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# For the past n days, the average (arithmetic mean) daily

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For the past n days, the average (arithmetic mean) daily [#permalink]  27 Dec 2012, 05:28
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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
[Reveal] Spoiler: OA
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Re: For the past n days, the average (arithmetic mean) daily [#permalink]  11 Aug 2015, 03:40
1
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$$\frac{50n + 90}{n+1} = 55$$

$$50n + 90 = 55n + 55$$

$$5n = 35$$

$$n = 7$$. Ans (E).
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Re: For the past n days, the average (arithmetic mean) daily [#permalink]  27 Dec 2012, 05:29
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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

(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.

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Re: For the past n days, the average (arithmetic mean) daily [#permalink]  18 Jul 2014, 10:52
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Re: For the past n days, the average (arithmetic mean) daily [#permalink]  21 Oct 2014, 12:25
Bunuel 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.

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 [#permalink]  21 Oct 2014, 13:13
Expert's post
trambn wrote:
Bunuel 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.

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 [#permalink]  23 Oct 2014, 11:45
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 [#permalink]  23 Oct 2014, 21:16
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 [#permalink]  15 Jan 2015, 04:50
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 [#permalink]  10 Aug 2015, 22:28
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 [#permalink]  11 Aug 2015, 02:30
55(n+1) - 50n = 90

5n = 35
n = 7
Re: For the past n days, the average (arithmetic mean) daily   [#permalink] 11 Aug 2015, 02:30
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