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In a small snack shop, the average (arithmetic mean) revenue [#permalink]
22 Oct 2010, 05:49
Question Stats:
92% (01:53) correct
7% (00:45) wrong based on 56 sessions
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
Last edited by Bunuel on 09 Oct 2012, 03:55, edited 1 time in total.
Edited the question.
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Re: In a small snack shop......... [#permalink]
22 Oct 2010, 05:57
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, 05:58
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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]
22 Oct 2010, 06:00
Like I said, what Bunuel says is OA for me.
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Re: In a small snack shop......... [#permalink]
23 Oct 2010, 06:17
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]
08 Oct 2012, 22:33
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]
09 Oct 2012, 07:05
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]
15 Dec 2012, 08:47
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]
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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Re: In a small snack shop, the average (arithmetic mean) revenue
[#permalink]
16 Dec 2012, 04:08
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