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The average (arithmetic mean) age of the buildings on a cert

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The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 30 Aug 2018, 07:46
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The average (arithmetic mean) age of the buildings on a certain city block is greater than 40 years old. If four of the buildings were built two years ago and none of the buildings are more than 80 years old, which of the following could be the number of buildings on the block?

Indicate all such numbers.

A. 4
B. 6
C. 8
D. 11
E. 40
[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 31 Aug 2018, 21:03
How to solve it?
please explain
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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 12 Sep 2018, 00:50
explanation pls.??
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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 12 Sep 2018, 07:59
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\(40x= (x-4)(80)\)

\(-40x=-320\)

\(x= 7.8\)

x must be at least 8 to bring the average age up over 40.

Hope this helps.

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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 11 Jun 2019, 07:26
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if (x-4)80=40x, then (x-4)80 keeps average equal to 40 and there are the rest 4 buildings aged 2 years less than others .
Therefore, the value X ( integer as it is building number )decides the range of the number of buildings.
Thank you,
It helps immensely .
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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 26 Mar 2020, 15:23
I still don't get how the equation was set up
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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 27 Mar 2020, 00:41
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sandy wrote:
The average (arithmetic mean) age of the buildings on a certain city block is greater than 40 years old. If four of the buildings were built two years ago and none of the buildings are more than 80 years old, which of the following could be the number of buildings on the block?

Indicate all such numbers.

A. 4
B. 6
C. 8
D. 11
E. 40



let the number of buildings = n
number of buildings built 2yrs ago = 4 ;total age in yrs of 4 buildings = 2*4 = 8
remaining buildings = n-4 ; total age in years of (n-4) buildings = 80(n-4)
number of buildings = 8 + 80(n-4)/n >40
when n=4, 8<40
when n=6, 28<40
when n=8, 41>40
when n=11, 51.64>40
when n=40, 72.2>40
hence, ans is C,D,E
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Re: The average (arithmetic mean) age of the buildings on a cert [#permalink] New post 26 May 2020, 18:57
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I set up the equation differently:

The average age of any set of buildings would be defined as: [total age in years of all the buildings] / [# of buildings]

In this case, we have 4 buildings built 2 years ago (4*2 = 8 years), and we can pretend that the rest of the buildings are 80 years old to solve for the extreme case. Let B be the REST of the buildings. So:

[total age in years of all the buildings] = 8 + 80B.

The total number of buildings will be 4 + B.

Solve for the inequality:

[8+80B]/[B+4] > 40
You'll get
B>3.8

You have to go back and add the remaining 4 buildings back in...

B>3.8+4 --> B>7.8

So, B > 7.8, which makes the answer choices C D and E
Re: The average (arithmetic mean) age of the buildings on a cert   [#permalink] 26 May 2020, 18:57
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The average (arithmetic mean) age of the buildings on a cert

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