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

Expert's post

00:00

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?

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?

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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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?

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

Answer: 13/4

Cheers, Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails