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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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Question Stats:

71% (01:16) correct 29% (01:37) wrong based on 75 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.

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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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Re M09-15  [#permalink]

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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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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]

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New post 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.

Answer is A.
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New post 23 Jan 2019, 13:18
Bought gmat club tests so that I could have explanations I could get grasp for such problems. May be it's my current level of quant, but I wasn't able to get the explanations for this problem at all.

Can anyone please help explain this step by step how this can be best solved?
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New post 23 Jan 2019, 21:36
nust2017 wrote:
Bought gmat club tests so that I could have explanations I could get grasp for such problems. May be it's my current level of quant, but I wasn't able to get the explanations for this problem at all.

Can anyone please help explain this step by step how this can be best solved?


The question is not that hard. The explanation provided above is clear and simple. So, if you cannot get it AT ALL, then it might be a good idea to go through basics and only then practice the questions.


16. Distance/Rate Problems



For more check below:
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Re: M09-15  [#permalink]

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New post 23 Jan 2019, 23:19
Bunuel wrote:
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.


formula to find avg speed when distance travelled is same:
2ab/ (a+b)

so
here we are given onward speed @ 60 kmph
and
#1
return speed @40kmph
avg speed : 2 * 60*40 / 100 = 48 kmph
sufficeint
#2

time of drive given ; in sufficient

IMO A
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Re: M09-15  [#permalink]

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New post 23 Jan 2019, 23:40
Bunuel wrote:
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.


MH = Mother's house, T = Town

T -> MH, speed was 60 km/h

MH -> T, speed was reduced.

Statement 1, gives reduced speed

Average speed = 2 * a * b/ (ab), we get the value sufficient

2) Just knowing the time wont help us.

Answer A
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Re: M09-15   [#permalink] 23 Jan 2019, 23:40
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