{4}{7}\right)^2 = 0 \Rightarrow c = \frac{4}{7}

a−b+7c=−1−1+7(47)=−2+4=2a - b + 7c = -1 - 1 + 7\left(\frac{4}{7}\right) = -2 + 4 = 2