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# In a small snack shop, the average (arithmetic mean) revenue

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

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com
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Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]