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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]
Tough problem, but good explanation!
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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]
1
I am trying to solve it with a weighted average but it doesn't work. Can you please take a look at why my method doesn't work?

First, I find the time it takes to drive all 40 miles knowing that the average is 60.
40/60 = 2/3

Now I find the time it takes to pass the first 20 miles.
20/50 = 2/5

Then I find the time it takes to pass the rest 20 miles.

2/3 - 2/5 = 4/15

Now calculate using the weighted average:
(2/5)*50 + (4/15)*X = 60

20 + (4/15)*X = 60

(4/15)*X = 40

(1/15)*X = 10

X = 150

Where is my mistake here?
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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]
1
Paul121 wrote:
I am trying to solve it with a weighted average but it doesn't work. Can you please take a look at why my method doesn't work?

First, I find the time it takes to drive all 40 miles knowing that the average is 60.
40/60 = 2/3

Now I find the time it takes to pass the first 20 miles.
20/50 = 2/5

Then I find the time it takes to pass the rest 20 miles.

2/3 - 2/5 = 4/15

Now calculate using the weighted average:
(2/5)*50 + (4/15)*X = 60

20 + (4/15)*X = 60

(4/15)*X = 40

(1/15)*X = 10

X = 150

Where is my mistake here?




I think that I figured out my mistake. It should be:
(2/5)/(2/3)*50 + (4/15)/(2/3)*X = 60
Instead of:
(2/5)*50 + (4/15)*X = 60
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Re: A driver completed the first 20 miles of a 40-mile trip at a [#permalink]
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