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At 1:00 pm, Devi begins driving from Townville to Villageton

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At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 08 Nov 2016, 09:32
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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

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Re: At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 08 Nov 2016, 10:22
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GMATPrepNow wrote:
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

*Kudos for all correct solutions

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


Distance travelled by Devi in 20 min =\(\frac{36}{60} * 20\) = 12 miles
Time is 1.20 pm at this stage.
Relative speed between Mark and Devi = 51-36 = 15 miles/hr
Since the difference in distance between Mark and Devi is 12 miles, Mark will catch up when the difference is removed.
So, \(\frac{60}{15} * 12\) = 48 min.

Time when Mark catch up = 1.20 pm + 48 min = 2.08 pm.

If u liked my solution, do not forget to give kudos! :)
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Re: At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 08 Nov 2016, 09:57
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GMATPrepNow wrote:
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

*Kudos for all correct solutions

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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Re: At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 09 Nov 2016, 13:06
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GMATPrepNow wrote:
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

*Kudos for all correct solutions



how much time will it take to Mark to catche up to Devi?

the distance the two guys made are \(D=36*T\) and \(D=51*(T-\frac{1}{3})\)

we have to find when these two distances are equal ====> \(36*T=51*(T-\frac{1}{3})\)

T=17/15 hours = 1h 8 min

as the first guy start at 1:00 the time when the meet will be 2:08

answer E
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Re: At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 21 Apr 2017, 10:20
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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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Re: At 1:00 pm, Devi begins driving from Townville to Villageton  [#permalink]

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New post 29 Nov 2019, 01:41
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 270 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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Re: At 1:00 pm, Devi begins driving from Townville to Villageton   [#permalink] 29 Nov 2019, 01:41
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At 1:00 pm, Devi begins driving from Townville to Villageton

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