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 By: acelow (offline)  Monday, August 05 2013 @ 07:48 AM CDT (Read 960 times)
acelow

Mr Siew drove from town X to town Y at 60km/h. If he increased his speed by 12km/h, he would reach town Y 1 hour earlier. Find the speed he must travel after the first 120km so that he can reach town Y 4/5 hour earlier.
(Ans: 75km/h)

Pls help...thank you very much.

Newbie

Registered: 05/02/09
Posts: 14

 By: echeewh (offline)  Monday, August 05 2013 @ 08:30 AM CDT
echeewh

Hey <acelow>

Following is my worked solution:

Apply <Common Distance> method, time ratio is the reverse of speed ratio.

s1 : s2
60 : 72
5 : 6

t1 : t2
6 : 5

1u time = 1 hr
t2 = 5u --> 5 hrs

Distance (XY) = 72 × 5 = 360 km

Time taken (XY) at s1 = 60 km/h:
360 ÷ 60 = 6 hrs

Remaining distance = 360 - 120 = 240 km
Remaining time (duration) to reach (4/5)hr earlier = 5 1/5 - 2 = 3 1/5 hrs

Remaining speed = 240 ÷ (16/5) = 75 km/h

========

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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