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Rebecca runs at a constant rate on the treadmill and it take

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Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 14 Jun 2013, 10:06
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Rebecca runs at a constant rate on the treadmill and it takes her 2 hours longer to reach a certain milestone than the time it takes Jeff to reach that same milestone. If Jeff also moves at a constant rate and the two are allowed to combine their distances traveled to reach 25% beyond the initial milestone in 3 hours, how many hours would it take for Rebecca alone to reach the equivalent distance of 2 milestones?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

I do not have the OA. Sorry for the inconvenience....
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 14 Jun 2013, 10:21
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R*(t+2)=1 milestone
J*t=1 milestone

(R+J)*3=\frac{5}{4} milestones

From the upper equations get the rate for R and J
R=\frac{1}{t+2} and J=\frac{1}{t}, now substitute them into the lower equation

(\frac{1}{t+2}+\frac{1}{t})*3=\frac{5}{4}

Solve for t and you get t=4

So Rebecca alone to reaches 1 milestone in 4+2=6 hours => two milestones in 12
IMO E
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 19 Jun 2013, 15:57
Zarrolou wrote:
R*(t+2)=1 milestone
J*t=1 milestone

(R+J)*3=\frac{5}{4} milestones

From the upper equations get the rate for R and J
R=\frac{1}{t+2} and J=\frac{1}{t}, now substitute them into the lower equation

(\frac{1}{t+2}+\frac{1}{t})*3=\frac{5}{4}

Solve for t and you get t=4

So Rebecca alone to reaches 1 milestone in 4+2=6 hours => two milestones in 12
IMO E



Don't we have to take 25% of combined distance ( 1+1) = 5/2 ?
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 19 Jun 2013, 21:05
targetbschool wrote:
Don't we have to take 25% of combined distance ( 1+1) = 5/2 ?


The wording may be a bit confusing, but "the two are allowed to combine their distances traveled to reach 25% beyond the initial milestone in 3 hours" is supposed to be equivalent to:

In 3 hours, if R and J travel at their constant rates, their combined distance is 25% beyond the initial milestone.

So indeed it should be 3 * (R + J) = 5/4 * milestone.
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 20 Jun 2013, 15:56
mattce wrote:
targetbschool wrote:
Don't we have to take 25% of combined distance ( 1+1) = 5/2 ?


The wording may be a bit confusing, but "the two are allowed to combine their distances traveled to reach 25% beyond the initial milestone in 3 hours" is supposed to be equivalent to:

In 3 hours, if R and J travel at their constant rates, their combined distance is 25% beyond the initial milestone.

So indeed it should be 3 * (R + J) = 5/4 * milestone.



Even I thought wordings are confusing.... Not sure how will I tackle such Q in actual Exam? :(

Thanks!
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 20 Jun 2013, 21:09
targetbschool wrote:
Zarrolou wrote:
R*(t+2)=1 milestone
J*t=1 milestone

(R+J)*3=\frac{5}{4} milestones

From the upper equations get the rate for R and J
R=\frac{1}{t+2} and J=\frac{1}{t}, now substitute them into the lower equation

(\frac{1}{t+2}+\frac{1}{t})*3=\frac{5}{4}

Solve for t and you get t=4

So Rebecca alone to reaches 1 milestone in 4+2=6 hours => two milestones in 12
IMO E




Don't we have to take 25% of combined distance ( 1+1) = 5/2 ?


The text is a bit complicated, however the equation ( 1+1) = 5/2 cannot be the one we are looking for as 2\neq{\frac{5}{2}}.
We have to write this condition with the speed*time=space relationship
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 07 Aug 2013, 13:56
Rebecca runs at a constant rate on the treadmill and it takes her 2 hours longer to reach a certain milestone than the time it takes Jeff to reach that same milestone. If Jeff also moves at a constant rate and the two are allowed to combine their distances traveled to reach 25% beyond the initial milestone in 3 hours, how many hours would it take for Rebecca alone to reach the equivalent distance of 2 milestones?

Milestone = d

Rebecca = t+2
Jeff = t

1.25 = r*3

Rebecca: d=r*(t+2)
Jeff: d = r*(t)

r*t=1.25d
(d/t+2 + d/t)*3 = 1.25d

Could someone explain to me why 1.) you use 1 and 5/4 instead of d and 2.) how to simplify? Thanks!










(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Re: Rebecca runs at a constant rate on the treadmill and it take   [#permalink] 07 Aug 2013, 13:56
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