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## Difficult Math Problem #102 - Rate

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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.

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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.

gabriel Legendary Member
Joined
20 Dec 2006
Posted:
986 messages
Followed by:
1 members
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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
Joined
27 Jun 2006
Posted:
354 messages
Followed by:
5 members
11
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.

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