Alternatively, you can use the ratio method to solve. This concept is very useful and ensures that carelessness/error is minimised. Hope that helps.soseducation wrote:but i know schools are teaching it using the algebra methodiwork wrote:This is a distance-speed problem that can be solved without algebra. It could be solved by simply understanding the problem and worked from another angle. (In fact, it was trained in Heuristic Maths or Maths Olympiad program which I had taught in schools.)belnanna wrote:Can someone please advise on this question? Dont really understand how to solve. ><

Thank you so much in advance!

Mr Teo traveled at 40km/h from town P to town Q. If he increases his speed by 30km/h ,he will take 6 hours lesser to reach his destination.

What was the distance between town P and town Q?

You have to realize few points

1. the speed increase by 30km/h

2. the time taken reduced by 6 hours.

point 2 => the extra 6 hours will cut the travel distance by 6 x 40 =240km

point 1 => this distance will have to covered by the extra speed of 30km/h in 240/30 = 8 hours.

So the distance between the 2 towns will be 70km/h x 8 h = 560km.

S1 : S2 = 4 : 7

Hence, T1 : T2 = 7 : 4 (concept: faster speed, shorter time taken)

Difference = 3u

3u = 6 hours

1u = 2 hours

Distance (using S1) = 40 km/h x 14 hours = 560 km

Check (using S2) = 70 km/h x 8 hours = 560 km