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For the past n days, the average (arithmetic mean) daily
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27 Dec 2012, 06:28
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For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ? (A) 30 (B) 18 (C) 10 (D) 9 (E) 7
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Re: For the past n days, the average (arithmetic mean) daily
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27 Dec 2012, 06:29




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Re: For the past n days, the average (arithmetic mean) daily
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11 Aug 2015, 04:40
\(\frac{50n + 90}{n+1} = 55\) \(50n + 90 = 55n + 55\) \(5n = 35\) \(n = 7\). Ans (E).
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Re: For the past n days, the average (arithmetic mean) daily
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21 Oct 2014, 13:25
Bunuel wrote: Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 (average production for n days) * n = (total production for n days) > 50n=(total production for n days); (total production for n days) + 90 = (average production for n+1 days) * (n+1) > 50n + 90 = 55 * (n+1) > n=7. Or as 40 extra units increased the average for n+1 days by 5 units per day then 40/(n+1)=5 > n=7. Answer: E. Could you explain why there is n+1 days? I don't quite get the logic here. Thanks in advance.



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Re: For the past n days, the average (arithmetic mean) daily
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21 Oct 2014, 14:13
trambn wrote: Bunuel wrote: Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 (average production for n days) * n = (total production for n days) > 50n=(total production for n days); (total production for n days) + 90 = (average production for n+1 days) * (n+1) > 50n + 90 = 55 * (n+1) > n=7. Or as 40 extra units increased the average for n+1 days by 5 units per day then 40/(n+1)=5 > n=7. Answer: E. Could you explain why there is n+1 days? I don't quite get the logic here. Thanks in advance. The past n days plus today, so one more day = n + 1 days.
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Re: For the past n days, the average (arithmetic mean) daily
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23 Oct 2014, 12:45
For past n days, average daily production is 50. Today's average production of 90 unit raises average for (n+1)days by 55.
Weight Average formula.
(n*daily average+1*today's average) = 55 (n+1)
n*55 + 1*90 = 55 (n+1)
n=7
OA is E



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Re: For the past n days, the average (arithmetic mean) daily
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23 Oct 2014, 22:16
Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 For "N" days avg production was 50 Units Today production of 90 Units raises avg to 55 Units So 90 Units of today raises each of N days avg unit by 5 Units + 55 Units of today. Considering that if 55 Units were produced today then remaining 35 Units of today add up to raise value from 50 Units to 55 Units. Implies that if 5 units each are given to "N" days then the avg will be 55 Units So, 35 / 5 = 7 Therefore "N" = 7 Ans E



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For the past n days, the average (arithmetic mean) daily
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15 Jan 2015, 05:50
Sort of same as above, but let me explain:
Start by writing down the formula for the mean: M=S/N
From stem: M=50 N=n days S= unknown, but using the formula for the mean we have 50=S/n, S=50n.
So, it becomes: 50 = (50n)/n, which btw makes sense if you look at it matheatically too.
Now, we know that the new M=55 and the new n=one more than before, so new n=n+1. The new S is 90 plus the S we had before, so 50n+90.
Side by side:
50 = (50n)/n VS 55= (50n+90)/n+1
Use 55= (50n+90)/n+1 to solve for n:
55 = (50n+90)/n+1 50n+90 = 55n + 55 90  55 = 5n 35 = 5n n = 7 ANS E



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Re: For the past n days, the average (arithmetic mean) daily
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10 Aug 2015, 23:28
I thought about this like a weighted average problem. Since 90 (the last day; weight of 1) is +35 above 55 (the new average) and 50 (all the previous days; weight of n) is 5 below 55. In order to reach the new average, the sum of everything above and below 55 must equal to 0, so...
+35(1) + 5n = 0 n = 7
(E)



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Re: For the past n days, the average (arithmetic mean) daily
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11 Aug 2015, 03:30
55(n+1)  50n = 90
5n = 35 n = 7



