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Dear AbdurRakib Another good question. I am happy to respond.

Statement #1: this gives us the length of OP, and we have the slope of PQ, but we have no idea how high Q goes. Because we don't know this, we can't compute the slope of OQ. This statement, alone and by itself, is insufficient.

Now, forget statement #1.

Statement #2: this gives us Q = (5,4). Well, we know O is at the origin, (0,0). We know this, because this is one thing we are allowed to assume from a coordinate geometry diagram: if a point looks like it is at the origin, and especially this point is given the name "O," then we can be sure that it is exactly at the origin. Thus, we know the coordinates of O & Q, so we can find the slope, m = 4/5. With this statement, we can directly compute an answer to the prompt question. This statement, alone and by itself, is sufficient.

First statement insufficient, second sufficient. Answer = (B)

Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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05 Jul 2017, 06:49

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Lets Assume P (a,b) and Q (x,y) It is given that Slope of OP \(b/a=1/2\) \(a=2b\)

Slope of OQ is \(y/x\)...Required..

Slope of PQ will be \(y-b/x-a = 2\)

y-b=2(x-2b) y=2x-3b

(1) OP = 2\sqrt{5} OP = \sqrt{(a)^2 + (b^2)} 20 = a^2 + b^2 We already know a =2b 20 = 5b^2 b=+2 or -2 then a would be +4 or -4 If we use (4,2) to solve for (x,y) to reach to slope then y-2/x-4 =2 y-2=2x-8 y=2x-6 We do not have another equation to solve for (x,y)

If we use (-4,-2) to solve for (x,y) to reach slope then y+2/x+4 =2 y+2=2x+8 y=2x+6 We do not have another equation to solve for (x,y)

Hence Not Sufficient

(2) Q(5,4)--> x=5, y=4

y/x = 4/5

Bunuel Can u explain the approach, my wife is saying something is wrong
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My friend, on GMAT PS geometry diagrams, there are lots of special cases that we can't assume---we can assume parallel or right angles or two equal sides simply because something looks that way. "Almost" doesn't count in geometry.

Nevertheless, there are a few basic assumptions we can make. If a line looks straight, we can assume that it is straight, that is has no hidden bend or curve to it. Also, if a point is portrayed as "on" a line, we can assume that it is in fact, "on" the line and not simply near the the line. In this diagram, point O is shown at the juncture of the x- and y-axes. We absolutely can assume that it is on both of those lines, and therefore, must be the intersection, the origin. Also, conventionally, the origin is often labeled O. For a problem to have a point labeled O near the origin but not exactly on it would a level of deviousness that we simply don't see on the GMAT.

Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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03 Sep 2017, 23:27

Slope OP=1/2 Slope PQ=2 Find slope OQ Assume p =(px,py) Q= (qx,qy) And as O is origin O= (0,0).

So slope OP py/px=1/2 slope PQ = (qy-py)/(qx-px) =2 Find slope qy/qx

1) it gives distance of OP =2root5 by distance formula px^2+py^2 =(2root5)^2 As py/px=1/2 => px=2py. Putting this value in above equation => 5py^2=20 => py=2 => px=4

Putting these values in slope PQ => (qy-2)/(qx-4)=2 Not enough detail to find value of qx and qy or there ratios.

Not sufficient.

2) Direct co-ordinate of Q is given (5,4), so we cna find slope of line OQ. Slope OQ = (5-0)/(4-0) = 5/4

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