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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]

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14 Jun 2013, 11:06
5
14
00:00

Difficulty:

95% (hard)

Question Stats:

33% (02:47) correct 67% (02:51) wrong based on 167 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  [#permalink]

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14 Jun 2013, 11:21
4
5
$$R*(t+2)=1 milestone$$
$$J*t=1 milestone$$

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

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

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19 Jun 2013, 16:57
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]

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19 Jun 2013, 22:05
1
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]

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20 Jun 2013, 16:56
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]

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20 Jun 2013, 22:09
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]

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07 Aug 2013, 14:56
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]

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19 Jan 2015, 10:41
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.

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

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20 Jan 2015, 02:34
4
1
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.

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

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

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11 Feb 2015, 06:12
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  [#permalink]

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19 Jun 2015, 23:41
1
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  [#permalink]

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26 Nov 2015, 16:37
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, 5/12-1/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  [#permalink]

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20 Oct 2018, 13:41
Let distance =1
rate of jef =1/t-2
rate of rebecca=1/t
so as per equation d=r*t
(1/t-2 + 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   [#permalink] 20 Oct 2018, 13:41
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