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Greg is training for a marathon by running to and from work each day,

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Greg is training for a marathon by running to and from work each day,  [#permalink]

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New post 02 Apr 2017, 12:33
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

78% (01:03) correct 22% (01:03) wrong based on 45 sessions

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Greg is training for a marathon by running to and from work each day, a distance of 12 miles each way. He runs from home to work at an average speed of 6 miles per hour and returns at an average speed of 4 miles per hour. What is Greg’s average speed, in miles per hour, for the round trip?

A. 5.5
B. 5.0
C. 4.8
D. 2.5
E. 2.4

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Greg is training for a marathon by running to and from work each day,  [#permalink]

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New post 02 Apr 2017, 12:42
SajjadAhmad wrote:
Greg is training for a marathon by running to and from work each day, a distance of 12 miles each way. He runs from home to work at an average speed of 6 miles per hour and returns at an average speed of 4 miles per hour. What is Greg’s average speed, in miles per hour, for the round trip?

A. 5.5
B. 5.0
C. 4.8
D. 2.5
E. 2.4


for two given speeds x and y the avg speed is calculated as 2xy/(x+y)

so Greg’s average speed = 2*6*4/(6+4) = 48/10 = 4.8 mph

Hence option C is correct
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Re: Greg is training for a marathon by running to and from work each day,  [#permalink]

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New post 04 Oct 2018, 19:40
SajjadAhmad wrote:
Greg is training for a marathon by running to and from work each day, a distance of 12 miles each way. He runs from home to work at an average speed of 6 miles per hour and returns at an average speed of 4 miles per hour. What is Greg’s average speed, in miles per hour, for the round trip?

A. 5.5
B. 5.0
C. 4.8
D. 2.5
E. 2.4


We can use the formula: average rate = total distance/total time:

average rate = 24/(12/6 + 12/3) = 24/5 = 4.8

Answer: C
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Re: Greg is training for a marathon by running to and from work each day,  [#permalink]

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New post 05 Oct 2018, 03:23
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It is common to solve a question such as this using logic that average speed should be average of two speeds, especially when the distance covered is the same. Right? Actually no.

The best way to solve such a question would be using the formula: (2*speed 1*speed 2) / (speed 1+ speed 2).

However, a slightly crude but effective technique that one can use ONLY if one is running out of time is to choose the option which is slightly less than the average of the speeds. This trick works only if the distance traveled is the same and if options give you a clear winner as it does in this case.

A) Greater than average: Incorrect
B) Equal to average: Incorrect
C) Slightly lesser than average: Correct
D) Way lesser than average: Incorrect
E) Way lesser than average: Incorrect

Hope this helps!
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Re: Greg is training for a marathon by running to and from work each day,  [#permalink]

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New post 05 Oct 2018, 14:51
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Hi All,

We're told that Greg is training for a marathon by running to and from work each day, a distance of 12 miles each way; he runs from home to work at an average speed of 6 miles per hour and returns at an average speed of 4 miles per hour. We're asked for Greg’s average speed, in miles per hour, for the round trip. This question can be solved in a couple of different ways. Sometimes the answer choices are 'spread out' in such a way that you can avoid almost all of the 'math' and use a 'logic shortcut' to get to the correct answer.

To start, since Greg is running the SAME distance in each direction, it will take him MORE time travel that distance at 4 miles/hour than it will take him to travel at 6 miles/hour. This makes this a 'Weighted Average' scenario - meaning that his average speed will be CLOSER to 4 miles/hour than it will be to 6 miles/hour. We can eliminate Answers A and B.

Since his speed never drops below 4 miles/hour, there's no way for his average speed to be less than that - so we can eliminate Answers D and E. There's only one answer remaining...

Final Answer:

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Re: Greg is training for a marathon by running to and from work each day, &nbs [#permalink] 05 Oct 2018, 14:51
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