Question

P and Q start from opposite ends, with Q starting an hour later and at twice the speed. When P has covered 1/6th the distance, Q has also covered the same. Where do they meet?

• A. Closer to A
• B. Exactly between A and B
• C. Closer to B
• D. They never meet

Hint:

When one person starts earlier but slower, meeting point shifts toward the one with the earlier start.

Answer 1

The Correct Option is A

Let total distance between A and B = \( D \)
Let P’s speed = \( v \), so Q’s speed = \( 2v \)
Let the time when they meet be \( t \) hours after P started. Q starts 1 hour later, so Q has traveled for \( t - 1 \) hours. Distance covered by P: \( vt \)
Distance covered by Q: \( 2v(t - 1) \) Given: \[ vt = 2v(t - 1) vt = 2vt - 2v vt - 2vt = -2v -vt = -2v t = 2 \text{ hours} \] So, at time of meeting: - Distance covered by P = \( v \times 2 = 2v \) - Since \( D \) is constant, if \( P \) covered 2v and Q covered 2v, total = 4v - But that’s less than full D, so meeting happens before halfway. \[ \boxed{\text{Closer to A}} \]