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# M09-15

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Math Expert
Joined: 02 Sep 2009
Posts: 39723

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16 Sep 2014, 00:39
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Difficulty:

25% (medium)

Question Stats:

71% (01:43) correct 29% (00:56) wrong based on 97 sessions

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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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Math Expert
Joined: 02 Sep 2009
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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.

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Joined: 27 Mar 2014
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18 Sep 2016, 13:18
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 25 Jan 2016
Posts: 8
Location: United States (NJ)
GPA: 2.34
WE: Analyst (Commercial Banking)

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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.
Math Expert
Joined: 02 Sep 2009
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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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Joined: 01 Nov 2016
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08 May 2017, 11:24
The distance/rate eqation is: distance = rate * time. We are trying to find rate, so change it to: distance/time = rate

D/T1 = 60
D/T2 = R2
(2D)/(T1+T2) = Average Rate

Just to make the problem easier let's just say that the distance is 100 miles. This will make T1 = 5/3. Average rate will equal 200/(5/3 + T2). We can find average rate as long as we find T2, or the amount of time it took for the evening trip.

1) R2 = 40.
If R2 = 40 and D = 100 then you can find T2. T2 will be 5/2. With T2 you can find average rate which will be 200/(5/3+5/2) = 200/(13/3) = 600/13 which is about 46.15 mph

2) T1 = 2
With T1 we can find the actual D but we still don't know T2. Insufficient.

Re: M09-15   [#permalink] 08 May 2017, 11:24
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# M09-15

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