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# Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios

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Math Expert
Joined: 02 Sep 2009
Posts: 58402
Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios  [#permalink]

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18 Dec 2018, 00:10
00:00

Difficulty:

5% (low)

Question Stats:

95% (01:18) correct 5% (00:57) wrong based on 44 sessions

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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?

A. 4/13
B. 9/13
C. 13/9
D. 13/4
E. 17/4

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Re: Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios  [#permalink]

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18 Dec 2018, 03:06
Bunuel 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?

A. 4/13
B. 9/13
C. 13/9
D. 13/4
E. 17/4

Total radio A = 6000 *2 = 12000

Total radio B = 10000 *4 = 40000

Average = 52000 / 16000 = 13 /4.

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Re: Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios  [#permalink]

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18 Dec 2018, 08:25
Bunuel 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?

A. 4/13
B. 9/13
C. 13/9
D. 13/4
E. 17/4

$$Required \ Average = \frac{6000*2 + 10000*4}{(6000 + 10000)}$$

Or, $$Required \ Average = \frac{52000}{16000}$$

Or, $$Required \ Average = \frac{13}{4}$$, Thus Answer must be (D)
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Re: Town A has 6,000 citizens and an average (arithmetic mean) of 2 radios   [#permalink] 18 Dec 2018, 08:25
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