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Cars P & Q are approaching each other on the same highwa

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by Mike@Magoosh » Fri Nov 21, 2014 4:15 pm
Cars P & Q are approaching each other on the same highway. Car P is moving at 49 mph northbound and Car P is moving at 61 mph southbound. At 2:00 pm, they are approaching each other and 120 mi apart. Eventually they pass each other. At what clock time are they moving away from each other and 45 miles apart?
(A) 3:06 pm
(B) 3:30 pm
(C) 3:54 pm
(D) 5:21 pm
(E) 6:15 pm


For a set of practice problems on Motion, including the OA & OE for this particular problem, see:
https://magoosh.com/gmat/2014/gmat-pract ... on-motion/

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by Chicagosummer » Fri Nov 21, 2014 5:41 pm
The combined speed of the cars is 110 m/ hr
The have to cut a distance of 165 m
It will take 1.5 hours
Hence the time will be 3:30 (B)
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by Mathsbuddy » Mon Nov 24, 2014 6:32 am
A sketch can help (see graph).
Then using y = mx + c for each car:
Distance between P and Q = pt - (qt + 120) = 45
So, t = (45 + 120)/(p-q) = 165/(49 - -61) = 165/110 = 1.5
Therefore clock time = 2pm + 1.5h = 3.30pm
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by GMATinsight » Tue Nov 25, 2014 3:39 am
Nice Graph!!!

But GMAT doesn't allow us leisure of time that we can be that creative.

Relative Distance to be travelled = 120 + 45 = 165 Miles

Relative Speed (Opp.Dir) = 49 + 61 = 110 Miles/Hr

Time = Distance / Speed = 165 / 110 = 1.5 Hour from the state of Journey i.e. they meet at 3:30 PM

Answer: Option B
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by Mathsbuddy » Wed Nov 26, 2014 7:09 am
GMATinsight wrote:Nice Graph!!!

But GMAT doesn't allow us leisure of time that we can be that creative.

Relative Distance to be travelled = 120 + 45 = 165 Miles

Relative Speed (Opp.Dir) = 49 + 61 = 110 Miles/Hr

Time = Distance / Speed = 165 / 110 = 1.5 Hour from the state of Journey i.e. they meet at 3:30 PM

Answer: Option B
Thanks for your response,

True, GMat time is very limited. However, I am offering an alternative method of understanding. With a mental image of the graphs and a comprehension of the equation of a straight line, it is very simple and quick to create the equation given. I like the simplicity of your approach of finding the relative velocity by adding them. Cheers!
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