Andrew left Town X for Town Y which was 500 km apart. He travelled at an average speed of 90 km/h for 3/5 of the journey. He then increased his speed by 30 km/h for the rest of the journey and reached Town Y at 2 pm. Richard also left Town X for Town Y at the same time as Andrew and he drove at an average speed of 100 km/h for the whole journey.

a) What time did Andrew leave Town X?

b) How far apart were they at 1 pm?

Solution

5 units ---- 500 km

1 unit ---- 500 km divided by 5 = 100 km

Every 1/5 of journey ----- 100 km

(a)

Andrew’s Time (1st 3/5 of journey) ---- distance divided by speed

= 300 km divided by 90 km/h

= 3 and 1/3 hours

Andrew’s Time (last 2/5 of journey)

= 200 km/h divided by 100 km/h

= 1 and 2/3 hours

Total time taken by Andrew ----

(3 and 1/3 hours) + (1 and 2/3 hours) = 5 hours

He reached Town Y at 2 pm.

5 hours before 2 pm is 9 am.

Answer: Andrew left Town X at 9 am.

(b)

At 1 pm, both of them travelled for 4 hours.

Richard’s distance ----- speed x time

= 100 km/h x 4 h

= 400 km

Andrew’s distance at 1 pm was ----

For the 1st 3 and 1/3 hours, he travelled 3/5 of journey ---- 300 km.

For the next 2/3 hour he travelled ---- speed x time

= 120 km/h x 2/3 h

= 80 km

Andrew’s total distance in 4 hours ----

300 km + 80 km

= 380 km

They were apart by ----

400 km (Richard’s distance) – 380 km (Andrew’s distance)

= 20 km

Answer: They were 20 km apart.