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Speed question - NYC to Lake House

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Speed question - NYC to Lake House

by sal_xcool » Thu Feb 24, 2011 9:38 pm
Andrew rides his bike at 15 miles per hour, and it takes him 5 hours to ride from New York City to his lake house.
Andrew's wife, Andrea, leaves New York at the same time, and drives on the same road to the lake house at an
average speed of 50 miles per hour. Thirty minutes after arriving at the lake house, Andrea realizes she forgot
her computer, and begins to drive back to New York. How long after leaving the lake house will she pass her
husband, Andrew, on the road?

A. 1/3 of an hour
B. 5/13 of an hour
C. 1/2 of an hour
D. 9/13 of an hour
E. 19/26 of an hour

OA D

Ok, I have 1:30 mins max to solve this question??
Anybody has developed a proven technique for solving the speed/rate questions? what's the best strategy to tackle these type of problems?
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by Rich@VeritasPrep » Thu Feb 24, 2011 9:51 pm
Hey sal_xcool,

First of all, the average time per question is more like 2 minutes, so no need to cut yourself off at 1:30! Also, keep in mind that 2 minutes is an AVERAGE. There are going to be some easier problems that you might solve in less than a minute, and some more difficult ones that take you closer to 3 minutes. Therefore, saying you have a specific amount of time to solve any one given question doesn't really make sense. And it puts unneeded pressure on yourself!!

Hope this helps.

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by GMATGuruNY » Thu Feb 24, 2011 9:54 pm
sal_xcool wrote:Andrew rides his bike at 15 miles per hour, and it takes him 5 hours to ride from New York City to his lake house.
Andrew's wife, Andrea, leaves New York at the same time, and drives on the same road to the lake house at an
average speed of 50 miles per hour. Thirty minutes after arriving at the lake house, Andrea realizes she forgot
her computer, and begins to drive back to New York. How long after leaving the lake house will she pass her
husband, Andrew, on the road?

A. 1/3 of an hour
B. 5/13 of an hour
C. 1/2 of an hour
D. 9/13 of an hour
E. 19/26 of an hour

OA D

Ok, I have 1:30 mins max to solve this question??
Anybody has developed a proven technique for solving the speed/rate questions? what's the best strategy to tackle these type of problems?
Since Andrew rides at 15mph for 5 hours, d = r*t = 15*5 = 75 miles.
Time for Andrea to arrive at the lake house = d/r = 75/50 = 3/2 hours.
30 minutes later = 2 hours total.
So when Andrea leaves the lake house, Andrew has traveled r*t = 15*2 = 30 miles.
Thus, distance remaining between Andrew and the lake house = 75-30 = 45 miles.
Combined rate for Andrew and Andrea as they travel toward each other = 15+50 = 65 miles per hour. (We combine their rates because they are working together to cover the distance between them, and when people work together, we add their rates.)
Time for Andrew and Andrea to travel 45 miles = d/r = 45/65 = 9/13 hours.

The correct answer is D.
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by sal_xcool » Thu Feb 24, 2011 10:20 pm
GMATGuruNY wrote:
sal_xcool wrote:Andrew rides his bike at 15 miles per hour, and it takes him 5 hours to ride from New York City to his lake house.
Andrew's wife, Andrea, leaves New York at the same time, and drives on the same road to the lake house at an
average speed of 50 miles per hour. Thirty minutes after arriving at the lake house, Andrea realizes she forgot
her computer, and begins to drive back to New York. How long after leaving the lake house will she pass her
husband, Andrew, on the road?

