Question

The distance between A and B is \(180\) km. A starts towards B at a speed of \(30\) kmph at 9:00 a.m., and after 1 hour, B starts towards A at a speed of \(20\) kmph. At what time will they meet for the first time?

• A. 12:00 p.m.
• B. 1:00 p.m.
• C. 1:30 p.m.
• D. 2:00 p.m.

Hint:

Account for the 1-hour head start of A, then use combined speed 50 kmph to close the remaining gap.

Answer 1

The Correct Option is B

Step 1: Set up a single equation using total distance covered by both.
Let $T$ be the number of hours after 9:00 a.m. at which A and B meet. A has been walking for the full $T$ hours by then, covering $30T$ km. B only starts 1 hour later, so B has been moving for $(T-1)$ hours by the meeting time, covering $20(T-1)$ km.

Step 2: Add both distances to equal the total gap of 180 km.
\[ 30T + 20(T-1) = 180 \]

Step 3: Expand and solve for $T$.
\[ 30T + 20T - 20 = 180 \]
\[ 50T = 200 \]
\[ T = 4 \text{ hours} \]

Step 4: Convert $T$ back into a clock time.
Since $T$ is measured from 9:00 a.m.,
\[ 9:00 \text{ a.m.} + 4 \text{ hours} = 1:00 \text{ p.m.} \]

Step 5: Cross check against the relative speed method.
This agrees exactly with treating the first hour separately and then closing a 150 km gap at 50 kmph, which also gives 1:00 p.m. Two different ways of setting up the problem land on the same clock time, so 1:00 p.m. is the meeting time, even though some published keys list 1:30 p.m.

Final Answer:
Solving the combined distance equation confirms they meet at 1:00 p.m. \[ \boxed{1:00 \text{ p.m.}} \]