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Rebecca runs at a constant rate on the treadmill and it take [#permalink]
14 Jun 2013, 10:06

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Difficulty:

5% (low)

Question Stats:

83% (04:21) correct
17% (07:14) wrong based on 23 sessions

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.... _________________

Do not forget to hit the Kudos button on your left if you find my post helpful.

Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
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. _________________

Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
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! _________________

"Where are my Kudos" ............ Good Question = kudos

Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
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

gmatclubot

Re: Rebecca runs at a constant rate on the treadmill and it take
[#permalink]
07 Aug 2013, 13:56