Difficult Math Problem #102 - Rate

I wonder if there is enough information.

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.

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.

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....