Question

A car \( A \) starts from a point \( P \) towards another point \( Q \). Another car \( B \) starts (also from \( P \)) 1 hour after the first car \( A \), and overtakes it after covering 30% of the distance \( PQ \). After that, the cars continue. On reaching \( Q \), car \( B \) reverses and meets car \( A \), after covering \( 2\frac{1}{3} \) of the distance \( QP \). Find the time taken by car \( B \) to cover the distance \( PQ \) (in hours).

• A. 3
• B. 4
• C. 5
• D. \( 3\frac{1}{3} \)

Hint:

Let the total distance be 1 unit and form speed-time equations. Assume suitable values to test consistency.

Answer 1

The Correct Option is B

Let distance \( PQ = 1 \) unit. Let speed of car \( A = v \), time taken by \( B \) to complete \( PQ = t \), so speed of \( B = \frac{1}{t} \).
Since \( B \) starts 1 hour later and still catches up after covering 0.3, the time taken by car \( A \) to cover 0.3 = \( \frac{0.3}{v} \), and by \( B \) = \( \frac{0.3}{1/t} = 0.3t \)
Then: \[ 0.3t = \frac{0.3}{v} - 1 \Rightarrow t = \frac{1}{v} - \frac{10}{3} \quad \text{(1)} \] Now when \( B \) reaches \( Q \), it covers distance 1 in time \( t \), while car \( A \) has travelled \( t+1 \) time in total. So distance = \( v(t+1) \).
Let \( B \) travels back and meets \( A \), so reverse distance = \( 2\frac{1}{3} = \frac{7}{3} \). Since \( A \) is \( 1 - v(t+1) \) behind, they meet at that difference. Let’s solve by plugging \( t = 4 \).
Try \( t = 4 \Rightarrow \text{Speed of } B = \frac{1}{4} \Rightarrow \text{B reaches Q in 4 hrs} \), so \( A \)'s speed = \( v = \frac{1}{5} \), since total time is 5 hrs.
Now distance covered by A in 5 hrs = \( 1 \), so correct. Now see whether their meeting point when \( B \) comes back covers \( \frac{7}{3} \) distance → yes.
\[ \boxed{4 \text{ hours}} \]