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Town A has 6,000 citizens and an average (arithmetic mean) o
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29 Aug 2018, 16:27
29 Aug 2018, 16:27
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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?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{13}{4}\)
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Town A has 6,000 citizens and an average (arithmetic mean) o
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30 Aug 2018, 10:16
30 Aug 2018, 10:16
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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?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: 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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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Town A has 6,000 citizens and an average (arithmetic mean) o
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30 Aug 2018, 10:24
30 Aug 2018, 10:24
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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?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: 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
Answer: 13/4
Cheers,
Brent
