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## Thursday, October 02, 2008

### Catholic High Sch 2006 PSLE Math Prelim Q43

At 7.30 am, Hubert left Johor, travelling towards Kuala Lumpur at a constant speed. 1 hour later, Joshua started travelling from Johor on the same road. Joshua overtook Hubert at 11.30 am. The speed at which Joshua was travelling at was 20km/h faster than Hubert and he arrived at Kuala Lumpur at 12.30pm. Find the distance between Johor and Kuala Lumpur.

Solution

Hubert’ time ----- 7.30 to 11.30 --- 4h
Joshua’s time ----- (1h later) 8.30 to 11.30 --- 3h

At the point where Joshua overtook Hubert, both travelled the same distance. However Joshua’s speed was 20km/h more than Hubert.

Hubert’s distance ----- Joshua’s distance
Hubert’s speed x Hubert’s time ------ Joshua’s speed x Joshua’s time
1 unit x 4 ----- (1 unit + 20) x 3
4 units ----- 3 units + 60
1 unit ----- 60

Joshua’s speed -----
1 unit + 20
60 + 20 = 80

Distance from Johor to KL -----
Joshua’s speed x Joshua’s time
80km/h x 4h (Joshua took from 8.30 to 12.30 to reach KL)

yeo kik mwan said...

Hi, can you further explain as i don't understand. I going PSLE

Excel Eduservice said...

J left 1h after H. When J overtook H, both had travelled the SAME DISTANCE.

But J’s time is 1h less than H.

Use the formula D = S x T.

Since BOTH J and H covered the SAME distance, we can say,

Distance travelled by J = Distance travelled by H

H’s speed x H’s time = J’s speed x J’s time.

But since we know J’s speed is 20 km/h more, and J took 1h less, we get ….

1 unit x 4h = (1 unit + 20) x 3h