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Shannon and Maxine work in the same building and leave work [#permalink]

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15 May 2012, 10:39

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Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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16 May 2012, 03:03

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Expert's post

alexpavlos wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Say Maxine lives \(d\) miles from office, then Shannon lives \(60-d\) miles from office.

If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon --> \(\frac{d}{2r}=\frac{60-d}{2r}+\frac{2}{3}\) --> \(3d=90+2r\);

If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine --> \(\frac{d}{r}=\frac{60-d}{2r}+2\) --> \(3d=60+4r\);

Solve \(3d=90+2r\) and \(3d=60+4r\) for \(d\) --> \(d=40\).

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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16 May 2012, 09:49

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Expert's post

alexpavlos wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Here is the ratios approach to the problem:

Shannon drives at the speed of 2R in both the cases so she takes the same time. In the first case Maxine reaches home 40 mins after Shannon. In the second case, Maxine reaches 2 hrs after Shannon. Why did Maxine take 1 hr 20 mins extra in the second case? Because she drove at half the speed. Speed1: Speed 2 = 2:1 Time 1: Time 2 = 1:2 ( since distance stays the same) The difference between Time1 and Time 2 is 1 hr 20 mins = 80 mins. So Time 1 must be 1hr 20 mins i.e. time taken by Maxine when she drives at speed 2R. Time taken by Shannon must be 1 hr 20 mins - 40 mins = 40 mins (because she reaches 40 mins early)

When their speeds were same in the first case, Time taken by Maxine : Time taken by Shannon = 80 mins :40 mins = 2:1 Distance traveled by Maxine : Distance traveled by Shannon = 2:1

Total distance is 60 miles so Maxine lives 40 miles away and Shannon lives 20 miles away from office. _________________

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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31 Dec 2013, 10:24

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alex1233 wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Nice question +1

We have that X/24 - (60-X)/2R = 40

Also X/R - (60-X)/2R = 120

So we get that 2x - 60 = 80R 3x - 60 = 240R

Get rid of R

120 = 3x x = 40

Hence answer is D

Hope it helps Cheers! J

Y'all also, one can do the following

(X/2R)-(60-X/2R) = 2/3

(X/R)-(60-X/2R)=2

Therefore we have the following pair of equations:

6x-180=4r (1) 3x-60=4r (2)

Therefore we can equate both and obtain x=40.

Hence D

Hope it helps Cheers J

Last edited by jlgdr on 09 Apr 2014, 08:53, edited 1 time in total.

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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15 Feb 2014, 00:00

VeritasPrepKarishma wrote:

alexpavlos wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Here is the ratios approach to the problem:

Shannon drives at the speed of 2R in both the cases so she takes the same time. In the first case Maxine reaches home 40 mins after Shannon. In the second case, Maxine reaches 2 hrs after Shannon. Why did Maxine take 1 hr 20 mins extra in the second case? Because she drove at half the speed. Speed1: Speed 2 = 2:1 Time 1: Time 2 = 1:2 ( since distance stays the same) The difference between Time1 and Time 2 is 1 hr 20 mins = 80 mins. So Time 1 must be 1hr 20 mins i.e. time taken by Maxine when she drives at speed 2R. Time taken by Shannon must be 1 hr 20 mins - 40 mins = 40 mins (because she reaches 40 mins early)

When their speeds were same in the first case, Time taken by Maxine : Time taken by Shannon = 80 mins :40 mins = 2:1 Distance traveled by Maxine : Distance traveled by Shannon = 2:1

Total distance is 60 miles so Maxine lives 40 miles away and Shannon lives 20 miles away from office.

Hi Karishma,

You mentioned time difference between time1 and time 2 is 80 mins... How dod you take the value for time1 as 80 mins? Please explain the basis.

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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28 Jan 2015, 10:20

VeritasPrepKarishma wrote:

alexpavlos wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Here is the ratios approach to the problem:

Shannon drives at the speed of 2R in both the cases so she takes the same time. In the first case Maxine reaches home 40 mins after Shannon. In the second case, Maxine reaches 2 hrs after Shannon. Why did Maxine take 1 hr 20 mins extra in the second case? Because she drove at half the speed.

