Shannon and Maxine work in the same building and leave work

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 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.

ThNks

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

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)

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

Shouldn’t the speed of max should be 2R, in second eqn too,
But u kept it R??? When maxine cycles from 2R speed,she reaches 45 min earlier and when she cycles from R speed, she reaches 2 hours earlier???how

Shouldn’t in second equation Shannon should be R and Maxine be 2R

Shouldn’t the speed of max should be 2R, in second eqn too,
But u kept it R??? When maxine cycles from 2R speed,she reaches 45 min earlier and when she cycles from R speed, she reaches 2 hours earlier???how

Shouldn’t in second equation Shannon should be R and Maxine be 2R

I solved this question as following:

Maxine traveled the distance from her work to her house "DM" twice, once by car at a rate 2R mph and the second time by bike at a rate R mph. Since the distance is the same, time and rate will vary inversely:
(t+2/3)/t+2 = R/2R = 1/2 (t is the time Shannon is taking to drive to her house and 2/3 hours=40 min)
From the above, we can solve for t, hence t=2/3 hours

At first, both Shannon and Maxine drove by car to their houses at the rate 2Rmph. Since the rate in this case is the same, time and distance will vary directly:
t/(t+2/3) = DS/DM (DS and DM are Shonnon and Maxine respective distances to their houses from work).
Replacing t by 2/3, we get DS/DM=1/2
Therefore,
DS:DM:TOT
1:2:3
(knowing that the actual total distance is 60, the unkown multiplier would hence be 20)
DS:DM:TOT
20:40:60

So DM= 40 and correct answer is D

Dear KarishmaB
I figured that this question can be solved algebraically as well as suggested by Bunuel above but I have been unsuccessful trying to solve this question using relative speed. Can this question be solved using relative speed to start with? If not can you please explain why?
Thank you so much!

This is not a relative speed question because they are not covering the 60 miles "together."
They are each covering their own distance in their own time. Shannon arrived early and stopped while Maxine kept going.
Had they covered a certain distance together in the same time, relative speed would have come into play.

was able to solve it algebraically but took 6mins !

Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM

Not sure how much of my working makes sense but given that I struggle with Algebra (mental blocks), I just played around with the figures based on my understanding of the concept.