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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
[Reveal] Spoiler: OA

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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
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 09 Sep 2014, 06:24
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 19 Jan 2015, 09:41
Rebecca: Rate: a/t, time = t, distance = a
Jeff: Rate = a/(t+2), time = t, distance = a
Combine rate, reach 1.15 distance, in 3 hour
[a/t + a/(t+2)] x 3 = 1.25 a
=> t = 4
Rebecca: Rate = a/t, time = x?, distance = 2a
x = 2t = 8.

Answer: C
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 20 Jan 2015, 01:34
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Icerockboom wrote:
Rebecca: Rate: a/t, time = t, distance = a
Jeff: Rate = a/(t+2), time = t, distance = a
Combine rate, reach 1.15 distance, in 3 hour
[a/t + a/(t+2)] x 3 = 1.25 a
=> t = 4
Rebecca: Rate = a/t, time = x?, distance = 2a
x = 2t = 8.

Answer: C


Actually, Jeff is more faster than Rebecca, his rate should be \(\frac{1}{t-2}\)

\((\frac{1}{t-2} + \frac{1}{t}) * 3 = 1.25\)

t = 6

Time required by Rebecca = 2 * 6 = 12

Answer = E = 12
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink] New post 11 Feb 2015, 05:12
How do you derive so easily the solution for t from this 2nd order equation ?? It takes me at least 3 minutes :cry:
Re: Rebecca runs at a constant rate on the treadmill and it take   [#permalink] 11 Feb 2015, 05:12
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