A. 1/3 of an hour
B. 5/13 of an hour
C. 1/2 of an hour
D. 9/13 of an hour
E. 19/26 of an hour

OA D

Ok, I have 1:30 mins max to solve this question??
Anybody has developed a proven technique for solving the speed/rate questions? what's the best strategy to tackle these type of problems?
Since Andrew rides at 15mph for 5 hours, d = r*t = 15*5 = 75 miles.
Time for Andrea to arrive at the lake house = d/r = 75/50 = 3/2 hours.
30 minutes later = 2 hours total.
So when Andrea leaves the lake house, Andrew has traveled r*t = 15*2 = 30 miles.
Thus, distance remaining between Andrew and the lake house = 75-30 = 45 miles.
Combined rate for Andrew and Andrea as they travel toward each other = 15+50 = 65 miles per hour. (We combine their rates because they are working together to cover the distance between them, and when people work together, we add their rates.)
Time for Andrew and Andrea to travel 45 miles = d/r = 45/65 = 9/13 hours.

The correct answer is D.


Thank you for your awesome explanation :) it' def. helps me out.
I didn't see that I could add the speeds. Can you give me example where you can't combine the speeds or rates?
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by Stuart@KaplanGMAT » Fri Feb 25, 2011 8:24 pm
sal_xcool wrote: I didn't see that I could add the speeds. Can you give me example where you can't combine the speeds or rates?
Here's the general rule for multiple travelling objects:

if two objects are moving in OPPOSITE directions, ADD the speeds; and

if two objects are moving in the SAME direction, SUBTRACT the speeds.

(Opposite = directly towards each other or directly away from each other.)

This rule applies to all rates, not just speed; for example, on the GMAT we've seen questions involving 2 people reading the same book that ask when the second person will be on the same page as the first person - since they're reading in the same direction (start to finish), you subtract their rates to come up with the "catch up" rate.
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by ankur.agrawal » Fri Feb 25, 2011 10:05 pm
I think there is a flaw in the question.

No where in the question it is mentioned that Andrea drives back to Newyork at the same average speed.

How do we assume that ?

sal_xcool wrote:Andrew rides his bike at 15 miles per hour, and it takes him 5 hours to ride from New York City to his lake house.
Andrew's wife, Andrea, leaves New York at the same time, and drives on the same road to the lake house at an
average speed of 50 miles per hour. Thirty minutes after arriving at the lake house, Andrea realizes she forgot
her computer, and begins to drive back to New York. How long after leaving the lake house will she pass her
husband, Andrew, on the road?

A. 1/3 of an hour
B. 5/13 of an hour
C. 1/2 of an hour
D. 9/13 of an hour
E. 19/26 of an hour

OA D

Ok, I have 1:30 mins max to solve this question??
Anybody has developed a proven technique for solving the speed/rate questions? what's the best strategy to tackle these type of problems?
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by sanju09 » Sat Feb 26, 2011 12:41 am
sal_xcool wrote:Andrew rides his bike at 15 miles per hour, and it takes him 5 hours to ride from New York City to his lake house.
Andrew's wife, Andrea, leaves New York at the same time, and drives on the same road to the lake house at an
average speed of 50 miles per hour. Thirty minutes after arriving at the lake house, Andrea realizes she forgot
her computer, and begins to drive back to New York. How long after leaving the lake house will she pass her
husband, Andrew, on the road?

A. 1/3 of an hour
B. 5/13 of an hour
C. 1/2 of an hour
D. 9/13 of an hour
E. 19/26 of an hour

OA D

Ok, I have 1:30 mins max to solve this question??
Anybody has developed a proven technique for solving the speed/rate questions? what's the best strategy to tackle these type of problems?

The distance from New York City to the lake house = 15 × 5 = 75 miles. These 75 miles could be covered by Andrea in 1.5 hours. Thirty minutes after Andrea arriving at the lake house, Andrew would have covered 15 × 2 = 30 miles. Hence the moment Andrea begins to drive back to New York, the couple are 75 - 30 = 45 miles apart from each other (here is where I smell little ambiguity in the wordings, it should have been worded like "Andrea realizes she forgot her computer, and begins to drive back with her same average speed to New York", but I am taking it for granted here), which is meant to be covered by the two with the relative speed of 15 + 50 = 65 mph, and hence it would take Andrea [spoiler]45/65 = 9/13 hours[/spoiler] to pass her husband, Andrew, on the road.

[spoiler]D[/spoiler]
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