SSC CHSL 2024 PYQ — Question 70

SSC CHSL - 11 July 2024 - Shift 4 • Quantitative Aptitude • Q70

If 158+[{45of512÷(1115)}×1 712]\frac{15}{8} + \left[\left\{\frac{4}{5} of \frac{5}{12} \div \left(1\frac{1}{15}\right)\right\} \times 1\,\frac{7}{12}\right]  is expressed in the form of ab\frac{a}{b} where a and b are integers and relatively prime to each other, the value 4b - a is:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 313.

a/b = 158+[{45of512÷(1115)}×1 712]\frac{15}{8} + \left[\left\{\frac{4}{5} of \frac{5}{12} \div \left(1\frac{1}{15}\right)\right\} \times 1\,\frac{7}{12}\right]

⇒  a/b = 158+[{13÷1615}×1912]\frac{15}{8} + \left[\left\{\frac{1}{3} \div \frac{16}{15}\right\} \times \frac{19}{12}\right]

⇒  a/b = 158+[{13×1516}×1912]\frac{15}{8} + \left[\left\{\frac{1}{3} \times \frac{15}{16}\right\} \times \frac{19}{12}\right]

⇒  a/b = 158+[516×1912]\frac{15}{8} + \left[\frac{5}{16} \times \frac{19}{12}\right]

⇒  a/b = 158+[516×1912]\frac{15}{8} + \left[\frac{5}{16} \times \frac{19}{12}\right]

⇒ a/b = 15/8 + 95/192 = 455/192

a = 455 and b = 192

So, 4b - a = 4 × 192 - 455 = 768 - 455 = 313

Therefore, the value of 4b - a is 313.