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Charlie drives 30 miles each way to work and back.

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Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 27 Feb 2017, 00:54
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Charlie drives 30 miles each way to work and back. If on his drive to work he averages 40 miles per hour, but on his return trip he can only manage an average of 20 miles per hour, what is his average rate for the day, in miles per hour?

(A) 22.5

(B) 26.67

(C) 30

(D) 32.5

(E) 33.33
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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 27 Feb 2017, 02:30
When distance s same bt speed varies..
Then avg speed s 2ab/(a+b)
So option B

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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 27 Feb 2017, 21:09
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Hi vikasp99,

This question is an example of a classic 'Weighted Average' prompt. The answer choices are sufficiently 'spaced out' that we can actually answer it without doing any serious math (although you could certainly calculate the exact answer if you wanted to).

IF.... we spent the exact same amount of TIME driving at the two speeds (40 mph and 20 mph), then the average speed would be (20+40)/2 = 30 mph. However, since we're driving the same DISTANCE at each of those two speeds, we'll end up spending MORE TIME at 20mph than we will at 40mph (but not that much more time though). Thus, the average has to be a "little closer" to 20 than it is to 40. There's only one answer that matches....

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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 28 Feb 2017, 20:28
vikasp99 wrote:
Charlie drives 30 miles each way to work and back. If on his drive to work he averages 40 miles per hour, but on his return trip he can only manage an average of 20 miles per hour, what is his average rate for the day, in miles per hour?

(A) 22.5

(B) 26.67

(C) 30

(D) 32.5

(E) 33.33


60 total miles/2.25 total hours=26.67 mph average rate
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Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 18 Mar 2017, 22:30
Average Speed = Total Distance/Total time = 60/(30/40+30/20) = 26.67.

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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 20 Mar 2017, 04:14
vikasp99 wrote:
Charlie drives 30 miles each way to work and back. If on his drive to work he averages 40 miles per hour, but on his return trip he can only manage an average of 20 miles per hour, what is his average rate for the day, in miles per hour?

(A) 22.5

(B) 26.67

(C) 30

(D) 32.5

(E) 33.33


time taken to go work = 30/40 = 0.75 hrs
time taken for coming back = 30/20 = 1.5 hrs

total time taken = 2.25 hrs

distance traveled = 30 +30 = 60

average speed for the day = 60/2.25 = 26.67mph
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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 22 Mar 2017, 08:37
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vikasp99 wrote:
Charlie drives 30 miles each way to work and back. If on his drive to work he averages 40 miles per hour, but on his return trip he can only manage an average of 20 miles per hour, what is his average rate for the day, in miles per hour?

(A) 22.5

(B) 26.67

(C) 30

(D) 32.5

(E) 33.33


We can use the formula for average rate:

average rate = total distance/total time

We know that the total distance is 30 x 2 = 60 miles. Since time = distance/rate, the time to work is 30/40 = 3/4 hour and the time from work is 30/20 = 3/2 hours. Thus,

average rate = (30 + 30)/(3/4 + 3/2)

average rate = 60/(3/4 + 6/4) = 60/(9/4) = 240/9 = 26.67

Answer: B
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Re: Charlie drives 30 miles each way to work and back.  [#permalink]

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New post 22 Mar 2017, 10:19
vikasp99 wrote:
Charlie drives 30 miles each way to work and back. If on his drive to work he averages 40 miles per hour, but on his return trip he can only manage an average of 20 miles per hour, what is his average rate for the day, in miles per hour?

(A) 22.5

(B) 26.67

(C) 30

(D) 32.5

(E) 33.33


\(Average \ speed = \frac{2*40*20}{(20+40)}\)

So, \(Average \ speed = \frac{2*40*20}{(20+40)}\)

Or, \(Average \ speed = \frac{1600}{60}\)

Or, \(Average \ speed = 26.67\)

Thus, answer must be (B) \(26.67\)
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Re: Charlie drives 30 miles each way to work and back. &nbs [#permalink] 22 Mar 2017, 10:19
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