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

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megafan
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A car travels from Town A to Town B at an average speed [#permalink] New post   05 Jan 2013, 18:08
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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
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PrashantPonde
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   05 Jan 2013, 18:47
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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, 02:07
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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, 01:22
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\(Average(speed) = \frac{{Total (Distance) }}{{ Total (Time)}}\)

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] New post   26 Jun 2017, 16:11
Pick a multiple of 40 and 50 as the distance. 200 is a good number.
Hence the average speed is 400/9 = 44 4/9, D
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   13 Aug 2017, 13:06
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


This is really like an approximation question and round trip question jumbled into one. Anyways, because this is a round trip question the distances for both trips are the same... mathematically that means

distance 1 = 40 (t)
distance 2= 50 (t)

We can just plug in a random value like 200

200= 40t
t=5

200=50t
t=4

Total Distance= Avgspeed * Total Time
400= avgsped* 9
400/9= 44.4444444.. round to closet value

44 4/9

D
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   24 Aug 2017, 22:11
A MOTOR TRAVEL WITH AN AVERAGE SPEED OF 49/HOUR FOR 3 HOURS THEN AN avreage speed of 99/ hour at spped of 39/ 2 hours
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   05 Dec 2017, 03:22
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egmat wrote:
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
When distance traveled in each segment is constant, then average speed simply be \(\frac{2xy}{x+y}\), where x and y are speeds.\(\frac{2*40*50}{40+50}\) = \(44\frac{4}{9}\)
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A car travels from Town A to Town B at an average speed [#permalink] New post   03 Nov 2019, 07:19
When the distance is constant on both directions of travel, then the formula 2*s1*s2/s1+s2 can be used where the s1 and s2 are the speeds. So here the solution would be 2*40*50/40+50 ===> 400/9. Converting into a mixed fraction will give us 44 4/9
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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   27 Aug 2020, 02:46
Given: 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.
Asked: What is the average speed in miles per hour for the round-trip?

Total distance = 2x
Total time = x/40 + x/50 = 9x/200
Average speed = 2x*200 / 9x = 400/9 = 44 4/9 mph

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Re: A car travels from Town A to Town B at an average speed [#permalink] New post   22 Oct 2023, 03:35
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