While driving from Corvallis to Portland, Matt obeys the spe

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While driving from Corvallis to Portland, Matt obeys the speed limit. Matt's friend Vadim likes to drive faster than the speed limit and believes that he could leave well after Matt and still reach Portland first. The distance from Corvallis to Portland is 81 miles and both Matt and Vadim travel the same route at their respective constant speeds. If Matt travels at a constant speed of 36 mph and leaves 90 minutes before Vadim, what is the minimum constant speed that Vadim must exceed in order to arrive in Portland before Matt?

A. 72 mph
B. 81 mph
C. 96 mph
D. 108 mph
E. 162 mph

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OA=D

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by DavidG@VeritasPrep » Sun May 14, 2017 4:25 pm
ziyuenlau wrote:While driving from Corvallis to Portland, Matt obeys the speed limit. Matt's friend Vadim likes to drive faster than the speed limit and believes that he could leave well after Matt and still reach Portland first. The distance from Corvallis to Portland is 81 miles and both Matt and Vadim travel the same route at their respective constant speeds. If Matt travels at a constant speed of 36 mph and leaves 90 minutes before Vadim, what is the minimum constant speed that Vadim must exceed in order to arrive in Portland before Matt?

A. 72 mph
B. 81 mph
C. 96 mph
D. 108 mph
E. 162 mph

Source : Veritas Prep
OA=D
First, let's figure out how long It will take Matt to arrive at his destination. He's moving at 36 mph and he needs to cover 81 miles. 36t = 81
t = 81/36 = 2 1/4 hours.

Matt's got a 90 minute, or 1 1/2 hour head start, so Vadim needs to complete the trip in 2 1/4 - 1 1/2 = 3/4 of an hour. He needs to cover 81 miles in 3/4 of an hour. R * (3/4) = 81 --> R = 81 * (4/3) = 108 mph. The answer is D
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by elias.latour.apex » Mon May 15, 2017 11:24 am
This problem can be solved effectively using a parts solution. To start, we determine the Greatest Common Factor between 36, Matt's speed, and 81, the total distance. The GCF is 9.

Therefore, let 1 part = 9 miles. Fundamentally, this is no different from counting beer by the six-pack or eggs by the dozen.

Matt travels at 4 parts per hour, and he has a 1.5 hour head start, so he has already traveled 6 parts in a 9-part journey.

Vadim must cover the entire 9-part journey in the amount of time that Matt takes to cover the remaining 3 parts of his journey. Thus, Vadim must travel at least three times as quickly as Matt.

So he must travel at least 12 parts per hour. Since 1 part = 9 miles, he must travel at least 12x9 = 108 mph.

Or we can simply multiply 36 mph, Matt's speed, by 3 to get 108 mph.
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by Scott@TargetTestPrep » Fri May 19, 2017 5:00 am
ziyuenlau wrote:While driving from Corvallis to Portland, Matt obeys the speed limit. Matt's friend Vadim likes to drive faster than the speed limit and believes that he could leave well after Matt and still reach Portland first. The distance from Corvallis to Portland is 81 miles and both Matt and Vadim travel the same route at their respective constant speeds. If Matt travels at a constant speed of 36 mph and leaves 90 minutes before Vadim, what is the minimum constant speed that Vadim must exceed in order to arrive in Portland before Matt?

A. 72 mph
B. 81 mph
C. 96 mph
D. 108 mph
E. 162 mph
We are given that Matt travels at a rate of 36 mph and leaves 90 minutes, or 1.5 hours, before Vadim. We can let Vadim's time = t, so then Matt's time = t + 1.5.. Since rate x time = distance, we can create the following equation and solve for t:

36(t + 1.5) = 81

36t + 54 = 81

36t = 27

t = 27/36 = ¾

So, we know Vadim must reach Portland in ¾ of an hour or less. Since rate = distance/time, his minimum speed is 81/(¾) = 81 x 4/3 = 108 mph.

Answer: D

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