SSC CHSL 2024 PYQ — Question 52

The correct answer is option 4 i.e.  $33\frac{1}{3}$%.

Let, the original speed = s

Original time = t

Distance = st

New time (t₁) = (75/100) × t = (3/4)t

Let new speed = s₁

So, st = s₁(3/4)t

⇒ s₁ = (4/3)s

Increase in speed = (4/3)s - s = (1/3)s

Percentage increase = [(s₁ - s)/s] × 100 = [(1/3)s/s] × 100 = (1/3) × 100% = $33\frac{1}{3}$%

Therefore, the racer should increase the speed by $33\frac{1}{3}$%.