Tuesday, September 27, 2011

Challenging Practice Q3 WS 9 Solution


A and B are at starting and finishing points respectively of a 12km cross-country route.
At 07 00, Chad jogs from A to B at 6km/h
At 07 30, Wei Liang jogs from B to A at 3km/h
i) What time will they both meet?
ii) How far would Chad have jogged when he meets Wei Liang?

Let's take Chad's distance travelled as d and Wei Liang's as 12-d.
Chad's time will be t-7 and Wei Liang's is t-7.5

Since Speed = Distance/Time
Chad's Speed is 6 = d/t-7
Wei Liang's is 3= 12-d / t-7.5
Thus, Chad's distance is 6t-42 and Wei Liang's is 3(T-7.5)

When we substitute them, 3(t-7.5) = 12-(6t-42)
When we further simplify by groupig them, 9t = 76.5
Thus t=8.5

So they both meet at 08 30.

From here we can solve ii. 
If he had travelled 6km in an hour, he would have travelled 9km by 08 30

Thus the ans is

i) 0830
ii) 9km

by Saishwar, Jun Wei and Yong Hong

4 comments:

  1. great effort and just a feedback
    inconsistent use of T or t.
    note that for variable just stick to lower case.

    ReplyDelete
  2. Good effort but can u clarify why chad's distance as d and Wei lang's distance 12-d?

    ReplyDelete
  3. quite complicated but still understandable (:

    ReplyDelete
  4. A good effort indeed but numbers are a little confusing to understand.

    ReplyDelete