Question

Two persons P and Q start from A and travel towards B which is at a distance of 100 km. P's speed exceeds that of Q by 2 kmph. After reaching B, P returns back and meets Q at a distance of 10 km from B. Then the speed of Q (in kmph) is

• A. 8
• B. 9
• C. 10
• D. 12

Hint:

For objects traveling for the same time, the ratio of distances covered is equal to the ratio of their speeds.

Answer 1

The Correct Option is B

Concept: This problem involves calculating speeds based on travel time. Since both P and Q travel for the same duration until they meet, their distances traveled are proportional to their speeds.

Step 1: Calculate the distance traveled by P and Q.
Total distance AB = 100 km. P reaches B (100 km) and returns 10 km to meet Q, so P travels: \[ \text{Dist}_P = 100 + 10 = 110 \text{ km} \] Q travels from A towards B and is 10 km from B when they meet, so Q travels: \[ \text{Dist}_Q = 100 - 10 = 90 \text{ km} \]

Step 2: Relate the speeds using the time proportionality.
Since time is constant for both until the meeting point: \[ \frac{\text{Dist}_P}{\text{Speed}_P} = \frac{\text{Dist}_Q}{\text{Speed}_Q} \] Let \( v \) be the speed of Q. Then speed of P is \( v + 2 \). \[ \frac{110}{v+2} = \frac{90}{v} \]

Step 3: Solve for \( v \).
\[ 110v = 90(v + 2) \] \[ 110v = 90v + 180 \] \[ 20v = 180 \implies v = 9 \text{ kmph} \] \centerline{{9 kmph}}