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# Rebecca runs at a constant rate on the treadmill and it takes her 2 ho

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Math Expert
Joined: 02 Sep 2009
Posts: 60515
Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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04 Dec 2019, 02:14
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12
00:00

Difficulty:

95% (hard)

Question Stats:

25% (02:51) correct 75% (02:48) wrong based on 44 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

Are You Up For the Challenge: 700 Level Questions

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Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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04 Dec 2019, 05:24
1
1
Time taken by Rebecca to reach 1 milestone= T(R)
Time taken by Jeff to reach 1 milestone= T(J)

Time taken by both= T(R+J)

T(R+J)= 3*4/5= 2.4 Hours

As T(R)>T(J)> T(R+J), Eliminate option A, B and C

Option D
T(R+J)= $$\frac{3*5}{8}$$=$$\frac{15}{8}$$ < 2.4
Eliminate it

Option E
T(R+J)= $$\frac{4*6}{10}$$= 2.4 (Answer)

Bunuel wrote:
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

Are You Up For the Challenge: 700 Level Questions
Manager
Joined: 09 Nov 2015
Posts: 160
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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07 Dec 2019, 22:53
1
Let the time taken by Rebecca(R) to complete 1 milestone (MS) be T hours. Then Jeff(J) takes (T-2) hours to complete 1 MS.
In 3 hrs, R completes 3/T MS and J completes 3/(T-2) MS so the sum of the distances covered by each in 3 hrs is: [3/T+3/(T-2)] hrs which is equal to
1.25 MS. Therefore:
3/T+3/(T-2)=5/4....> 5T(T-6)-4(T-6)=0....> (5T-4)(T-6)=0. So T is equal to 4/5 or 6. Since Jeff takes 2 hrs less than Rebecca to complete 1 MS, T cannot be 4/5 hrs because that would give a negative value for Jeff's time. So T=6 which means R will take 6*2=12 hrs to complete 2 MS.

ANS: E
Intern
Joined: 23 May 2019
Posts: 14
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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11 Dec 2019, 13:05
3/T+3/(T-2)=5/4

Why the speed is not included in the formula ?
Intern
Joined: 29 Sep 2019
Posts: 9
Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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28 Dec 2019, 21:25
2
1
Bunuel wrote:
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

Hi Bunuel, would you please release an official solution for this one? Tried to understand other's explanation but still very lost.. thanks!
Intern
Joined: 28 Aug 2019
Posts: 9
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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29 Dec 2019, 01:03
1
IMHO E
Idk how much correct I was when arriving at the solution, still lost a little.

Let the time taken for Jeff = t
Let the time taken for Rebecca = t+2
Distance till first milestone be = 100km

Distance ___________100km___________ | __25km__ | __25km__ | __25km__ |

Jeff ---------------t hrs-------------------|---t+3 hrs--|---t+6 hrs--|--t+9 hrs----|
Rebecca ---------------t+2 hrs----------------|---t+5 hrs--|---t+8 hrs--|--t+11 hrs----|

Eliminate a,b,c,d for t for rebecca > 11
Math Expert
Joined: 02 Aug 2009
Posts: 8344
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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29 Dec 2019, 01:28
1
Bunuel wrote:
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

Are You Up For the Challenge: 700 Level Questions

The first method would be to make use of choices..

The choices give us time of R to cover 2 milestone, so half of each choice will give us time taken by R to cover 1 milestone, and THAT half -2 will give us time taken by J..

(A) 4
So R takes 4/2 or 2 hrs and J takes 2-2 =0...NOT possible

(B) 6
So R takes 6/2 or 3 hrs and J takes 3-2 =1, so J alone will cover 3/3 or 3 milestones in 3 hrs, but we are looking a combined distance of 1.25 milestone in 3 hrs...NOT possible

(C) 8
So R takes 8/2 or 4 hrs and J takes 4-2 =2, so J alone will cover 3/2 or 1.5 milestones in 3 hrs, but we are looking a combined distance of 1.25 milestone in 3 hrs...NOT possible

(D) 10
So R takes 10/2 or 5 hrs and J takes 5-2 =3, so J alone will cover 1 milestones in 3 hrs and R will cover 3/5 or 0.6 milestones. Combined they cover 1.6, but we are looking a combined distance of 1.25 milestone in 3 hrs...NOT possible

(E) 12
Only choice left, so our answer, but let us check
So R takes 12/2 or 6 hrs and J takes 6-2 =4, so J alone will cover 3/4 or 0.75 milestones in 3 hrs and R will cover 3/6 or 0.5 milestones in 3 hrs. Combined they cover 0.75+0.5=1.25 milestones..... Our answer

E
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Math Expert
Joined: 02 Aug 2009
Posts: 8344
Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho  [#permalink]

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29 Dec 2019, 01:44
Bunuel wrote:
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

Are You Up For the Challenge: 700 Level Questions

Second method could be..

R takes R hrs for 1 milestone, so in 3 hours, distance covered will be 3/R milestone
Similarly J will take R-2 hrs for 1 milestone, so in 3 hours, distance covered will be 3/(R-2) milestone
These 2 should equal 1.25 milestone.

So, $$\frac{3}{R}+\frac{3}{R-2}=1.25=\frac{5}{4}..............................\frac{3R-6+3R}{R(R-2)}=\frac{5}{4}$$
=> $$4*6(r-1)=5*R(R-2)$$
Either use choices to see.....E gives R as 12/2=6.....$$4*6*(6-1)=5*6*(6-2).........4*6*5=5*6*4$$..YES
OR solve further
$$24R-24=5R^2-10R...........5R^2-34R+24=0.....5R^2-30R-4R+24=0.....(R-6)(5R-4)=0$$
So, R can be 6 or 4/5, but R cannot be 4/5 as J will become negative then..
Thus R=6, and twice of R or 12 hrs is required to cover 2 milestones..

E
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Re: Rebecca runs at a constant rate on the treadmill and it takes her 2 ho   [#permalink] 29 Dec 2019, 01:44
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