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# In a certain building, 1/5 of the offices have both a window

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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
Hi Bunuel!

I just did this problem on a Kaplan CAT and used the same approach as you did. However, i chose 100 at first as the total number of offices, not 300 like you did. Eventually i got a decimal for the # of offices, so i decided to choose 300 total offices then. My question is: How did you decide to immediately use 300 and not 100 as total # of offices? I got the answer right but it cost me some seconds at first...

cheers!
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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
Let us assume that the number of offices in the building is N.
Then number of offices containing both bookshelves and window is N/5.
Also, number of office containing only windows = x and only bookshe.. = y.
Hence x+y+n/5=n
or x+y = 4n/5.

From 1)x= (4/5)y. Using this and original equation both x and y can be obtained in terms of n and x/y will be the answer to the original question.

From 2)(3/10)(y+n/5) = n/5.
Using this and original eq. above we can again solve x and y in terms of n.

hence D is the right answer.
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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
Bunuel wrote:
In a certain building, 1/5 of the offices have both a window and bookshelves. If the rest of the offices in the building have either a window or bookshelves but not both, what is the ratio of the number of offices with a window but not bookshelves to the number of offices with bookshelves but not a window?

Say there are total of 300 offices. Then given that:
Attachment:
Stem.png
We need to find the ratio of the yellow boxes.

(1) The number of offices with a window is 4/5 the number with bookshelves. Say the number of the offices with bookshelves is x and the number of the offices with a window is 4/5*x. Then the number of the offices with bookshelves but not a window is x-60. Also, the total number of the offices with no window is x-60. Thus, 4/5*x+(x-60)=300. We can find x, thus we can find the value of the yellow boxes. Sufficient.
Attachment:
1.png

(2) 3/10 of the offices with bookshelves also have a window. Again, say the number of the offices with bookshelves is x, then we are given that 60=3/10*x. We can find x (200), thus we can find the number of the offices with both a window and bookshelves (3/10*x=60) and then find the number of the offices with bookshelves but not a window (200-60=140). Also, we can find the total number of the offices with no bookshelves (300-200=100), which is the same as the number of the offices with a window but no bookshelves. So, we have the values of both yellow boxes. Sufficient.
Attachment:
2.png

Hope it's clear.

Hi Bunuel,
Could you please explain following statement

we can find the total number of the offices with no bookshelves (300-200=100), which is the same as the number of the offices with a window but no bookshelves

Thanks
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In a certain building, 1/5 of the offices have both a window [#permalink]
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Jp27 wrote:
In a certain building, 1/5 of the offices have both a window and bookshelves. If the rest of the offices in the building have either a window or bookshelves but not both, what is the ratio of the number of offices with a window but not bookshelves to the number of offices with bookshelves but not a window?

(1) The number of offices with a window is 4/5 the number with bookshelves.

(2) 3/10 of the offices with bookshelves also have a window.

Bunuel - Could you please show me how to solve this problem in 2 set matrix? Many thanks for all your responses so far.

Cheers

GIVEN:
TOTAL = W + B - Both + Neither
1 = W + B - 1/5 + 0
W + B = 6/5

THE QUESTION:
W/B = ?

(1) W = 4/5 B -> Two unknowns and two linear equations we can solve with plug in method -> W + B = 6/5 -> 4/5B + B=6/5 -> B= 2/3-> W= 8/15 ->W/B = 4/5 SUFFICIENT;

(2) 1/5=3/10 B -> B=2/3 -> W+2/3=6/5 -> W = 8/15 -> W/B = 4/5 SUFFICIENT;

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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
As Both = 1/5 (*)
=> w-only + b-only = 4/5 (**)

(1) The number of offices with a window is 4/5 the number with bookshelves.

=> (w-only + both) / ( b-only + both) = 4 / 5 (***)
From (*), (**) & (***), we surely can find w-only, b-only & both
=> suff.

(2) 3/10 of the offices with bookshelves also have a window

=> b-only / both = 7/3 (****)
From (*), (**) & (****), we surely can find w-only, b-only & both
Suff.

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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
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Re: In a certain building, 1/5 of the offices have both a window [#permalink]
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