A delivery van travels at a constant speed of 60 km/h to cover a certain route in 1 hour and 45 minutes. If traffic increases the time taken by 15 minutes, by what percentage must the van's average speed decrease to complete the route?

Correct — B. The route length does not change, so find it first.

Time: 1 h 45 min = 1.75 h
Distance: 60 × 1.75 = 105 km

New time: 1.75 + 0.25 = 2 h
New speed: 105 ÷ 2 = 52.5 km/h

Fall in speed: 60 − 52.5 = 7.5 km/h
Percentage: 7.5 ÷ 60 × 100 = 12.5% → option (b)

Why the others are wrong

• (a)
  10.5% — 10.5% leaves 60 × 0.895 = 53.7 km/h, and 105 km at that speed takes about 1.96 h, a little under the 2 hours the traffic imposes.
• (c)
  15% — 15% leaves 51 km/h, and 105 ÷ 51 ≈ 2.06 h, about three and a half minutes longer than the 2 hours available.
• (d)
  20% — 20% leaves 48 km/h, and 105 ÷ 48 = 2.1875 h, more than 11 minutes over the 2 hours available.

Over a fixed distance, speed × time stays constant, so speed and time move in inverse proportion.

The time here goes from 1.75 h to 2 h, a ratio of 7 : 8. Speed must move the opposite way, 8 : 7, from 60 to 60 × 7⁄8 = 52.5 km/h.

Key facts

• Distance = speed × time, so over a fixed distance speed and time are inversely proportional.
• If the time over the same distance rises from t₁ to t₂, speed falls by (t₂ − t₁) ⁄ t₂ × 100%.
• 45 minutes = 0.75 hour, so 1 h 45 min = 1.75 h.

Common traps

• Dividing the 15-minute delay by the original 1 h 45 min: 0.25 ⁄ 1.75 ≈ 14.29% is the rise in time, not the fall in speed.
• Writing 1 h 45 min as 1.45 h instead of 1.75 h.

No directly related past PYQ was found.