Question

Two cars P and Q start from points A and B simultaneously. They meet 40 km from B. After meeting, they exchange speeds and return. They meet again 20 km from A. Find distance AB.

• A. 130 km
• B. 100 km
• C. 120 km
• D. 110 km

Hint:

Use ratio of speeds from first meeting, and apply swapped ratios for second meeting using time = distance/speed logic.

Answer 1

The Correct Option is C

Let distance AB = \( D \) First meet: P and Q meet 40 km from B \(\Rightarrow\) Distance travelled by Q = 40 km \(\Rightarrow\) Distance travelled by P = \( D - 40 \) km So, ratio of speeds: \[ P : Q = (D - 40) : 40 \] Now, after exchanging speeds, they return. Second meet: They meet 20 km from A \(\Rightarrow\) Distance travelled by P (returning from B) = 20 km \(\Rightarrow\) Distance travelled by Q (returning from A) = \( D - 20 \) But now P is running at Q's speed and Q is at P's speed: \[ \text{Time taken to reach 2nd meeting point is same:}
\frac{20}{Q} = \frac{D - 20}{P} \Rightarrow \frac{20}{Q} = \frac{D - 20}{P} \Rightarrow 20P = (D - 20)Q \] Use earlier speed ratio: \[ \frac{P}{Q} = \frac{D - 40}{40} \Rightarrow P = Q \frac{D - 40}{40} \] Substitute into equation: \[ 20 Q \frac{D - 40}{40} = (D - 20)Q \Rightarrow \frac{20(D - 40)}{40} = D - 20 \Rightarrow \frac{D - 40}{2} = D - 20 \Rightarrow D - 40 = 2D - 40 \Rightarrow D = 120 \] Final Answer: \( \boxed{120 \text{ km}} \)