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# In the rectangular coordinate system above, if OP < PQ, is

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In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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Updated on: 11 Sep 2017, 23:33
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In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8)

(2) The coordinates of point Q are (13,0)

Attachment:

Triangle.png [ 12.98 KiB | Viewed 17905 times ]

Originally posted by eybrj2 on 14 Mar 2012, 15:21.
Last edited by Bunuel on 11 Sep 2017, 23:33, edited 3 times in total.
Edited the question and added the diagram.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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14 Mar 2012, 16:30
19
13

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

Attachment:

Triangle.png [ 15.88 KiB | Viewed 19047 times ]

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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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30 Mar 2012, 18:51
3
thanks Bunuel.
stmt 1 shows us that the smaller triangle is 1/2*6*8 and the other triangle must be larger than the smaller one. adding the two triangles together you get a number larger than 48.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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09 Jul 2013, 16:37
Bunuel wrote:

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunuel,

Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 ....
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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09 Jul 2013, 16:47
Mountain14 wrote:
Bunuel wrote:

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

Hi Bunuel,

Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 ....

No, that's not possible. We know that the coordinates of point P are (6,8). PS is altitude, thus the coordinates of point S are (6,0), so OS=6.

Hope it' clear.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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13 Jul 2013, 02:05
Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12

Thanks.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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13 Jul 2013, 02:54
Kriti2013 wrote:
Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12

Thanks.

Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12).

Hope it's clear.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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30 Sep 2013, 06:28
Hi Bunuel,
Is there any other way(as fast as the given solution) to solve this without drawing the perpendicular from P to X-axis?
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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12 Oct 2013, 10:03
Hi Bunnel,

From 1 and 2

$$\frac{1}{2}*OS*PS$$ =$$\frac{1}{2}*6*8$$ = 24
$$\frac{1}{2}*SQ*PS$$ =$$\frac{1}{2}*7*8$$ = 28

1 + 2 = 52

Whats wrong here?
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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12 Oct 2013, 11:12
GMAT40 wrote:
Hi Bunnel,

From 1 and 2

$$\frac{1}{2}*OS*PS$$ =$$\frac{1}{2}*6*8$$ = 24
$$\frac{1}{2}*SQ*PS$$ =$$\frac{1}{2}*7*8$$ = 28

1 + 2 = 52

Whats wrong here?

Nothing wrong but what are you trying to say?
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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12 Oct 2013, 20:44
Bunuel wrote:
GMAT40 wrote:
Hi Bunnel,

From 1 and 2

$$\frac{1}{2}*OS*PS$$ =$$\frac{1}{2}*6*8$$ = 24
$$\frac{1}{2}*SQ*PS$$ =$$\frac{1}{2}*7*8$$ = 28

1 + 2 = 52

Whats wrong here?

Nothing wrong but what are you trying to say?

Hi,

The answer is A but dont we need both the statements to conclude this
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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13 Oct 2013, 03:41
GMAT40 wrote:
Bunuel wrote:
GMAT40 wrote:
Hi Bunnel,

From 1 and 2

$$\frac{1}{2}*OS*PS$$ =$$\frac{1}{2}*6*8$$ = 24
$$\frac{1}{2}*SQ*PS$$ =$$\frac{1}{2}*7*8$$ = 28

1 + 2 = 52

Whats wrong here?

Nothing wrong but what are you trying to say?

Hi,

The answer is A but dont we need both the statements to conclude this

To conclude WHAT? The exact area? If so, then yes.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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21 Oct 2013, 13:27
Bunuel wrote:
Attachment:
Triangle.png
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

I didn't understand how are we assuming that OS is equal 6? and how are we concluding t either that OS=QS?Can you pls explain?
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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21 Oct 2013, 22:36
Punyata wrote:
Bunuel wrote:
Attachment:
Triangle.png
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

I didn't understand how are we assuming that OS is equal 6? and how are we concluding t either that OS=QS?Can you pls explain?

Please read the thread: in-the-rectangular-coordinate-system-above-if-op-pq-is-129092.html#p1244697

The coordinates of point P, which is just above S, are (6,8), thus the coordinates of point S are (6, 0).
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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20 Nov 2013, 23:40
Hi Bunuel,

Option (2)
Isn't it possible to get the answer, if OPQ>48, from...

