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Average walking rate and logic behind it

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Average walking rate and logic behind it  [#permalink]

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New post 13 Jun 2016, 13:06
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Hi,

I'm having a hard time understanding the reason for something.

"If Lucy walks to work at a rate of 4 mph, and she walks home by the same route at a rate of 6 mph, what is Lucy's average walking rate for the round trip ?"

So the problem never establishes a specific distance. Apparently, her average walking rate doesn't depend on the distance (I tried with some different values for the distance and yeah I got the same answer every time). But why ? What's the logic behind that ? I can't seem to get it.

Please help ! :shock:
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Re: Average walking rate and logic behind it  [#permalink]

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New post 13 Jun 2016, 13:29
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The main idea is that the person traveling a certain distance at 4 mph will take 50% longer than a person traveling the same distance at 6 mph. So, the distance is irrelevant.
This video covers average speeds in greater detail:


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Brent
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Re: Average walking rate and logic behind it  [#permalink]

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New post 13 Jun 2016, 16:50
jbyx78 wrote:
Hi,

I'm having a hard time understanding the reason for something.

"If Lucy walks to work at a rate of 4 mph, and she walks home by the same route at a rate of 6 mph, what is Lucy's average walking rate for the round trip ?"

So the problem never establishes a specific distance. Apparently, her average walking rate doesn't depend on the distance (I tried with some different values for the distance and yeah I got the same answer every time). But why ? What's the logic behind that ? I can't seem to get it.

Please help ! :shock:


Hi! There are two ways.

1st Way- Let's suppose distance = d, so the total distance going to work and coming back = d+d= 2d
Total time= Time to go to work +time to come back = d/4+d/6

Average speed= total distance/total time

2d/d/4+d/6= 2*4*6/10= 4.5 m/h

2nd way- when the distance travelled is same, we can apply direct formula 2S1S2/S1+S2= 2*4*6/4+6= 4.5m/h

Hope it helps.
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Re: Average walking rate and logic behind it  [#permalink]

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New post 13 Jun 2016, 20:01
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jbyx78 wrote:
Hi,

I'm having a hard time understanding the reason for something.

"If Lucy walks to work at a rate of 4 mph, and she walks home by the same route at a rate of 6 mph, what is Lucy's average walking rate for the round trip ?"

So the problem never establishes a specific distance. Apparently, her average walking rate doesn't depend on the distance (I tried with some different values for the distance and yeah I got the same answer every time). But why ? What's the logic behind that ? I can't seem to get it.

Please help ! :shock:


Since you are asking about logic...
the reason is that the Distance being travelled is in terms of same variable say D, which finally gets eliminated

you are travelling the same distance say D in two different speeds ..
so total time = D/4 + D/6 = 5D/12..

and your total distance becomes 2D..... but time is 5D/12..
speed = \(2D/\frac{5D}{12}= 24/5 = 4.8mph\)...
so you see here that the Distance D is eliminated being common to numerator and denominator.....

Of course direct method is as also given by Divyadisha = 2ab/(a+b) = 2*4*6/10 = 4.8mph...
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Re: Average walking rate and logic behind it  [#permalink]

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New post 13 Jun 2016, 20:25
jbyx78 wrote:
Hi,

I'm having a hard time understanding the reason for something.

"If Lucy walks to work at a rate of 4 mph, and she walks home by the same route at a rate of 6 mph, what is Lucy's average walking rate for the round trip ?"

So the problem never establishes a specific distance. Apparently, her average walking rate doesn't depend on the distance (I tried with some different values for the distance and yeah I got the same answer every time). But why ? What's the logic behind that ? I can't seem to get it.

Please help ! :shock:


The issue is that distance eventually cancels out. Let us do it by using algebra

Suppose the distance from Lucy's house to office is "d". Obviously she covers same distance when she returns from her office to her house.

Time taken to reach office: (d/4) hours [assume d is in miles]

Time taken to return: (d/6) hours

Total Distance covered: 2d
Total time: \(\frac{d}{4}+\frac{d}{6}\) = \(\frac{5d}{12}\)

Average speed: 2d/ (5d/12) = [(2 * 12)/ 5]--> Note that d cancels out
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Re: Average walking rate and logic behind it  [#permalink]

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New post 14 Jun 2016, 13:58
Imagine that you have two jobs. At one job, you work for two days a week at $60 an hour, and at the other job, you work for three days a week at $40 an hour. You'll make a certain amount of money per week, and you can also work out the average amount that you make per hour over the whole week. After one week, you'll have made a certain average number of dollars per hour (somewhere between 40 and 60). After two weeks, that average will be the same, as long as your hours stay the same. After three weeks, same thing. You won't suddenly start making more or less money on average, no matter how many weeks you work for in total, unless your hours change.

In this analogy, the $60/hour job corresponds to walking faster, and the $40/hour job corresponds to walking slower. As long as the amount of time you spend walking faster and the amount of time you spend walking slower stay in the same proportion to each other, your average speed will always stay the same. It doesn't matter whether you walk for a total of one minute, ten minutes, or a week. What matters is that out of every minute (or hour, or whatever), you spend a fixed part of it walking more slowly on average, and a fixed part of it walking more quickly.

A second aspect of this problem that's hard to understand intuitively, is why the result comes out closer to 4mph than to 6mph. It seems like it should be 5mph, logically. The reason it isn't is because you actually spend significantly more time walking at 4mph, so the 4mph affects the average more. To take it to an extreme, imagine taking a 3,000 mile trip in an airplane going one way at 500mph, then getting off the plane and returning on foot at 3mph. You couldn't logically say that you averaged (3000+3)/2 = 1501.5 mph, because that would be a very fast average speed. In fact, your average speed would be very slow, which makes intuitive sense, since the entire trip would clearly take a very long time.
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Re: Average walking rate and logic behind it   [#permalink] 14 Jun 2016, 13:58

Average walking rate and logic behind it

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