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# M15-32

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Math Expert
Joined: 02 Sep 2009
Posts: 58098

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16 Sep 2014, 00:57
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Difficulty:

65% (hard)

Question Stats:

65% (02:05) correct 35% (01:42) wrong based on 75 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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Math Expert
Joined: 02 Sep 2009
Posts: 58098

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16 Sep 2014, 00:57
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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: $$total \ time=\frac{x}{80}+\frac{x}{24}+\frac{x}{48}=\frac{18x}{240}$$;

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

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Joined: 15 Feb 2018
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24 Feb 2018, 09:58
I think this is a high quality question, thanks
Intern
Status: Current Student
Joined: 27 Mar 2014
Posts: 26
Concentration: Entrepreneurship, Finance

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24 Feb 2018, 18:41
$$Rate*time=distance$$
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
M15-32   [#permalink] 24 Feb 2018, 18:41
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# M15-32

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