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Re: For the past n days, the average (arithmetic mean) daily
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26 Feb 2016, 11:44
n/new one = 35/5 hence n/new one = 7/1
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Re: For the past n days, the average (arithmetic mean) daily
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13 Jul 2016, 07:45
Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 We start with the formula for calculating the average. average = sum/quantity We are given that the initial quantity is n days and that the average for those n days is 50 units. We can use this to determine a sum for the n days. Since sum = average x quantity, we know: sum = (50) (n) = 50n We are next given that today, the production was 90 units. This means we can add “1” to the current quantity and “90” to the current sum. We are also given that this new average is 55 units per day. Substituting this information into the average formula we get: 55 = (50n + 90)/(n + 1) 55(n + 1) = 50n + 90 55n + 55 = 50n + 90 5n = 35 n = 7 days The answer is E.
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Re: For the past n days, the average (arithmetic mean) daily
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Re: For the past n days, the average (arithmetic mean) daily
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01 Apr 2017, 09:44
Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 Total production for past n days= 50n The new average after today's production= 55(n+1) (past n days+today's production) 55n+5550n= 90 5n=35 n=7



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Re: For the past n days, the average (arithmetic mean) daily
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01 Apr 2017, 23:56
Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 \(\frac{50n + 90}{n + 1}= 55\) \(50n + 90 = 55n + 55\) \(5n = 45\) So, \(n = 9\) Thus, answer must be (D) 9
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Re: For the past n days, the average (arithmetic mean) daily
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02 Apr 2017, 06:25
Abhishek009 wrote: Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 \(\frac{50n + 90}{n + 1}= 55\) \(50n + 90 = 55n + 55\) \(5n = 45\) So, \(n = 9\) Thus, answer must be (D) 9Hi Abhishek There is a calculation error. 9055=35 and not 45 Thus the answer is 7.



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Re: For the past n days, the average (arithmetic mean) daily
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17 May 2017, 11:43
So we have average for n days = 50 in 1 day we added production of 90 units and the average became 55, average = sum/number or a=s/n s=a*n or = 50n so the sum after adding 90 units in 1 day => s=55(n+1) we can also write new sum as follows: 50n + 90 and it must be equal 55(n+1) => 50n+90 = 55(n+1) => 50n+90 = 55n + 55=> 35=5n => n=7. Answ. E.



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Re: For the past n days, the average (arithmetic mean) daily
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24 Jul 2017, 09:21
My 50 cents here. I guess Using the Differential way (Or thinking of average in terms of Sums) is the quickest way to answer this questions. Also the scope of error is less. Pardon me if the explanation looks abstruse Given: Average of n days is 50. Production of 90 on (n+1)th day "pulls" the average of n+1 days to 55. In terms of overall sum of n+1 what this means is  for n days we lost 50  55 = 5 per day on the total sum needed to make the average of 55. Meaning, over n days we lost a total of 5n that would be needed to make the average of 55. The 90 on the (n+1)th day makes up for all the Lost sum and brings the average to 55. Meaning, a value of 9055 = +35 will offset all the sum that we lost in last n days. Mathematically we can simply write 5n+35 = 0 => n = 7. Hope this makes sense and als helps
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Re: For the past n days, the average (arithmetic mean) daily
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19 Feb 2018, 13:35
Hi All, While I'm a fan of the algebraic approach for this type of question (because it's just about using the "average formula" and doing a bit of algebra), you have other ways to get to the correct answer. You could TEST THE ANSWERS. Here's how: One of the 5 answers must be the number of days that will "fit" the given information in the question stem and lead to the end result that's written. Let's start with answer D… If N = 9 then…. 9(50) + 90 = 540 over 10 days 540/10 = average = 54 The prompt says that the average IS 55 though, so we're really close, but D is incorrect. It's got to be C or E. Most people would say that Answer C looks easier to deal with, so we can test that one next…. If N=10 then… 10(50) + 90 = 590 over 11 days 590/11 = average = a little over 53…. This isn't the answer EITHER. The average appears to be getting smaller, so E is probably the correct answer. If N=7 then…. 7(50) + 90 = 440 over 8 days 440/8 = average = 55 THAT is the MATCH. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: For the past n days, the average (arithmetic mean) daily
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21 May 2018, 03:53
Walkabout wrote: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n ?
(A) 30 (B) 18 (C) 10 (D) 9 (E) 7 WKT the sum/n = Average Given for n days the average production was 50 there sum/n = 50 or sum = 50n ...(1) If we include todays production the number of days will increase by 1 hence the n+1 Sum would increase by 90 new average is 65 therefor sum/n=average would imply sum+90/n+1 = 55 ...(.2 ) Substitue 1 in 2 and solve 50n + 90 = 55n + 55 0r 5n = 35 implies n = 7




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