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# Jeff is painting two murals on the front of an old apartment

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Senior Manager
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Jeff is painting two murals on the front of an old apartment [#permalink]

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28 Apr 2006, 20:56
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Jeff is painting two murals on the front of an old apartment building that he is renovating. One mural will cover the quadrilateral face ABCD while the other will cover the circular face (shown to the right, with radius XY). Assuming that the thickness of the coats of paint is negligible, will each mural require the same amount of paint? Note: Figures are not drawn to scale.

(1) AB = BC = CD = DA and $$AB=XY\sqrt{\pi}$$

(2) AC = BD and $$AC=XY\sqrt{2\pi}$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Apr 2014, 04:45, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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28 Apr 2006, 21:22
I wud Say A

From I we know that the quad is a square.

Hence the Area = AB^2 == Area of the circle (pi*XY^2).

From B. You could have multiple quadirlaterals!

If the two diagnals are perpendicular, then the given equation could be used to determine (the quad will then be a square or a kite). But since nothing is provided abt their angle of intersection, we cannot assume that.

Hence, A remains!

Whats the OA?
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28 Apr 2006, 21:31
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Is it C?

1. ABCD could be square or a rhombus.
If ABCD is a square, then area of Circle = area of ABCD
If ABCD is a rhombus with sat 60-120-60-120 degress,
Area of ABCD < AB^2 = XY^*pi
INSUFF

2. AC = BD = XY*SQRT(2pi).
ABCD could be a rectangle or a square or a rhombus.
IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle
If ABCD is a square, both areas are equal.
Rhombus.. unequal.

Combine (1) & (2), ABCD has to be a square with the same area as that of the circle.
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28 Apr 2006, 21:33
sm176811 wrote:
I wud Say A

From I we know that the quad is a square.

Hence the Area = AB^2 == Area of the circle (pi*XY^2).

From B. You could have multiple quadirlaterals!

If the two diagnals are perpendicular, then the given equation could be used to determine (the quad will then be a square or a kite). But since nothing is provided abt their angle of intersection, we cannot assume that.

Hence, A remains!

Whats the OA?

it's quadirlaterals not square... it can be rhombus!
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28 Apr 2006, 21:34
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giddi77 wrote:
Is it C?

1. ABCD could be square or a rhombus.
If ABCD is a square, then area of Circle = area of ABCD
If ABCD is a rhombus with sat 60-120-60-120 degress,
Area of ABCD < AB^2 = XY^*pi
INSUFF

2. AC = BD = XY*SQRT(2pi).
ABCD could be a rectangle or a square or a rhombus.
IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle
If ABCD is a square, both areas are equal.
Rhombus.. unequal.

Combine (1) & (2), ABCD has to be a square with the same area as that of the circle.

for 2 AC=BD can not be rhombus, rhombus has equal sides not diagonal!
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28 Apr 2006, 21:35
Agreed... It could be Rhombus or a square..

Hence C!
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28 Apr 2006, 21:36
B could also generate a trapezoid!
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28 Apr 2006, 21:36
chiragr wrote:
for 2 AC=BD can not be rhombus, rhombus has equal sides not diagonal!

Oops... take out rhombus. Still it can be a rectangle or square. Hence B is INSUFF?
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28 Apr 2006, 22:02
C is the correct answer, the only reason I have this question is following explaination

"
Statement 2 tells us that the diagonals are equal--thus telling us that ABCD has right angle corners (The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees.) Statement 2 also gives us a numerical relationship between the diagonal of ABCD and the radius of the circle. If we assume that ABCD is a square, this relationship would allow us to determine that the area of the square and the area of the circle are equal. However, once again, we cannot assume that ABCD is a square."

The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees.

I can not really visualize this! Can any one show me the light?
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28 Apr 2006, 22:34
chiragr wrote:
The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees.
I can not really visualize this! Can any one show me the light?

chiragr-buddy,
I think there is no light to show here.. There was a blackout in Manhattan when this question was created and explained by 99%tile students
They probably didn't know that a regular trapezium also has equal diagonals. If the diagonals are equal and also bisect each other, then the corners have to be 90 degrees.

Still it's a very good question. We should remember that Sides equal => the quadrilateral could be a nice Square or an ugly Rhombus!
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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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28 Apr 2006, 23:18
now I can sleep without nightmares
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04 May 2006, 06:23
giddi77 wrote:
Is it C?

1. ABCD could be square or a rhombus.
If ABCD is a rhombus with sat 60-120-60-120 degress,
Area of ABCD < AB^2 = XY^*pi

Please explain how the degrees will be 60-120-60-120?
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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27 Apr 2014, 11:08
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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01 Jul 2015, 22:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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05 Jul 2016, 03:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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25 Aug 2016, 02:37
Bunuel Please let us know what would be the ans?
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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29 Aug 2016, 21:05
giddi77 wrote:
Is it C?

1. ABCD could be square or a rhombus.
If ABCD is a square, then area of Circle = area of ABCD
If ABCD is a rhombus with sat 60-120-60-120 degress,
Area of ABCD < AB^2 = XY^*pi
INSUFF

2. AC = BD = XY*SQRT(2pi).
ABCD could be a rectangle or a square or a rhombus.
IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle
If ABCD is a square, both areas are equal.
Rhombus.. unequal.

Combine (1) & (2), ABCD has to be a square with the same area as that of the circle.

If the diagonals are equal it cant be a rhombus. It would be either a square or a rectangle.

Combining 1 and 2, we get C.
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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30 Dec 2016, 05:34
Is there any alternate way or easier way to attack this problem?
I really this this cannot be solved within 2 minutes.
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]

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01 Jan 2017, 04:09
St1 - ABCD is a rhombus. If it were a square, the condition would have implied paint is equal coz area would be equal. However, we dont know that. All we know is its a rhombus, and the area of the rhombus could be anything relative to the circle depending on the ratio of its diagonal to its side -> INSUFF

St2 - ABCD is a ||m. Again, a parallelogram has area = base x height. But we only know info about diagonals. In other words, for the same length of diagonals, you could have infinitely many areas of ||ms - so not enough info - INSUFF

Both combined - we know sides of ABCD are equal and so are the diagonals. Ratio of diagonal to side is sqrt(2) -> This means that ABCD is now a square. In this case area = piXY^2 = area of circle

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Re: Jeff is painting two murals on the front of an old apartment   [#permalink] 01 Jan 2017, 04:09
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