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## Difficult Math Problem #102 - Rate

This topic has 4 member replies
800guy Master | Next Rank: 500 Posts
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#### Difficult Math Problem #102 - Rate

Mon Feb 26, 2007 12:42 pm
Jane gave Karen a 5 meter head start in a 100 meter race and Jane was beaten by 0.25 meters. In how many meters more would Jane have overtaken Karen?

oa coming after some people answered with explanations. from diff math doc.

gabriel Legendary Member
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Thu Mar 01, 2007 8:05 am
Okay guys .... if u can make use of the proportianality between distance , speed and time under different conditions this becomes much easier problem to solve....

for this particular problem the ratio of distance covered by jane and karen for a particular period of time is 99.75/95..... this ratio will be in proportion to (x+0.25)/x... solve for x and the answer wuld be x=5 ... therfore the answer is 5+.25= 5.25....

just let me know if any one needs a detailed explanation for the problem....

800guy Master | Next Rank: 500 Posts
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Wed Feb 28, 2007 7:15 pm
oa from diff math doc:

Jane gave Karen a 5 m head start means Karen was 5 m ahead of Jane. So after the lead, Karen ran 95m and Jane ran 99.75 m when the race ended.

Let speed of Karen and Jane be K and J respectively and lets say after X minutes Jane overtakes Karen.

1st condition: 95/K = 99.75/J
2nd condition JX - KX = 0.25

Solving for JX, we get JX=21/4.
Hence Jane needs to run 5.25m more (or total of 105m) to overtake Karen.

ssiva Junior | Next Rank: 30 Posts
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Mon Feb 26, 2007 5:39 pm
I wonder if there is enough information.

jayhawk2001 Community Manager
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Mon Feb 26, 2007 10:05 pm
I'm sure there is a simpler way but here goes ...

J = speed of Jane
K = speed of Karen
t = time when they finish the race

Jane finishes 100-0.25 meters in t time, so
Jt = 100 - .25 = 99.75

Karen finishes 100 min in t time, so
5 + Kt = 100 or Kt = 95

Hence
t = (99.75 - 95) / (J-K) = 4.75 / (J-K)

J-K = 4.75 / t

Lets assume at time t1, both Jane and Karen are equal (i.e. after time t1,
Jane starts to take the lead)

Jt1 = Kt1 + 5

Hence

t1 = 5 / (J-K)
= (5 / 4.75) * t

We are asked J(t1-t) which is the time it takes for Jane to gain lead

(t1-t)J = ((5 / 4.75)t - t)J
= (0.25 / 4.75) * JT
= 0.25 / 4.75 * 99.75
= 5.25 meters

So, it takes 5.25 meters extra for Jane to beat Karen.

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