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On a graph the four corners of a certain quadrilateral

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On a graph the four corners of a certain quadrilateral [#permalink]

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New post 04 Jan 2013, 03:06
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On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6
(2) The quadrilateral is a rectangle.
[Reveal] Spoiler: OA
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 04 Jan 2013, 04:41
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trex16864 wrote:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6
(2) The quadrilateral is a rectangle.


There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:
Attachment:
Quadrilateral.PNG
Quadrilateral.PNG [ 9.43 KiB | Viewed 2760 times ]
As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 04 Jan 2013, 03:54
My solution:

Given the information in the question we can already conclude, that the quadriliteral is a rectangle with a width of 5. We need to figure out the length of the rectangle in order to find the area.

(1) with
I) \(a + c = 12\)
II) \(b + c = 6\)

we can I) - II) => \((a-b) = 6\)

Length = Distance [(a;5) (b;5)] = Distance [(a;0) (b;0)] =
= SQR [ \((a-b)^2 + (5-5)^2\) ]
= SQR [ \(6^2 + 0^2\) ]

= 6

=> \(Area = 6 * 5 = 30\) sufficient

(2) We already know that we have a rectangle. The statement gives us no further information about the length. Not sufficent

Hence A

Last edited by trex16864 on 04 Jan 2013, 05:04, edited 2 times in total.
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 01 Jun 2013, 01:16
Bunuel wrote:
trex16864 wrote:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6
(2) The quadrilateral is a rectangle.


There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:
Attachment:
Quadrilateral.PNG
As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.



Hi,

Sorry to re-open the thread. But Can you please explain how does the question stem tells us that it is a rectangle?

Thank you
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 01 Jun 2013, 04:17
Genfi wrote:
Bunuel wrote:
trex16864 wrote:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6
(2) The quadrilateral is a rectangle.


There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:
Attachment:
Quadrilateral.PNG
As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.



Hi,

Sorry to re-open the thread. But Can you please explain how does the question stem tells us that it is a rectangle?

Thank you


Mark (a,5), (b,5), (a,0) and (b,0) on the plane and you'll see that you'll get a rectangle for any values of a and b (a>b).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 01 Jun 2013, 16:27
The b > a threw me off...
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 17 May 2016, 06:13
Bunuel wrote:
trex16864 wrote:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6
(2) The quadrilateral is a rectangle.


There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be:
On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:
Attachment:
Quadrilateral.PNG
As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.


Thanks Bunel, I couldn't agree more, due to this a<b issue, I was getting the error as "negative" and marked E instead, I would my answer had correct now - not gonna spoil my metrics for an incorrect question, Author - Please edit and correct the question, it's misleading.
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Re: On a graph the four corners of a certain quadrilateral [#permalink]

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New post 06 Jun 2016, 11:01
Same here, the part on a<b made me choose E. Could anyone please fix this issue?
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Re: On a graph the four corners of a certain quadrilateral   [#permalink] 06 Jun 2016, 11:01
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