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A car travels from Town A to Town B at an average speed

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A car travels from Town A to Town B at an average speed [#permalink] New post 05 Jan 2013, 17:08
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A car travels from Town A to Town B at an average speed of 40 miles per hour, and returns immediately along the same route at an average speed of 50 miles per hour. What is the average speed in miles per hour for the round-trip?

(A) \(45\frac{4}{9}\) mph

(B) 44 mph

(C) 45 mph

(D) \(44\frac{4}{9}\) mph

(E) \(44\frac{1}{9}\) mph
[Reveal] Spoiler: OA

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Re: A car travels from Town A to Town B at an average speed [#permalink] New post 05 Jan 2013, 17:47
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Using "Pick numbers" strategy, assume total distance as 200 miles. (LCM of 40 & 50)

Time taken to travel from A to B = 200/40 = 5 hours
Time taken to travel from B to A = 200/50 = 4 hours

\(Average(speed) = \frac{{Total (Distance) }}{{ Total (Time) }}= \frac{(200+200)}{(5+4)} = \frac{400}{9} = 44 \frac{4}{9}\)

Hence Choice(D).
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post 08 Jan 2013, 01:07
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megafan wrote:
A car travels from Town A to Town B at an average speed of 40 miles per hour, and returns immediately along the same route at an average speed of 50 miles per hour. What is the average speed in miles per hour for the round-trip?

(A) \(45\frac{4}{9}\) mph

(B) 44 mph

(C) 45 mph

(D) \(44\frac{4}{9}\) mph

(E) \(44\frac{1}{9}\) mph


Hi,

This is a classic question on Speed, time, & distance.

1. When time traveled in each segment is constant, then average speed is simple mean of speeds.

2. When distance traveled in each segment is constant, then average speed is reciprocal of simple mean of reciprocal of speeds. It is basically called Harmonic mean.

So this question falls in the category of 2.

=> So, average speed = Reciprocal of mean of reciprocals of 40 & 50.

=> Average speed = Reciprocal of mean of 1/40 & 1/50.

=> Average speed = Reciprocal of (1/40 + 1/50)/2 = Reciprocal of (5+4)/400 = 44 4/9 .

-Shalabh
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post 14 Jan 2015, 00:22
\(Average(speed) = \frac{{Total (Distance) }}{{ Total (Time)}}\)

Let the distance between Town A & Town B = 1

Setting up the equation:

Average speed \(= \frac{1+1}{\frac{1}{40} + \frac{1}{50}} = \frac{400}{9} = 44\frac{4}{9}\)
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Re: A car travels from Town A to Town B at an average speed   [#permalink] 14 Jan 2015, 00:22
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