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At 1:00 pm, Devi begins driving from Townville to Villageton at a speed of 36 miles per hour. 20 minutes later, Mark begins driving from Townville to Villageton at a speed of 51 miles per hour. What will the time be when Mark catches up to Devi?

(A) 1:48 pm
(B) 1:54 pm
(C) 2:04 pm
(D) 2:05 pm
(E) 2:08 pm

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Here's our video solution to this question: https://www.gmatprepnow.com/module/gmat ... /video/916

when Mark catches up with Devi, both would have traveled the same distance.
Speed= Distance/Time
Let the distance traveled by Mark be x miles and time taken be t minutes
then the distance traveled by Devi is 12+x because Devi would have traveled 20 minutes more (36 miles in 1 hour so 12 miles in 20 minutes)
Equating distance traveled 12+x=x==> 36(t+20)= 51t ==> t=(36*20)/15= 48 minutes
So if Mark started at 1:20 PM so he will catch up with Devi in 48 minutes more i.e., at 2:08 PM
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Another wording of the same solution:
36 miles per hour means 36 miles every 60 minutes, which is equivalent to 12 miles every 20 minutes
Therefore Davi did 12 miles after 20 minutes. It was then 01:20 when Mark started.
Relative speed is 15 miles per hours (15 = 51 - 36). In other words, Mark cacth-up 15 miles every 60 min, which is same as 1 mile every 4 min.
Hence mark needs 48 minutes to catch-up 12 miles (48=12*4)

48 munites after 01:20 (time when Mark started), is 02:08

Answer is E.
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This is a fairly straightforward question on the concept of Relative Speed.

Relative Speed is the speed of two objects that are moving simultaneously.

    In case of two objects that are moving in the same direction, the relative speed is the difference of the speeds.
    If the objects are moving in opposite direction, the relative speed is the sum of the speeds.

But, the most important thing you need to remember is that you can calculate the relative speed only when both the objects are moving simultaneously.

In this question, Devi and Mark both leave from Townsville and travel towards Villageton and are hence travelling in the same direction. Therefore, the relative speed will be the difference in their speeds.

However, the relative speed comes into effect only at 1:20 PM since Devi had been travelling alone between 1 PM and 1:20 PM.
20 minutes is 1/3rd of a hour; at 36 mph, distance travelled by Devi = 36 * \(\frac{1}{3}\) = 12 miles ( Distance = Speed * Time).

This means that, when Mark starts his journey, the distance between him and Devi is 12 miles. The relative speed = 51 – 36 = 15 mph. A schematic representation of this situation is shown below:

Attachment:
29th Nov 2019 - Reply 3.jpg
29th Nov 2019 - Reply 3.jpg [ 26.47 KiB | Viewed 4878 times ]

At this pace, time taken by Mark = \(\frac{12 }{ 15}\) = \(\frac{4}{5}\) hours = 48 minutes (Time = Distance / Speed)
Observe that we have taken Relative Speed in place of speed since both Mark and Devi are travelling simultaneously.
48 minutes from 1:20 PM would be 2:08 PM. The correct answer option is E.

Note that the relative speed of 15 mph means that Mark gains 15 miles over Devi in every 1 hour; it can also mean that Devi loses 15 miles to Mark in every 1 hour. Therefore, to gain/lose 12 miles, it will take Mark/Devi, a time of 48 minutes.

Hope that helps!
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Devi leaves at 1 PM and gets a 20 minute head start before Mark Leaves.

Then Mark Leaves.

When Mark catches up to Devi, they will have traveled the same distance.

Distance of Devi = Distance of Mark


D of Devi = 36 mph * (T + (20/60)) h

D of Mark = 51 mph * (T)


Where T is the time it takes for Mark to catch up to Devi after he leaves at 1 : 20 PM

36 (T + 1/3) = 51 (T)

36T + 12 = 51T

12 = 15T

T = 12 / 15 = 48 / 60 in hours units

This is equivalent to 48 minutes.

After Mark leaves at 1 : 20, it takes him 48 min to catch up.

This will be at

2 : 08
(E)

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