SSC Selection Post 2022 PYQ — Question 59

The correct answer is option 3 i.e. 105.

Let the equal distance = d km

Total time taken = ($\frac{d}{2} + \frac{d}{4} + \frac{d}{5} + \frac{d}{7} + \frac{d}{12}$)

Take LCM of 2, 4, 5, 7, 12 = 420

So,

Total time = $d\left( \frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{7} + \frac{1}{12} \right)$ = $d \left( \frac{210 + 105 + 84 + 60 + 35}{420} \right)$ = $d \left( \frac{494}{420} \right)$

According to the question,

⇒ $\frac{494d}{420}$ = $1485\over60$

⇒ d = 21 km approx

Total distance = 21 × 5 = 105 km