A bullet train covers a fixed distance in 30 minutes at an average speed of 240 km/h. Due to track maintenance, it needs to be diverted, increasing the distance by 20%. If the train needs to arrive at its destination on time (i.e., in 30 minutes), what should its new average speed be in km/h?

Correct — C. Find the original distance, stretch it by 20%, then divide by the same time.

Time = 30 minutes = 1⁄2 hour
Original distance = 240 × 1⁄2 = 120 km
New distance = 120 × 1.2 = 144 km
New speed = 144 ÷ 1⁄2 = 288 km/h → option (c)

Shortcut: with the time fixed, speed rises in the same ratio as distance, so 240 × 1.2 = 288.

Why the others are wrong

• (a)
  280 km/h — 280 km/h for half an hour covers 140 km, 4 km short of the 144 km diverted route, so the train would arrive late.
• (b)
  300 km/h — 300 km/h is 240 × 1.25, a 25% rise. It covers 150 km in half an hour, which suits a 25% detour, not the 20% one here.
• (d)
  320 km/h — 320 km/h is 240 × 4⁄3, a rise of about 33%. In half an hour it covers 160 km, 16 km more than the 144 km route.

Distance = speed × time. When the time is held fixed, speed and distance move in the same ratio: stretch the route by 20% and the speed must rise by exactly 20%.

The distance itself is never needed for the answer — 240 × 1.2 gives it in one line. Working it out (120 km, then 144 km) is only a check.

Watch the units: 30 minutes must become 1⁄2 hour before it meets a speed in km/h.

Key facts

• At a fixed time, speed is directly proportional to distance.
• A 20% increase means multiplying by 1.2: 240 × 1.2 = 288.
• In 30 minutes a train covers half its speed in km: 240 km/h gives 120 km.

Common traps

• Dividing the 144 km by 30 without converting minutes to hours, which gives 4.8 — a speed in km per minute, not km/h.
• Dividing only the extra 24 km by the half hour, which gives 48 km/h — the increase in speed, not the new speed.

No directly related past PYQ was found.