Question

Two persons A and B start walking from P to reach Q at speeds of 5 kmph and 4 kmph respectively. If A arrives at Q, 30 minutes before B, then the distance between P and Q in km is

• A. \(6\)
• B. \(8\)
• C. \(10\)
• D. \(12\)

Hint:

In speed-time problems, when distance is same, directly equate the difference of times.

Answer 1

The Correct Option is C

Concept: Use: \[ \text{Time}=\frac{\text{Distance}}{\text{Speed}} \] Difference in arrival times is given.

Step 1: Assume distance.
Let the distance between \(P\) and \(Q\) be: \[ x \text{ km} \] Time taken by A: \[ \frac{x}{5} \] Time taken by B: \[ \frac{x}{4} \]

Step 2: Use time difference.
A arrives \(30\) minutes earlier: \[ 30 \text{ minutes}=\frac12 \text{ hour} \] So: \[ \frac{x}{4}-\frac{x}{5}=\frac12 \]

Step 3: Simplify.
Take LCM \(20\): \[ \frac{5x-4x}{20}=\frac12 \] \[ \frac{x}{20}=\frac12 \] \[ x=10 \] Thus, the distance between \(P\) and \(Q\) is: \[ \boxed{10 \text{ km}} \]