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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 ?

Re: For the past n days, the average (arithmetic mean) daily [#permalink]
27 Dec 2012, 05:29

Expert's post

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This post was BOOKMARKED

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.

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:

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.

Re: For the past n days, the average (arithmetic mean) daily [#permalink]
21 Oct 2014, 13:13

Expert's post

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. _________________

Re: For the past n days, the average (arithmetic mean) daily [#permalink]
23 Oct 2014, 21:16

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

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...

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