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In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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02 Jul 2016, 09:38
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In the xyplane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ? (1) r = 7 (2) p = 4 and q = 3
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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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02 Jul 2016, 09:51
I am getting D.
St 2 is correct for obvious reasons. St 1 Given r=7 Also, AO>AB
[(P0)^2+(Q0)^2]^1/2>[(PR)^2+Q^2]^1/2 P^2+Q^2>(PR)^2+Q^2 P^2>(PR)^2 P^2>P^2+R^22PR 2PR>R^2 2P>R 2P>7 P>3.5 Area(AOB)= 1/2b.h =1/2X7XP since P>3.5 Area(AOB)>12.25 Thus A is sufficient.
Let me know what mistake I made..



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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02 Jul 2016, 10:16
Saurabh I think you considered the height of the triangle in incorrectly Sent from my iPhone using GMAT Club Forum mobile app



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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02 Jul 2016, 10:20
By using Heron's Formula Area of triangle is 1/2(x1(y2y3)+x2(y3y1)+x3(y1y2)) Consider statement 1 which is not enough Now consider statement 2 which still does not gives us all three coordinates of vertices. Thus, C is the answer. Sent from my iPhone using GMAT Club Forum mobile app



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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02 Jul 2016, 11:56
comprehensive soln anyone folks?



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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07 Jul 2016, 23:31
Ans should be B. The most important clue that we have is AO>AB This restricts how far r can go. AO from coordinates will be 5 units in length. So maximum value of AB < 5. And all points are in first quadrant only so maximum value of r can be 8 (excluding 8). So maximum area of tringle could be 1/2 * 8 * 3 = 12 (excluding 12) Let me know if its not clear or if I am wrong somewhere. Sent from my SMN910H using Tapatalk



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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07 Jul 2016, 23:56
To add to my solution above, Algebraically we can show how r should be less than 8. We know that length of a line between two coordinates is equal to sqrt of ( (x1x2)^2 + (y1y2)^2 ) So from AO > AB we get: AO^2 > AB^2 (since both sides are positive ) So (40)^2 + (30)^2 > (r4)^2 + (30)^2 Simplifying : (r4)^2 < 16 This gives us 0 < r < 8 Let me know if its not clear or if I am wrong somewhere. Sent from my SMN910H using Tapatalk



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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25 Oct 2016, 07:20
saurabh87 wrote: In the xyplane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?
(1) r = 7
(2) p = 4 and q = 3 oh wow...took 3 minutes just to make sure i don't need 1... a truly 700 level question



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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15 Jul 2017, 07:28
shashanksagar wrote: Ans should be B. The most important clue that we have is AO>AB This restricts how far r can go. AO from coordinates will be 5 units in length. So maximum value of AB < 5. And all points are in first quadrant only so maximum value of r can be 8 (excluding 8). So maximum area of tringle could be 1/2 * 8 * 3 = 12 (excluding 12) Let me know if its not clear or if I am wrong somewhere. Sent from my SMN910H using Tapatalk Hey! Why is it excluding 8 here? maximum value of r can be 8 rite? area can also be 12, which is the maximum. Pls throw some light here! Thanks, Uma



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In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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15 Jul 2017, 09:01
saurabh87 wrote: In the xyplane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?
(1) r = 7
(2) p = 4 and q = 3 Detailed solution. We can clearly see the following things from the question statement: 1. All p,q,r >0 2. As AO > AB \(p^2 + q^2 > (pr)^2 + q^2\) Giving you, \(2*p*r > r^2\) 2 cases possible > Case 1: \(2prr^2 > 0\) > r>0 and 2p>r or Case 2:\(2prr^2 >0\) > r<0 and 2p<r . BUT this goes against #1 above as r MUST BE > 0. Hence ignore this case. Thus the only possible set of values are r>0 and 2p>r 3. Area of Triangle ABO = 0.5*r*q The question asks, is 0.5*r*q<12 or is r*q<24Per Statement 1: No information on q. NOT SUFFICIENT. Per Statement 2: q=3 and p=2 > 2p>r > r<8 Thus, the value of r*q MUST BE <24 as q=3 and r<8. By any combination of values of r and q, you will NEVER get a value >= 24. Hence, this statement is sufficient. Hope this helps. The question is very straightforward if you break it down into recognizable chunks.



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Re: In the xyplane, point O is located at the origin, point A has coordin [#permalink]
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20 Aug 2017, 22:20
shashanksagar wrote: Ans should be B. The most important clue that we have is AO>AB This restricts how far r can go. AO from coordinates will be 5 units in length. So maximum value of AB < 5. And all points are in first quadrant only so maximum value of r can be 8 (excluding 8). So maximum area of tringle could be 1/2 * 8 * 3 = 12 (excluding 12) Let me know if its not clear or if I am wrong somewhere. Sent from my SMN910H using Tapatalk While I understood the algebraic approach to deduce that r should be <8. The length of the 3rd side should be: greater than 1 and less than 9. Is there any other way to deduce that r<8? Bunuel: Please throw some light here.
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Re: In the xyplane, point O is located at the origin, point A has coordin
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