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

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

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New post Updated on: 09 Oct 2012, 02:55
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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?

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

Originally posted by pzazz12 on 22 Oct 2010, 04:49.
Last edited by Bunuel on 09 Oct 2012, 02:55, edited 1 time in total.
Edited the question.
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Re: In a small snack shop.........  [#permalink]

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New post 22 Oct 2010, 04:58
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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


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.

Answer: D.
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Re: In a small snack shop.........  [#permalink]

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New post 22 Oct 2010, 04:57
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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]

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New post 22 Oct 2010, 05:00
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Like I said, what Bunuel says is OA for me. :)
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Re: In a small snack shop.........  [#permalink]

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New post 23 Oct 2010, 05:17
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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

Thanks in advance
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Re: In a small snack shop.........  [#permalink]

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New post 08 Oct 2012, 21:33
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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
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 09 Oct 2012, 06:05
460
400 x 10= 4000
360 x 6 = 2160
4000-2160 = 1840/4=460
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 15 Dec 2012, 07:47
Bunuel's way of solving is definitely the most straight forward one; thanks!
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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

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New post 03 Jul 2013, 09:28
(400-360)*6/4=60
400+60=460
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In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 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  [#permalink]

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New post 04 Sep 2015, 04:37
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  [#permalink]

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New post 21 Jun 2016, 09:21
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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

(360 x 6 + 4r)/10 = 400

2,160 + 4r = 4,000

4r = 1,840

r = 460

Answer is D.
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In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 30 Jul 2016, 06:52
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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


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

Answer:

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

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New post 01 Apr 2017, 08:36
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

The average revenue for 4 days= 1840/4= 460
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 02 Jun 2017, 22:08
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
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 25 Sep 2017, 23:00
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  [#permalink]

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New post 18 Oct 2017, 10:51
360*6 +4x= 400*10
4x= 1840
x= 460
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Re: In a small snack shop, the average (arithmetic mean) revenue  [#permalink]

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New post 21 Feb 2018, 20:03
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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.

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

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New post 17 Aug 2018, 11:30
Hello,

To avoid calculations we can use deviation method

400=360+(4x/10)
4x= 400
x= 100
=360+100= 460
hence option D
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Re: In a small snack shop, the average (arithmetic mean) revenue   [#permalink] 17 Aug 2018, 11:30

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