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In a small snack shop, the average (arithmetic mean) revenue [#permalink]
22 Oct 2010, 04:49

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In a small snack shop, the average (arithmetic mean) revenue was $400 per day over a 10-day period. During this period, if the average daily revenue was $360 for the first 6 days, what was the average daily revenue for the last 4 days?

Re: In a small snack shop......... [#permalink]
22 Oct 2010, 04:57

1

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pzazz12 wrote:

In a small snack shop, the average (arithmetic mean) revenue was $400 per day over a 10-day period. During this period, if the average daily revenue was $360 for the first 6 days, what was the average daily revenue for the last 4 days?

A. $420 B. $440 C. $450 D. $460 E. $480

D. $460 is my answer.

400 = x/10 x = 4000

For first 6 days = 360 * 6 = $2160

therefore for the last 4 days revenue is $4000 - $2160 = 3840 hence, airthmatic mean = 3840/4 = $460

Please let me know the OA. Also, I am very bad in explaining things. I am waiting for Bunuel to respond. He is extremly good in explaining solution. _________________

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Re: In a small snack shop......... [#permalink]
22 Oct 2010, 04:58

1

This post received KUDOS

Expert's post

pzazz12 wrote:

In a small snack shop, the average (arithmetic mean) revenue was $400 per day over a 10-day period. During this period, if the average daily revenue was $360 for the first 6 days, what was the average daily revenue for the last 4 days?

A. $420 B. $440 C. $450 D. $460 E. $480

The average revenue was $400 per day over a 10-day period --> total revenue over 10-day period is 10*$400=$4,000;

The average daily revenue was $360 for the first 6 days --> total revenue for the first 6 days is 6*$360=2,160;

Total revenue for the last 4 days is $4,000-2,160=$1,840;

Average daily revenue for the last 4 days is $1,840/4=$460.

Re: In a small snack shop......... [#permalink]
23 Oct 2010, 05:17

1

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I solved it in another way, but not sure whether its correct. Can some one help me by validating.. Let 'x' be the average daily revenue for last 4 days. [6(360)+4(x)]/10 = 400 solving, x = 460

Re: In a small snack shop......... [#permalink]
08 Oct 2012, 21:33

1

This post received KUDOS

weighted average

total 10 days, avg10 of this period is 400 first 6 days avg6 is 360 which is deltaAvg6 = 40 away from the overall avg10 last 4 day avg4 must have more weight to pull the avg10 to 400 (if 5 days-5 days then last 5 must have 440 so that avg10 is pulled back to 400, but this is 6/4, so we know last 4 days avg must have much more weight, so avg4 > 440, eliminate few choices)

since ratio of weight of 2 periods is 6/4 = 3/2, the avg4 must compensate for the loss of deltaAvg6 a 3/2 amount of deltaAvg6 so that the avg10 is pulled back to 400. therefore the last 4 days must have deltaAvg4 = 3/2 * deltaAvg6 = 3/2 * 40 = 60. so avg4 = 400 + deltaAvg4 = 400 + 60 = 460 --> D

Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
16 Dec 2012, 03:08

Ans: the total revenue for 10 days is 10x400=4000 and the total revenue for first 6 days is 6x360=2160, so the average for last 4 days will be (4000-2160)/4 = 460 (D). _________________

Re: In a small snack shop, the average (arithmetic mean) revenue [#permalink]
03 Oct 2014, 18:28

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In a small snack shop, the average (arithmetic mean) revenue [#permalink]
19 Aug 2015, 01:03

I've solved it using weighted average formula: So, 360 is 40 less then the average of 400 -> -40 * 6 Days; We have 4 Days and need to calculate the revenue =X

6*(-40) + 4*x=0 X=60, so it means that for the last 4 days the revenue was 60 above the average = 400+60 =460 (D)

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