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In a small snack shop, the average (arithmetic mean) revenue
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Updated on: 09 Oct 2012, 03:55

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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?

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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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.

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

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
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16 Dec 2012, 04:08

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).
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In a small snack shop, the average (arithmetic mean) revenue
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19 Aug 2015, 02:03

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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04 Sep 2015, 05:37

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?

Avg revenue per day = 400 Total revenue for 10 days = 4000

Avg Revenue for first 6 days = 360 Total revenue for first 6 days = 360 x 6 = 2160

Total revenue for 10 days = Total revenue for first 6 days + Total revenue for last 4 days 4000 = 2160 + X

X=1840, which is Total revenue for last 4 days

Avg revenue for last 4 days = \(\frac{1840}{4}\) = 460.
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Re: In a small snack shop, the average (arithmetic mean) revenue
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21 Jun 2016, 10:21

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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

This is a problem testing us on our knowledge of weighted averages. To solve, we can set up a weighted average equation:

(Revenue for first 6 days + Revenue for final 4 days)/10 days = 400

We know that the average revenue for the first 6 days was $360. We let r be the average daily revenue for the final four days, so we have

In a small snack shop, the average (arithmetic mean) revenue
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30 Jul 2016, 07:52

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Top Contributor

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

If the average (mean) daily revenue over 10 days is $400/day, then the TOTAL 10-DAY REVENUE = (10)($400) = $4000

The average daily revenue for the first 6 days is $360. So, the TOTAL 6-DAY REVENUE = (6)($360) = $2160

So, the TOTAL REVENUE for the last 4 days = $4000 - $2160 = $1840 So, the AVERAGE daily revenue for those 4 days = $1840/4 = $460

Re: In a small snack shop, the average (arithmetic mean) revenue
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01 Apr 2017, 09:36

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

Total revenue for 10 days= 400*10= 4000 Total revenue for 6 days= 360*6= 2160

Thus the total revenue for remaining 4 days would be 4000-2160=1840

Re: In a small snack shop, the average (arithmetic mean) revenue
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02 Jun 2017, 23:08

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

Total revenue 400* 10 = 4000 first 6 days = 360 *6 = 2160 rest of the four days = 1840 4x = 1840 X= 360

Re: In a small snack shop, the average (arithmetic mean) revenue
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26 Sep 2017, 00:00

The below is another method(easier) 400-360 = 40 for 6 days 40*6 = 240 this 240 need to be divide by remaining 4 days = 60 60 need to added to the average = 400+60 = 460 answer 460
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Re: In a small snack shop, the average (arithmetic mean) revenue
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21 Feb 2018, 21:03

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1

Hi All,

There are a number of ways to "do the math" on this type of question. Using the standard Average Formula will work perfectly here, and you'll be asked to use it a couple of times over the course of the Quant section, so you shouldn't try to avoid it.

That having been said, there's a weighted average "pattern" that can be used against this question:

The average of the 10 days = $400

The average for the first 6 days = $360. This means that - on each of the first 6 days - each day was $40 "behind" the average. To get the average "up" to $400, the remaining 4 days have to "make up" for that deficit…

6 x $40 = $240 "behind"

The last 4 days have to be above 400 by an amount that will make up for the $240.

$240/4 = $60. This means that each of the last 4 days needs to be $460 for the overall average to be $400.