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# Town A has 6,000 citizens and an average (arithmetic mean) o

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Town A has 6,000 citizens and an average (arithmetic mean) o [#permalink]  29 Aug 2018, 16:27
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Question Stats:

100% (00:53) correct 0% (00:00) wrong based on 6 sessions
Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios per citizen. Town B has 10,000 citizens and an average of 4 radios per citizen. What is the average number of radios per citizen in both towns combined?

[Reveal] Spoiler: OA
$$\frac{13}{4}$$

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Sandy
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Re: Town A has 6,000 citizens and an average (arithmetic mean) o [#permalink]  30 Aug 2018, 10:16
Expert's post
sandy wrote:
Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios per citizen. Town B has 10,000 citizens and an average of 4 radios per citizen. What is the average number of radios per citizen in both towns combined?

[Reveal] Spoiler: OA
$$\frac{13}{4}$$

One approach is to use weighted averages

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

There are 16,000 in total

Town A's proportion = 6,000/16,000 = 6/16 = 3/8
Town A's average = 2

Town B's proportion = 10,000/16,000 = 10/16 = 5/8
Town B's average = 4

Apply the weighted averages formula to get:
Average = (3/8)(2) + (5/8)(4)
= 6/8 + 20/8
= 26/8

NOTE: For Numeric Entry questions on the GRE, you are NOT required to express fractions in simplest terms.
So, the fraction 26/8 is considered correct.

Alternatively, you COULD reduce 26/8 to get 13/4, which is also correct.

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Brent Hanneson – Creator of greenlighttestprep.com

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Re: Town A has 6,000 citizens and an average (arithmetic mean) o [#permalink]  30 Aug 2018, 10:24
Expert's post
sandy wrote:
Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios per citizen. Town B has 10,000 citizens and an average of 4 radios per citizen. What is the average number of radios per citizen in both towns combined?

[Reveal] Spoiler: OA
$$\frac{13}{4}$$

Another approach is to first determine the total number of radios.

Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios per citizen.
Given: (total number of radios in Town A)/6000 = 2
Multiply both sides of the equation by 6000 to get: total number of radios in Town A = 12000

Town B has 10,000 citizens and an average of 4 radios per citizen
Given: (total number of radios in Town B)/10000 = 4
Multiply both sides of the equation by 10000 to get: total number of radios in Town B = 40000

What is the average number of radios per citizen in both towns combined?
TOTAL number of radios in BOTH towns = 12000 + 40000 = 52000
TOTAL number of people in BOTH towns = 6000 + 10000 = 16000

So, average number of radios per citizen = 52000/16000
= 52/16
= 26/8
= 13/4

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: Town A has 6,000 citizens and an average (arithmetic mean) o [#permalink]  31 Aug 2018, 20:43
We can use the weighted average approach:
nx+my/n+m
where n=6000,m=10000,x=2,y=4
Re: Town A has 6,000 citizens and an average (arithmetic mean) o   [#permalink] 31 Aug 2018, 20:43
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