Directions: Read the following passage carefully and answer the questions…

Concept

On a straight track, position from the start equals speed multiplied by the time spent travelling. Two bodies meet when their positions from the start are equal at the same clock time. When a body reaches a turning point and heads back toward the start, its position thereafter is the turning distance minus the speed times the time elapsed since it turned. Setting two such position expressions equal and solving for the clock time gives the meeting time.

Setting up speeds and start times

Let the base speed be such that P, Q and R move at 1, 1.25 and 0.5 units per hour (the given ratio x : 1.25x : 0.5x). Measure time from 8 am.

• P starts at 8 am, so P has been moving for the full elapsed time.

• Q starts at 9 am (1 hour later).

• R starts at 10 am (2 hours later).

Step 1 — locate where Q meets P

1. Q meets P four hours after Q sets out at 9 am, i.e. at 1 pm. By then Q has travelled 1.25 × 4 = 5 units.

2. At 1 pm, P has been moving since 8 am, that is 5 hours, covering 1 × 5 = 5 units.

3. Both are 5 units from the start, confirming they meet there at 1 pm. Q now turns and heads back toward the start at its same speed, 1.25 units per hour.

Step 2 — write the two positions after the turn

1. R keeps moving outward from the start. Measured from 8 am, R’s position is 0.5 × (t − 2), where t is hours after 8 am.

2. Q is returning from 5 units, so Q’s position is 5 − 1.25 × (t − 5).

3. Set them equal: 5 − 1.25(t − 5) = 0.5(t − 2).

4. Expand: 11.25 − 1.25t = 0.5t − 1, so 12.25 = 1.75t, giving t = 7.

Result

t = 7 hours after 8 am is 3 pm. At that moment both Q and R are 2.5 units from the start, so they meet at 3 pm.

Cross-check

From 1 pm to 3 pm Q travels back 1.25 × 2 = 2.5 units, landing at 5 − 2.5 = 2.5 units. R, moving from 10 am to 3 pm (5 hours), covers 0.5 × 5 = 2.5 units. The two positions match exactly, confirming the meeting time.