NandishSS 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.

Let x = the total number of offices.
Since 1/5 of the offices have both windows and bookshelves -- and every office must have either windows, bookshelves, or both -- the following matrix is yielded:


Statement 1: The number of offices with a window is 4/5 the number with bookshelves.
Let y = the number of offices with bookshelves, implying that the number of offices with windows = (4/5)y.
The following matrix is yielded:


Since the top row and the bottom row must each sum horizontally, we get:


Since the middle column must sum vertically, we get:
y - (1/5)x + 0 = x - (4/5)y
5y - x = 5x - 4y
9y = 6x
y/x = 6/9 = 2/3.

Let y=10 and x=15, with the result that y/x = 10/15 = 2/3.
Substituting y=10 and x=15 into the matrix just above, we get:


The resulting matrix indicates the following:
(windows but not bookshelves)/(bookshelves but not windows) = 5/7.
SUFFICIENT.

Statement 2: 3/10 of the offices with bookshelves also have a window.
In the matrix yielded by Statement 1, 10 offices have bookshelves and 3 offices have both, with the result that 3/10 of the offices with bookshelves also have a window.
Implication:
Statement 2 implies the SAME MATRIX as Statement 1.
Thus, Statement 2 must imply the same ratio as Statement 1:
(windows but not bookshelves)/(bookshelves but not windows) = 5/7.
SUFFICIENT.

The correct answer is D.

GMATGuruNY wrote:

NandishSS 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.

Let x = the total number of offices.
Since 1/5 of the offices have both windows and bookshelves -- and every office must have either windows, bookshelves, or both -- the following matrix is yielded:


Statement 1: The number of offices with a window is 4/5 the number with bookshelves.
Let y = the number of offices with bookshelves, implying that the number of offices with windows = (4/5)y.
The following matrix is yielded:


Since the top row and the bottom row must each sum horizontally, we get:


Since the middle column must sum vertically, we get:
y - (1/5)x + 0 = x - (4/5)y
5y - x = 5x - 4y
9y = 6x
y/x = 6/9 = 2/3.

Let y=10 and x=15, with the result that y/x = 10/15 = 2/3.
Substituting y=10 and x=15 into the matrix just above, we get:


The resulting matrix indicates the following:
(windows but not bookshelves)/(bookshelves but not windows) = 5/7.
SUFFICIENT.

Statement 2: 3/10 of the offices with bookshelves also have a window.
In the matrix yielded by Statement 1, 10 offices have bookshelves and 3 offices have both, with the result that 3/10 of the offices with bookshelves also have a window.
Implication:
Statement 2 implies the SAME MATRIX as Statement 1.
Thus, Statement 2 must imply the same ratio as Statement 1:
(windows but not bookshelves)/(bookshelves but not windows) = 5/7.
SUFFICIENT.

The correct answer is D.

Dear Mitch,

As the question ask for ratio, I have assumed that we have 100 offices but, during solving S1 & S2, the numbers yielded fractions which is illogical to have fraction of office . However, at end I got same ratio 5/7.
Is it correct to assume 100 from beginning?

Thanks

Mo2men wrote: Dear Mitch,

As the question ask for ratio, I have assumed that we have 100 offices but, during solving S1 & S2, the numbers yielded fractions which is illogical to have fraction of office . However, at end I got same ratio 5/7.
Is it correct to assume 100 from beginning?

Thanks

Since the question stem asks for a ratio, we can test any value for the total number of offices, as long as all of other conditions are satisfied (1/5 of the offices have both a window and bookshelves, the number of offices with a window is 4/5 the number of offices with bookshelves, and 3/10 of the offices with bookshelves also have a window).

In my solution above, I tested y=10 and x=15 for the following reasons:
The statements combined require that y/x = 2/3.
The prompt indicates that 1/5 of the offices have both a window and bookshelves, implying that the value of x -- which represents the total number of offices -- must be a multiple of 5.

Mo2men wrote: Dear Mitch,

As the question ask for ratio, I have assumed that we have 100 offices but, during solving S1 & S2, the numbers yielded fractions which is illogical to have fraction of office . However, at end I got same ratio 5/7.
Is it correct to assume 100 from beginning?

Thanks

Since the question stem asks for a ratio, we can test any value for the total number of offices, as long as all of other conditions are satisfied (1/5 of the offices have both a window and bookshelves, the number of offices with a window is 4/5 the number of offices with bookshelves, and 3/10 of the offices with bookshelves also have a window).

In my solution above, I tested y=10 and x=15 for the following reasons:
The statements combined require that y/x = 2/3, implying that y must be even and that x must be a multiple of 3.
The prompt indicates that 1/5 of the offices have both a window and bookshelves, implying that x -- which represents the total number of offices -- must be a multiple of 5.
Since x must be divisible by both 3 and 5, the least possible value for x is 15.

GMATGuruNY wrote: In my solution above, I tested y=10 and x=15 for the following reasons:
The statements combined require that y/x = 2/3, implying that y must be even and that x must be a multiple of 3.
The prompt indicates that 1/5 of the offices have both a window and bookshelves, implying that x -- which represents the total number of offices -- must be a multiple of 5.
Since x must be divisible by both 3 and 5, the least possible value for x is 15.

You are correct but the difference between me an you that you assumed the number after reaching x/y = 2/3. However, I assumed the 100 offices in the beginning and discovered that all number was dived by 3 i.e. (number/3). However, I continued working as it ratio and 3 will get canceled soon.

I think it it correct as long as it is ratio.. Do you personally see any problem in my approach??

Thanks in advance for your help

Mo2men wrote:

GMATGuruNY wrote: In my solution above, I tested y=10 and x=15 for the following reasons:
The statements combined require that y/x = 2/3, implying that y must be even and that x must be a multiple of 3.
The prompt indicates that 1/5 of the offices have both a window and bookshelves, implying that x -- which represents the total number of offices -- must be a multiple of 5.
Since x must be divisible by both 3 and 5, the least possible value for x is 15.

You are correct but the difference between me an you that you assumed the number after reaching x/y = 2/3. However, I assumed the 100 offices in the beginning and discovered that all number was dived by 3 i.e. (number/3). However, I continued working as it ratio and 3 will get canceled soon.

I think it it correct as long as it is ratio.. Do you personally see any problem in my approach??

Thanks in advance for your help

As stated in my post above:

Since the question stem asks for a ratio, we can test any value for the total number of offices, as long as all of other conditions are satisfied (1/5 of the offices have both a window and bookshelves, the number of offices with a window is 4/5 the number of offices with bookshelves, and 3/10 of the offices with bookshelves also have a window).

As long as you satisfied all of the given conditions, your approach is fine.