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Rebecca runs at a constant rate on the treadmill and it take
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14 Jun 2013, 10:06
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34% (02:47) correct 66% (02:54) wrong based on 166 sessions
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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
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Re: Rebecca runs at a constant rate on the treadmill and it take
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14 Jun 2013, 10:21
\(R*(t+2)=1 milestone\) \(J*t=1 milestone\) \((R+J)*3=\frac{5}{4} milestones\) 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
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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
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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
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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
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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
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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
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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
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20 Jan 2015, 01:34
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 \(\frac{1}{t2}\) \((\frac{1}{t2} + \frac{1}{t}) * 3 = 1.25\) 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
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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



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Re: Rebecca runs at a constant rate on the treadmill and it take
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19 Jun 2015, 22:41
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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Rebecca runs at a constant rate on the treadmill and it take
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26 Nov 2015, 15:37
d=distance milestone R+J's rate=(5d/4)/3=5d/12 J's rate=d/t R's rate =d/(t+2) eliminating d, 5/121/t=1/(t+2) t=4 R's time to reach 2d=2(4+2)=12 hours



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Re: Rebecca runs at a constant rate on the treadmill and it take
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20 Oct 2018, 12:41
Let distance =1 rate of jef =1/t2 rate of rebecca=1/t so as per equation d=r*t (1/t2 + 1/t)∗3=1.25 t = 6 Time required by Rebecca = 2 * 6 = 12




Re: Rebecca runs at a constant rate on the treadmill and it take
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20 Oct 2018, 12:41






