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M28-28

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The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)
[Reveal] Spoiler: OA

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Official Solution:

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If \(x=1\), then the difference is 63;

If \(x=2\), then the difference is 252;

In order the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible.


Answer: D
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Re: M28-28 [#permalink]

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New post 22 Oct 2014, 06:16
Bunuel wrote:
The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?
I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)



IMO D !!

After reading question, just check for ans choices

no option with none, thus answer choice can take any value.

Area of square - area of rhombus
\(Diagonal\)2 ...therefore if we consider scenario 1 as correct then only scenario which make sense is secnario 3 where value of secnario 1 is multiplied by 4...or we can say \(2\)2 thus...

Scenario 1 & Scenario 2 only make sense

D ...it was a calculated risk...a guess..i got it right

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Re: M28-28 [#permalink]

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New post 19 Jul 2016, 09:19
Bunuel wrote:
Official Solution:

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If \(x=1\), then the difference is 63;

If \(x=2\), then the difference is 252;

In order the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible.


Answer: D



Hi Bunuel,

What make you take 15x as diagonal of square and the rest as diagonals of rhombus from the given ratio?

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New post 19 Jul 2016, 09:25
avdgmat4777 wrote:
Bunuel wrote:
Official Solution:

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If \(x=1\), then the difference is 63;

If \(x=2\), then the difference is 252;

In order the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible.


Answer: D



Hi Bunuel,

What make you take 15x as diagonal of square and the rest as diagonals of rhombus from the given ratio?


Not sure that I understand what you mean...

The diagonals of square are equal and the diagonals of rhombus are not. We are given the the ration of three diagonals is 15:11:9, hence \(d_s:d_1:d_2=15x:11x:9x\).
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New post 19 Jul 2016, 10:29
Hi Bunuel,

What makes you take 15x as diagonal of square and the rest as diagonals of rhombus from the given ratio?[/quote]

Not sure that I understand what you mean...

The diagonals of square are equal and the diagonals of rhombus are not. We are given the the ration of three diagonals is 15:11:9, hence \(d_s:d_1:d_2=15x:11x:9x\).[/quote]

My doubt is why you choose specifically 15x (among 15, 11 and 9) as diagonal of square. Why not d1 and d2 be 15x and 11x respectively, and Ds be 9x?
If you consider this way, it could be "Area of rhombus - Area of square" since question stem mentions difference between area of square and area of rhombus.
I want to know the logic behind your selection.

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New post 19 Jul 2016, 10:41
avdgmat4777 wrote:
Hi Bunuel,

What makes you take 15x as diagonal of square and the rest as diagonals of rhombus from the given ratio?


Not sure that I understand what you mean...

The diagonals of square are equal and the diagonals of rhombus are not. We are given the the ration of three diagonals is 15:11:9, hence \(d_s:d_1:d_2=15x:11x:9x\).[/quote]

My doubt is why you choose specifically 15x (among 15, 11 and 9) as diagonal of square. Why not d1 and d2 be 15x and 11x respectively, and Ds be 9x?
If you consider this way, it could be "Area of rhombus - Area of square" since question stem mentions difference between area of square and area of rhombus.
I want to know the logic behind your selection.[/quote]

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively.
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New post 15 Aug 2016, 22:17
Bunuel wrote:
Official Solution:


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).




Hi Bunuel,

How do I know when I can just use ratios such as in this solution and assign some variable x (15x:11x:9x) and when I need to find LCM and calculate difference based on LCM solutions, e.g. LCM = 495, then proceed as you did, 495y/15, 495y/11, and 495y/9. Hope this is clear:)

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New post 14 Dec 2016, 07:45
Hii Bunuel
Given that the ratio of the diagonal is ds:d1:d2=15x:11x:9xds:d1:d2=15x:11x:9x, for some positive integer xx (where dsds is the diagonal of square S and d1d1 and d2d2 are the diagonals of rhombus R).

areasquare=d22areasquare=d22 and arearhombus=d1∗d22arearhombus=d1∗d22.

The difference is areasquare−arearhombus=(15x)22−11x∗9x2=63x2areasquare−arearhombus=(15x)22−11x∗9x2=63x2.

If x=1x=1, then the difference is 63;

If x=2x=2, then the difference is 252;

In order the difference to be 126, xx should be 2√2, which is not possible.



Why is it not possible for x to be √2? Its just a number not necessarily an integer?

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New post 14 Dec 2016, 08:53
Dreams17 wrote:
Hii Bunuel
Given that the ratio of the diagonal is ds:d1:d2=15x:11x:9xds:d1:d2=15x:11x:9x, for some positive integer xx (where dsds is the diagonal of square S and d1d1 and d2d2 are the diagonals of rhombus R).

areasquare=d22areasquare=d22 and arearhombus=d1∗d22arearhombus=d1∗d22.

The difference is areasquare−arearhombus=(15x)22−11x∗9x2=63x2areasquare−arearhombus=(15x)22−11x∗9x2=63x2.

If x=1x=1, then the difference is 63;

If x=2x=2, then the difference is 252;

In order the difference to be 126, xx should be 2√2, which is not possible.



Why is it not possible for x to be √2? Its just a number not necessarily an integer?


Please read carefully: The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?
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Re: M28-28 [#permalink]

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New post 18 Mar 2017, 07:45
Bunuel wrote:
Official Solution:

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If \(x=1\), then the difference is 63;

If \(x=2\), then the difference is 252;

In order the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible.


Answer: D


Why x= \(\sqrt{2}\) is not possible
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New post 18 Mar 2017, 07:50
daboo343 wrote:
Bunuel wrote:
Official Solution:

The length of the diagonal of square S, as well as the lengths of the diagonals of rhombus R are integers. The ratio of the lengths of the diagonals is 15:11:9, respectively. Which of the following could be the difference between the area of square S and the area of rhombus R?

I. \(63\)

II. \(126\)

III. \(252\)


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(III\) only
E. \(I\), \(II\) and \(III\)


Given that the ratio of the diagonal is \(d_s:d_1:d_2=15x:11x:9x\), for some positive integer \(x\) (where \(d_s\) is the diagonal of square S and \(d_1\) and \(d_2\) are the diagonals of rhombus R).

\(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

The difference is \(area_{square}-area_{rhombus}=\frac{(15x)^2}{2}-\frac{11x*9x}{2}=63x^2\).

If \(x=1\), then the difference is 63;

If \(x=2\), then the difference is 252;

In order the difference to be 126, \(x\) should be \(\sqrt{2}\), which is not possible.


Answer: D


Why x= \(\sqrt{2}\) is not possible


Please pay attention to the highlighted part.
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Re: M28-28 [#permalink]

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New post 17 Aug 2017, 06:16
This is really really a good sum
Forced me to think like crazy

Bunuel you are the best If i ever get a Q50 it will be because of you

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Re: M28-28   [#permalink] 17 Aug 2017, 06:16
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