Multiple choice

A car leaves City A at 7:30 AM and follows the same route every day. On one particular day, it travels at a steady speed of 48 km/h and arrives at its destination 2 hours later than its usual scheduled arrival time. The following day, the car covers three-fourths of the distance in half of the scheduled travel time, and covers the remaining distance at 32 km/h, reaching exactly on time. What is the car's usual scheduled arrival time?

  1. 5 : 30 PM

  2. 2 : 30 PM

  3. 1 : 30 PM

  4. 3 : 00 PM
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let D be distance and T be scheduled time. D/48 = T + 2. Also, (3/4)D / (T/2) + (1/4)D / 32 = T. Solving these equations: D = 48(T+2). Substitute: 1.5D/T + D/128 = T. 72(T+2)/T + 48(T+2)/128 = T. Solving for T gives T = 6 hours. 7:30 AM + 6 hours = 1:30 PM.

AI explanation

Let the total distance be D and the usual travel time be T; from the first day's trip, the actual time taken at 48 km/h is T + 2, so D = 48(T + 2). On the second day, the car covers three-fourths of the distance in T/2 hours, meaning the remaining one-fourth of the distance equals 12(T + 2)/4 or 3(T + 2). The car covers this remaining distance at 32 km/h in the remaining T/2 hours, leading to the equation 3(T + 2) = 32(T/2). Solving this gives T = 6 hours; since the car leaves at 7:30 AM, the usual scheduled arrival time is 1:30 PM.

Keep practicing — related questions