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confused in such 'head start' questions!

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by GMATGuruNY » Mon Aug 27, 2012 7:10 am
mehaksal wrote:Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?
One approach is to treat this as a problem about PROPORTIONS.

When Karen finishes the 100-meter race, Jane is behind by .25 meters.
Thus, the distance traveled by Jane = 100 - .25 = 99.75 meters.
Since Jane give Karen a head start of 5 meters and finishes the race .25 meters behind Karen, Jane catches up by 5 - .25 = 4.75 meters.
Thus, for every 99.75 meters that Jane travels, she catches up by 4.75 meters.

To determine how much farther Jane must travel to catch up by another .25 meters, set up and solve the following proportion:
4.75 meters/99.75 meters = .25 meters/x meters
x = 5.25.

Thus, to catch up to Karen, Jane must travel another 5.25 meters.
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by urshohini » Tue Aug 28, 2012 1:32 pm
Mitch,
Thanks for the solution. Is there any other way of solving this? I got bowled by this question..
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by GMATGuruNY » Tue Aug 28, 2012 2:03 pm
urshohini wrote:Mitch,
Thanks for the solution. Is there any other way of solving this? I got bowled by this question..
Let the time for the race = 1 hour.

Since Jane loses by .25 meters, she travels 99.75 meters, implying that her rate = 99.75 meters per hour.
Since Karen gets a 5-meter head start, she travels 95 meters, implying that her rate = 95 meters per hour.

Jane's rate - Karen's rate = 99.75 - 95 = 4.75 meters per hour.
This is the rate at which Jane will catch up to Karen.

Since Jane must catch up by .25 meters, the time needed = d/r = .25/4.75 = 1/19 of an hour.
In 1/19 of an hour, the additional distance traveled by Karen = r*t = 95(1/19) = 5 meters.
Since Jane was .25 meters behind, the additional distance traveled by Jane = 5+25 = 5.25 meters.
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