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M09-15

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M09-15 [#permalink]

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New post 16 Sep 2014, 00:39
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In the morning, John drove to his mother's house in the village at an average speed of 60 kilometers per hour. When he was going back to town in the evening, he drove more cautiously and his speed was lower. If John went the same distance in the evening as in the morning, what was John's average speed for the entire trip?


(1) In the evening, John drove at a constant speed of 40 kmh.

(2) John's morning drive lasted 2 hours.
[Reveal] Spoiler: OA

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Re M09-15 [#permalink]

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New post 16 Sep 2014, 00:39
Official Solution:


The average speed equals to \(\frac{\text{total distance}}{\text{total time}}\).

Say the distance between the town and the village is \(d\) kilometers, and the speed from the village to town is \(x\) kilometers per hour, then \(\frac{\text{total distance}}{\text{total time}}=\frac{d+d}{\frac{d}{60}+\frac{d}{x}}\). \(d\) can be reduced, and we get \(\text{speed}=\frac{2}{\frac{1}{60}+\frac{1}{x}}\). So, as you can see we only need to find the average speed from the village to town.

(1) In the evening, John drove at a constant speed of 40 kilometers per hour. Sufficient.

(2) John's morning drive lasted 2 hours. We know nothing about his evening drive. Not sufficient.


Answer: A
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New post 18 Sep 2016, 13:18
I think this is a high-quality question and I agree with explanation.
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Re: M09-15 [#permalink]

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New post 20 Mar 2017, 16:13
How do we reduce the distance in the equation from d to 1?

Since d is an unknown it can be 1 or 2 or 3 hours, etc.
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Re: M09-15 [#permalink]

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New post 21 Mar 2017, 04:02
Prostar wrote:
How do we reduce the distance in the equation from d to 1?

Since d is an unknown it can be 1 or 2 or 3 hours, etc.


We are reducing the fraction by common term. You can factor out d both from the denominator and numerator and then reduce.

\(\frac{d+d}{\frac{d}{60}+\frac{d}{x}}=\frac{2d}{d*(\frac{1}{60}+\frac{1}{x})}=\frac{2}{\frac{1}{60}+\frac{1}{x}}\).

Hope it's clear.
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Re: M09-15   [#permalink] 21 Mar 2017, 04:02
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