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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
6*360+4*x=400*10
4x=400*10-9*360
x=1000-540=460
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
use ratio idea

-----360----------400------------x---------

ratio 360:x = 6:4 (days)

so their differences between 400 should be the reverse

which would be 40 and 60

x=460

solved in 10 secs
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
Hello together,

i have found a different approach for solving the question, which was not mentioned yet.

The question states that for the first 6 of 10 days the average per day was $360.
-> The difference from an avg. of $360 per day to an avg. of $400 per day is $40. ($400 - $360 = $40)
-> For 6 days the total difference is $240. (6*$40 = $240)

In order to get an avg. of $400 for all 10 days, the "missing" $240 must be "compensated"/included in the values for the last 4 days.
-> Now we can calculate the daily difference for each of the last 4 days to the avg. of $400.

(A) $420-$400 = $20 => $20*4 = $80
(B) $440-$400 = $40 => $40*4 = $160
(C) $450-$400 = $50 => $50*4 = $200
(D) $460-$400 = $60 => $60*4 = $240 is the right answer
(E) $480-$400 = $80 => $80*4 = $320

=> Since the answer choices are ordered from least to greatest, you could quicken up the approach.
-> First try out answer choice (C). Since the sum of the differences of is $200 and $200 < $240, the avg. for the last 4 days must be > $450
-> Try out the next larger answer choice (D). Sum of the differences for (D) = $240, which is the right answer.
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
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Trog wrote:
Hello together,

i have found a different approach for solving the question, which was not mentioned yet.

The question states that for the first 6 of 10 days the average per day was $360.
-> The difference from an avg. of $360 per day to an avg. of $400 per day is $40. ($400 - $360 = $40)
-> For 6 days the total difference is $240. (6*$40 = $240)

In order to get an avg. of $400 for all 10 days, the "missing" $240 must be "compensated"/included in the values for the last 4 days.
-> Now we can calculate the daily difference for each of the last 4 days to the avg. of $400.

(A) $420-$400 = $20 => $20*4 = $80
(B) $440-$400 = $40 => $40*4 = $160
(C) $450-$400 = $50 => $50*4 = $200
(D) $460-$400 = $60 => $60*4 = $240 is the right answer
(E) $480-$400 = $80 => $80*4 = $320

=> Since the answer choices are ordered from least to greatest, you could quicken up the approach.
-> First try out answer choice (C). Since the sum of the differences of is $200 and $200 < $240, the avg. for the last 4 days must be > $450
-> Try out the next larger answer choice (D). Sum of the differences for (D) = $240, which is the right answer.


Hi Trog,

Yes - this method absolutely works here (it's sometimes referred to as "allegation", but it's essentially based on the same patterns as a Weighted Average) - and this approach is described in a post on Page 1 of this thread.

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Rich

Contact Rich at: Rich.C@empowergmat.com
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
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­Mixture in reverse. Same concept though:;

­
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
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