SarkariPYQ

SSC CGL 2020 PYQ — Question 90

SSC CGL Tier II - Quant - 15 Nov 2020 • Quantitative Aptitude • Q90

Back to paper

If  4553=1a+1b+1c−25,\frac{45}{53}=\frac{1}{a+\frac{1}{b+\frac{1}{c-\frac{2}{5}}}}, where a, b and c are positive integers, then what is the value of (4a - b + 3c)?

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. 5

 4553=1a+1b+1c−25\frac{45}{53}=\frac{1}{a+\frac{1}{b+\frac{1}{c-\frac{2}{5}}}}

⇒11+845=1a+1b+1c−25\Rightarrow \frac{1}{1+\frac{8}{45}}=\frac{1}{a+\frac{1}{b+\frac{1}{c-\frac{2}{5}}}}

⇒11+15+58=1a+1b+1c−25\Rightarrow \frac{1}{1+\frac{1}{5+\frac{5}{8}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c-\frac{2}{5}}}}

⇒11+15+12−25=1a+1b+1c−25\Rightarrow \frac{1}{1+\frac{1}{5+\frac{1}{2-\frac{2}{5}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c-\frac{2}{5}}}}

On comparing both sides

We get, a = 1, b = 5 and c = 2

Putting them in (4a - b + 3c) = (4 × 1 - 5 + 3 × 2) = (4 - 5 + 6) = 5