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M15-32

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M15-32  [#permalink]

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New post 15 Sep 2014, 23:57
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  65% (hard)

Question Stats:

65% (02:01) correct 35% (01:43) wrong based on 71 sessions

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If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?

A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 kmh

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Re M15-32  [#permalink]

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New post 15 Sep 2014, 23:57
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Official Solution:

If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?

A. 36 kmh
B. 40 kmh
C. 42 kmh
D. 44 kmh
E. 50 kmh


Say the total distance is \(3x\) kilometers, then: \(total \ time=\frac{x}{80}+\frac{x}{24}+\frac{x}{48}=\frac{18x}{240}\);

\(Average \ speed=\frac{total \ distance}{total \ time}=\frac{3x}{\frac{18x}{240}}=\frac{240*3}{18}=40\).


Answer: B
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Collection of Questions:
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Re: M15-32  [#permalink]

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New post 24 Feb 2018, 08:58
I think this is a high quality question, thanks
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M15-32  [#permalink]

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New post 24 Feb 2018, 17:41
\(Rate*time=distance\)
So, \(80*t=\frac{1}{3} => t1=\frac{1}{3}*80\)
Similarly, \(t2=\frac{1}{3}*24\) and \(t3=\frac{1}{3}*48\)
So, total time= \(t1+t2+t3 = \frac{1}{3}(\frac{1}{80}+\frac{1}{24}+\frac{1}{48}) = \frac{1}{3}*\frac{1}{8}(\frac{1}{10}+\frac{1}{3}+\frac{1}{6}) = \frac{1}{24}*\frac{18}{30} = \frac{1}{40}\)
Average speed = \(\frac{(total distance)}{(total time)}= \frac{1}{1/40}\) = 40 kph
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M15-32 &nbs [#permalink] 24 Feb 2018, 17:41
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