SSC CGL 2023 PYQ — Question 62

SSC CGL - 21 July 2023 - Shift 2 • Quantitative Aptitude • Q62

if 2a−5a−4b−5b+6c+5c=0\frac{2a-5}{a}-\frac{4b-5}{b}+\frac{6c+5}{c}=0, what is the value of  (1a−1b−1c)(\frac{1}{a}-\frac{1}{b}-\frac{1}{c})?

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 45\frac{4}{5}.

2a−5a−4b−5b+6c+5c=0\frac{2a-5}{a}-\frac{4b-5}{b}+\frac{6c+5}{c}=0

⇒ 2a−5a−4b−5b+6c+5c=0\frac{2a-5}{a}-\frac{4b-5}{b}+\frac{6c+5}{c}=0

2 - (5/a) - 4 + (5/b) + 6 + (5/c) = 0

⇒ 5a−5b−5c{5 \over a} - {5 \over b} - {5 \over c} = 8 - 4

⇒ (1a−1b−1c)(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}) = 454 \over 5