A person travels from city A to city B. If he travels at a speed of 50 km/h, he arrives 1 hour early. If he travels at a speed of 30 km/h, he arrives 1 hour late. What is the distance (in km) between city A and city B?

Correct — A. Arriving 1 hour early and 1 hour late puts the two travel times 2 hours apart.

Time at 30 km/h − time at 50 km/h = 2
d⁄30 − d⁄50 = 2
Over the common denominator 150: (5d − 3d)⁄150 = 2
2d = 300, so d = 150 km → option (a)

Why the others are wrong

• (b)
  120 km — 120 km takes 2.4 hours at 50 km/h and 4 hours at 30 km/h. The gap is 1.6 hours, short of the 2 hours between early and late.
• (c)
  200 km — 200 km takes 4 hours at 50 km/h and 6 2⁄3 hours at 30 km/h. That gap, 2 2⁄3 hours, is more than the 2 hours required.
• (d)
  180 km — 180 km takes 3.6 hours at 50 km/h and 6 hours at 30 km/h. The 2.4-hour gap overshoots the 2-hour difference.

Both trips cover the same distance, so only the time changes. One hour early and one hour late sit on either side of the scheduled time, so the two travel times differ by 1 + 1 = 2 hours.

Write each time as distance ÷ speed and set their difference to 2. Here the times come to 3 hours and 5 hours, so the scheduled time is 4 hours.

Key facts

• Time = distance ÷ speed.
• Early by a hours and late by b hours means the two travel times differ by a + b hours.
• Distance = S₁ × S₂ ⁄ (S₁ − S₂) × (difference in time).

Common traps

• Taking the gap as 1 hour: d⁄30 − d⁄50 = 1 gives 75 km, half the true distance
• Subtracting the slower trip's time from the faster one's, which gives a negative distance
• Leaving a gap given in minutes unconverted when the speeds are in km/h

No directly related past PYQ was found.