stats+applied problems

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stats+applied problems

by romitvsingh » Fri Nov 25, 2011 11:35 pm
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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by shankar.ashwin » Fri Nov 25, 2011 11:41 pm
We can use allegations to solve this.

50 and 90 are mixed to produce 55. Ratio in which they are mixed is;

(90-55)/(55-50) = 35/5 = 7/1 (We know 90 units is for 1 day, so we simplify the fraction to 7/1)

E IMO

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by chufus » Sat Nov 26, 2011 4:27 am
I think the solution is E. 7

Here is how

Number of Days = n
Average daily production = 50
So in n days total production = 50n
Today's production = 90
New average = 55

So:

(50n + 90)/(n+1) = 55

50n + 90 = 55n + 55

5n = 35

And hence n = 7

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by Abhishek009 » Sat Nov 26, 2011 9:07 am
romitvsingh 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 after n days is 50n

If avg production increases to 55 units , total production in n days is 55n


If today's production of 90 units raises the average to 55 units per day

Total Production of 55 days - Total Production of 50 days = 90

55 n - 50 n = 90

5n = 90

n = 18


Cross check and find out ...

55*18 - 50*18 = 90
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by GMATGuruNY » Sun Nov 27, 2011 4:32 am
romitvsingh 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
This is a weighted average problem.
A production rate of 50 is being combined with a production rate of 90 to yield an average production rate of 55.
We can use alligation, which dictates the following:

The proportion needed of each production rate is equal to the distance between the OTHER two production rates.

Proportion needed of the 50/day rate = |55-90| = 35.
Proportion needed of the 90/day rate = |55-50| = 5.
50/day rate : 90/day rate = 35:5 = 7:1.

Since 7 days at 50/day are needed for every 1 day at 90/day, n=7.

The correct answer is E.
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by LalaB » Sun Nov 27, 2011 5:17 am
Abhishek009 wrote:If today's production of 90 units raises the average to 55 units per day

Total Production of 55 days - Total Production of 50 days = 90

55 n - 50 n = 90
u forgot that u have not n ,but n+ 1 days

55(n+1)-50n=90
55n +55-50n=90
5n=35
n=7