1) OP<PQ
2) a^2+b^2=13^2 .... (a=OP, b=PQ)

I think there is no option of (a,b) to make ab>96 when OP<PQ because a^2+b^2=13^2 and ab=96 can not have any common answers.

a^2 +(96/a)^2 = 169 ...........ab=96
a^4-169*a^2+8716=0
A^2-169A+9216=0................A=a^2
169^2 -4*9216 = -6303............b^2-4ac <0 (ax^2+bx+c=0)

Thanks.

Bunuel wrote:
Attachment:
Triangle.png
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.
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Posts: 46264
Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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21 Nov 2013, 02:59
ohora wrote:
Hi Bunuel,

Option (2)
Isn't it possible to get the answer, if OPQ>48, from...

1) OP<PQ
2) a^2+b^2=13^2 .... (a=OP, b=PQ)

I think there is no option of (a,b) to make ab>96 when OP<PQ because a^2+b^2=13^2 and ab=96 can not have any common answers.

a^2 +(96/a)^2 = 169 ...........ab=96
a^4-169*a^2+8716=0
A^2-169A+9216=0................A=a^2
169^2 -4*9216 = -6303............b^2-4ac <0 (ax^2+bx+c=0)

Thanks.

Bunuel wrote:
Attachment:
Triangle.png
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.

This is not correct because we are not given that OPQ is a right triangle, thus you cannot write OP^2 + PQ^2 = OQ^2.

Hope it's clear.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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22 Nov 2013, 00:29
1
eybrj2 wrote:
Attachment:
Triangle.png
In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8)

(2) The coordinates of point Q are (13,0)

From St 1, we have the co-ordinates of P (6,8) ie. Pt S (6,0) and Pt O (0,0)

Now let's take SQ=x so basically the questions becomes is 1/2 (6+x)*8>48

or (6+x)>12 or x>6

We know OP =10 so PQ>10.....If now we take PQ= 10.5 and apply Pythagorus

(10.5)^2= 8^2+x^2 -------> x^2= 110.25-64 = 46.25 ie x =6.8 which is more than 6 and therefore sufficient

Option B,C AND E ruled out.

From St 2, we know only OQ =13 but don't know the height so not sufficient.

D ruled out so Ans is A
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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24 Sep 2014, 10:50
Although I understand the discussed solution I come to to a different solution due to a different interpretation of the question. Maybe some of you have a similar opinion...

In my opinion, the question at no point requires point Q to lie on the x-axis. The question only requires OP<PQ. Although the graph and question may be interpreted in a way that leads you to think that Q lies on the x-axis they certainly don't state this requirement explicitly.

If OP has a length of 10 and PQ has a length of 10+ but Q is located not on the x-axis but rather close to the origin the area of OPQ may be less than 48 despite having P located at (6,8)

I attached a graph of my example.

I know that the solutions manual offers the same solution as the forum agrees upon here. Is the solution manual wrong? Please point out any mistakes of mine.
Attachments

Screen Shot 2014-09-24 at 12.45.28 PM.png [ 46.79 KiB | Viewed 4595 times ]

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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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24 Jun 2015, 18:06
Bunuel wrote:
Kriti2013 wrote:
Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12

Thanks.

Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12).

Hope it's clear.

Hi Bunuel,

This is my first post.

Can you please explain how we got OS<SQ?

When I solved, I took the smallest value of Q possible i.e. Q (7,0) and got Area= 1/2 *7*6 = 21 so concluded that both (i) and (ii) are required.
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Re: In the rectangular coordinate system above, if OP < PQ, is [#permalink]

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25 Jun 2015, 04:20
1
robinpallickal wrote:
Bunuel wrote:
Kriti2013 wrote:
Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12

Thanks.

Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12).

Hope it's clear.

Hi Bunuel,

This is my first post.

Can you please explain how we got OS<SQ?

When I solved, I took the smallest value of Q possible i.e. Q (7,0) and got Area= 1/2 *7*6 = 21 so concluded that both (i) and (ii) are required.

Hi robinpallickal

You are probably overlooking the condition given in the question that OP < PQ

If $$OP < PQ$$
-> $$\sqrt{OS^2 + PS^2} < \sqrt{SQ^2 + PS^2}$$
-> $$OS < SQ$$ (as OS and SQ are both positive)

Therefore, you can't take the x-coordinate of Q as anything less than 12.
Hope that clears your doubt.
Re: In the rectangular coordinate system above, if OP < PQ, is   [#permalink] 25 Jun 2015, 04:20

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