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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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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?
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!
Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
09 Sep 2014, 06:24
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Re: Rebecca runs at a constant rate on the treadmill and it take [#permalink]
19 Jun 2015, 22:41
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Lapetiteflo wrote:
How do you derive so easily the solution for t from this 2nd order equation ?? It takes me at least 3 minutes
After you get here, (1/t+2 + 1/t)∗3=5/4, Take 3 to the other side, and it becomes (1/t+2 +1/t) = 5/12. Now look at the answer choices, t has to be a number divisible by 12. So either A, B or E. Replace t with A or 4. And you quickly see that 1/6 + 1/4 = 5/12. Therefore, t is 4. Don't over complicate things by trying to calculate everything. You can apply this strategy to a lot of questions and save time.
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