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m12, #28

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m12, #28 [#permalink]

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New post 13 Mar 2010, 17:38
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If the average price of items is $125 and the average price of another items is $205, what is the average price of all items?

1. M+N=20
2. M=2N

[Reveal] Spoiler:
B


Not the most difficult question but, Im curious how they get B here.

I can plug in to the weighted avg formula and cancel out terms to get a # but wont the avg change with real values for M,N? Hence E?
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Re: m12, #28 [#permalink]

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New post 14 Mar 2010, 03:17
TheSituation wrote:
If the average price of items is $125 and the average price of another items is $205, what is the average price of all items?

1. M+N=20
2. M=2N

[Reveal] Spoiler:
B


Not the most difficult question but, Im curious how they get B here.

I can plug in to the weighted avg formula and cancel out terms to get a # but wont the avg change with real values for M,N? Hence E?


stmt1: M+N = 20
not sufficient as we have to know avg of all the items we need to know the value (125M+205N)/(M+N)

stmt2: replace m by 2n in the above equation
(250 N+205N)/3N and u can find the value by canceling.
Avg would always be in terms of the above equation only cos even if the real values change, they will be in the proportion.
Like 20, 40 or 30, 60. so always the multiple N will be canceled out.


This is why B is the answer
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Re: m12, #28 [#permalink]

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New post 14 Mar 2010, 03:35
sidhu4u wrote:
TheSituation wrote:
If the average price of items is $125 and the average price of another items is $205, what is the average price of all items?

1. M+N=20
2. M=2N

[Reveal] Spoiler:
B


Not the most difficult question but, Im curious how they get B here.

I can plug in to the weighted avg formula and cancel out terms to get a # but wont the avg change with real values for M,N? Hence E?


stmt1: M+N = 20
not sufficient as we have to know avg of all the items we need to know the value (125M+205N)/(M+N)

stmt2: replace m by 2n in the above equation
(250 N+205N)/3N and u can find the value by canceling.
Avg would always be in terms of the above equation only cos even if the real values change, they will be in the proportion.
Like 20, 40 or 30, 60. so always the multiple N will be canceled out.


This is why B is the answer






I did the same calculation

stmt1 is insufficient

stmt2 - my doubt here is "can N be a non real number?" as 455N/3N gives 151.6 (N and M should be real numbers)
Ex - Average of 10 numbers is 100 then the sum is 1000 (100*10) but how can one find average of 9.5 numbers

Can someone explain ?
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Re: m12, #28 [#permalink]

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New post 14 Mar 2010, 04:02
The number of some entities or items, has to be whole number. It cannot be in decimal. ther ecan be 10 elements in a set not 9.5
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Re: m12, #28 [#permalink]

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New post 15 Mar 2010, 22:55
sidhu4u wrote:
The number of some entities or items, has to be whole number. It cannot be in decimal. ther ecan be 10 elements in a set not 9.5



Thanks for your efforts sidhu
but if whatever you are saying is correct then explain what is the average price of M+N, i AM EXPECTING A EXACT NUMBER if you say B is the answer
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Re: m12, #28 [#permalink]

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New post 11 Mar 2012, 09:15
125n + 205m / m+n

"stmt2: replace m by 2n in the above equation
(250 N+205N)/3N and u can find the value by canceling."

if you replace 2n where there is an "m" in the numerator & denom, doesn't it come out to - 125n + 205(2n) / n + 2n ... ? I'm confused.

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Re: m12, #28   [#permalink] 11 Mar 2012, 09:15
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