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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 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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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.
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Re: For the past n days, the average (arithmetic mean) daily [#permalink]

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

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 23 Oct 2014, 12:45
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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]

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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 [#permalink]

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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]

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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]

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New post 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 [#permalink]

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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]

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New post 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 [#permalink]

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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 [#permalink]

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New post 11 Dec 2016, 08:49
Here i my take on this one ->
Mean 1=50
Mean 2=55
Value added =100=50+40

Hence 40=5(n+1)
=>n=7

Hence E

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

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New post 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+55-50n= 90
5n=35
n=7

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

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New post 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 [#permalink]

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New post 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) 9


Hi Abhishek

There is a calculation error.
90-55=35 and not 45

Thus the answer is 7.

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

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New post 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 [#permalink]

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New post 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 :P

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 90-55 = +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   [#permalink] 24 Jul 2017, 09:21
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