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Math Expert V
Joined: 02 Sep 2009
Posts: 57244
M15-32  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 66% (02:01) correct 34% (01:43) wrong based on 73 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 V
Joined: 02 Sep 2009
Posts: 57244
Re M15-32  [#permalink]

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1
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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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Intern  B
Joined: 15 Feb 2018
Posts: 1
Re: M15-32  [#permalink]

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I think this is a high quality question, thanks
Intern  S
Status: Current Student
Joined: 27 Mar 2014
Posts: 26
Location: Bangladesh
Concentration: Entrepreneurship, Finance
M15-32  [#permalink]

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$$Rate*time=distance$$
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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