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Difficult Math Problem #102 - Rate

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #102 - Rate

Post 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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Post 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....

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post 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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gabriel Legendary Member
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Post 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....

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800guy Master | Next Rank: 500 Posts Default Avatar
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Post 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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