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racing problem!

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racing problem!

by Ozlemg » Mon Jul 04, 2011 1:07 pm
Hui challenged Feng to a marathon race (42 kilometers). Hui was so confident that he even gave Feng a 30-minute head start. Feng runs the race at a rate of 6 kilometers per hour. Hui runs the race at a rate of 7 kilometers per hour. If Hui and Feng start from the same point and follow the same path, how long will Hui run before he overtakes Feng?
A. 2.5 hours
B. 3 hours
C. 4.25 hours
D. 3.75 hours
E. 5 hours
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by Frankenstein » Mon Jul 04, 2011 7:04 pm
Hi,
Hui starts 30 mins after Feng.
In those 30 mins, Feng covers 6*(1/2) = 3km
Relative speed of Hui with respect to Feng is 7-6 =1kmph
So, for every hour, Hui covers 1 km more than Feng.
So, in order to overtake Fend, Hui needs to cover the extra 3 kms head start he gives.
So, it takes him 3 hours to overcome.

Hence, B
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by Jim@Knewton » Mon Jul 04, 2011 7:46 pm
Brilliant Frank! ...and Ozlemg, here is a here is a methodical quickie:

Let time(for Hui) to meet / overtake = t hours
Then Feng time = t hr + 30 minutes = (t + 0.5) hrs
Feng Distance = 6*(t+0.5) = 6t + 3 and Hui Distance = 7*t
=> 7t = 6t + 3 => t = 3

This method will help even if the numbers / question structure gets tougher! :-)
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by amit2k9 » Mon Jul 04, 2011 8:50 pm
1. t be the time for which both the runners run.

7t = 6t + 3 thus t=3 hrs.

2 since both the runners run in the same direction thus relative velocity = 7-6 = 1km/hr.

initial distance of separation = 6* 1/2 hr = 3 km

thus 3/1 = 3 hrs to cover the distance.

you may use whichever concept you feel is good for you.
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by gmatblood » Tue Jul 05, 2011 12:35 pm
arrange as a mapping!

Hui Feng
0 3 (distance in 30 mins)
7 9 (1st hour)
14 15 (2nd hour)
21 21 (3rd hour)

so it takes 3 hours to overcome!! :D
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