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SSC CHSL 2021 PYQ — Question 54

SSC CHSL - 15 April 2021 - Shift 1 • Quantitative Aptitude • Q54

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If 49a2 + 25b2 = 30 and ab = 1, a, b > 0, then the value of (7a + 5b) is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. 10.

(a+b)2=a2+b2+2ab(a + b)^2 = a^2 + b^2 + 2ab

By squaring the equation (7a + 5b);

⇒ (7a + 5b)2 = 49a2 + 25b2 + 2 × 7a × 5b 

⇒ (7a + 5b)2 = 49a2 + 25b2 + 2 × 7a × 5b 

⇒ (7a + 5b)2 = 49a2 + 25b2 + 70ab 

⇒ (7a + 5b)2 = 30 + 70 × 1

⇒ (7a + 5b)2 = 100

⇒ (7a + 5b) = √100 

⇒ (7a + 5b) = ± 10

⇒ (7a + 5b) = + 10,  (7a + 5b) ≠ - 10 ∵ a, b > 0