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A bakery sold an average(arithmetic mean) of 820 cookies per day in an

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A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 18 Oct 2016, 20:37
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A bakery sold an average(arithmetic mean) of 820 cookies per day in an operating period. On "good" days, the bakery sold an average of 980 cookies per day, and on "fair" days, the bakery sold an average of 640 cookies per day. If every day in the operating period was either "good" or "fair", what was the ratio of the number of "good" days to the number of "fair" days for the bakery's operating period?


a) 2:1
b) 3:2
c) 5:4
d) 7:6
e) 9:8

Please assist with above problem.
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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 18 Oct 2016, 21:31
2
Number of good days --> x
Number of bad days --> y

980x + 640y = 820(x + y)
160x = 180y
x/y = 9/8

Answer: E
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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 18 Oct 2016, 21:32
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1
Average No of cookies : 820

No of cookies sold on Good days = 980

No of cookies sold on fair days = 640

There are two methods to do it:

1. Allegation method - Here we can consider that we are mixing two things with concentration 980 and 640 and getting final concentration of 640.

Pls see attached doc for reference:



2. Method two is algebraic approach, consider no of good days as X and fair days as Y

So total no of cookies sold would be 980*X + 640*Y

we know average = 820 = Total no of cookies sold/Total no of days

= (980*X + 820*Y)/(X + Y) = 820

On solving this we will get 16X = 18Y

which is X/Y = 18/16 = 9/8

HTH
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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 09 Dec 2016, 02:05
Great Question this one.
Here is what i did in this Question-->

Let number of good days be G
Number of Bad days be B

Using=>

\(Mean =\frac{Sum}{#}\)



Sum of Cookies sold on Bad days =\(920G\)
Sum of Cookies sold on Fair days =\(640F\)

Now the Average of the entire operation => \(\frac{920G+640F}{G+F} =820\)

Solving it we get => \(16G=18F\)
hence \(G/F=9/8\)

Hence E

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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 09 Dec 2016, 10:55
W1
980

W2
640

Aavg
820

(W1-Aavg) / (Aavg-W2) = Ratio

(980-820) / (820 - 640)

160 / 180

8 / 9 Reverse it. 9:8
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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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New post 12 Dec 2016, 18:10
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alanforde800Maximus wrote:
A bakery sold an average(arithmetic mean) of 820 cookies per day in an operating period. On "good" days, the bakery sold an average of 980 cookies per day, and on "fair" days, the bakery sold an average of 640 cookies per day. If every day in the operating period was either "good" or "fair", what was the ratio of the number of "good" days to the number of "fair" days for the bakery's operating period?


a) 2:1
b) 3:2
c) 5:4
d) 7:6
e) 9:8


We are given that a bakery sold an average (arithmetic mean) of 820 cookies per day in an operating period, and that on good days the bakery sold an average of 980 cookies per day, and on fair days the bakery sold an average of 640 cookies per day.

If we let G = the number of good days, and F = the number of fair days, we can create the following equation:

820 = (980G + 640F)/(G + F)

820G + 820F = 980G + 640F

180F = 160G

9F = 8G

G/F = 9/8

Answer: E
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Re: A bakery sold an average(arithmetic mean) of 820 cookies per day in an  [#permalink]

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