Speed1: Speed 2 = 2:1 Time 1: Time 2 = 1:2 ( since distance stays the same) The difference between Time1 and Time 2 is 1 hr 20 mins = 80 mins. So Time 1 must be 1hr 20 mins i.e. time taken by Maxine when she drives at speed 2R. Time taken by Shannon must be 1 hr 20 mins - 40 mins = 40 mins (because she reaches 40 mins early)

When their speeds were same in the first case, Time taken by Maxine : Time taken by Shannon = 80 mins :40 mins = 2:1 Distance traveled by Maxine : Distance traveled by Shannon = 2:1

Total distance is 60 miles so Maxine lives 40 miles away and Shannon lives 20 miles away from office.

Hi Karisma,

You have an interesting approach. However, I am not sure if I understand you well here. Maxine reaches 2 hrs after Shanon. So Maxine should take 120 min more and not 1 hr 20 min (=80 min). Am I correct? If yes, how will your proposed solution change? Can you please explain?

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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28 Jan 2015, 10:25

jlgdr wrote:

alex1233 wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Nice question +1

We have that X/24 - (60-X)/2R = 40

Also X/R - (60-X)/2R = 120

So we get that 2x - 60 = 80R 3x - 60 = 240R

Get rid of R

120 = 3x x = 40

Hence answer is D

Hope it helps Cheers! J

Y'all also, one can do the following

(X/2R)-(60-X/2R) = 2/3

(X/R)-(60-X/2R)=2

Therefore we have the following pair of equations:

6x-180=4r (1) 3x-60=4r (2)

Therefore we can equate both and obtain x=40.

Hence D

Hope it helps Cheers J

Hi,

In the line "We have that X/24 - (60-X)/2R = 40" of your solution, can you please clarify from where did you get 24 in the denominator?

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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28 Jan 2015, 12:52

thorinoakenshield wrote:

jlgdr wrote:

alex1233 wrote:

Shannon and Maxine work in the same building and leave work at the same time. Shannon lives due north of work and Maxine lives due south. The distance between Maxine's house and Shannon's house is 60 miles. If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon. If Maxine rider her bike home at the rate of R per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine. How far does maxine live from work?

A. 20 B. 34 C. 38 D. 40 E. 46

Nice question +1

We have that X/24 - (60-X)/2R = 40

Also X/R - (60-X)/2R = 120

So we get that 2x - 60 = 80R 3x - 60 = 240R

Get rid of R

120 = 3x x = 40

Hence answer is D

Hope it helps Cheers! J

Y'all also, one can do the following

(X/2R)-(60-X/2R) = 2/3

(X/R)-(60-X/2R)=2

Therefore we have the following pair of equations:

6x-180=4r (1) 3x-60=4r (2)

Therefore we can equate both and obtain x=40.

Hence D

Hope it helps Cheers J

Hi,

In the line "We have that X/24 - (60-X)/2R = 40" of your solution, can you please clarify from where did you get 24 in the denominator?

TO

Should be 2R. (4 on key-koard is too close to R = typo)

Re: Shannon and Maxine work in the same building and leave work [#permalink]

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29 Jan 2015, 00:31

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Expert's post

thorinoakenshield wrote:

Hi Karisma,

You have an interesting approach. However, I am not sure if I understand you well here. Maxine reaches 2 hrs after Shanon. So Maxine should take 120 min more and not 1 hr 20 min (=80 min). Am I correct? If yes, how will your proposed solution change? Can you please explain?

TO

The solution I have given above is correct.

There are two cases: One in which both drive home - here Maxine reaches 40 mins after Shannon In the other case, Maxine bikes and Shannon drives - here Maxine reaches 120 mins after Shannon

So difference in time taken by Maxine in the two cases is 120 mins - 40 mins = 80 mins

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