Bunuel wrote:

jlgdr wrote:

priyamne wrote:

In the rectangular coordinate system below, if OABC is a parallelogram and OA > OC, is the area of OABC > 50 ?

(1) The coordinates of point C are (5, 5).

(2) The coordinates of point B are (0, 11).

See attached figure. CD is perpendicular to OB. Area of parallelogram OABC = OB x CD.

Statement 1:

OD = 5.

As, OA > OC, DB would have to be greater than 5. Hence OB > 5. Thus area of OABC > 50 (10 x 5)

Sufficient

Statement 2:

We know OB = 11, but have no idea about CD. Hence, we have no measure of area of OABC.

Insufficient

Solution: A

Regards

Mnemonic Education Pvt Ltd

ssingh@mnemoniceducation.comThanks Singh, could you please elaborate more on your explanation?

First, why OD = 5? is the parallelogram symmetric across the y-axis? Also, I don't get how DB would have to be greater than 5.

Statement 2 is more clear

Thanks a lot buddy!

Cheers!

J

Attachment:

Untitled.png

(1) The coordinates of point C are (5, 5).

Why OD = 5? The y-coordinate of D is the same as that of C, so it's 5 --> OD = 5.

Is the parallelogram symmetric across the y-axis? It's not a mirror image around y-axis but the areas around it are the same.

Also, I don't get how DB would have to be greater than 5. Since BC = OA and OA > OC, then BC > OC. Thus BD > (OD=5).

Hope it's clear.

Little imagination helped me, but there is a calculation also to show how OB>10.

As we know that the BC is greater than OC ( property of ||gram- opposites sides equal), and by option A we fixed the point C (5,5)- imagine a ladder BC reclining on wall ( y-axis in this case)- and the only way the ladder can recline is if DB is greater than 5.

Calculation - I dont know how to draw a figure on GMAT forum, so I will use the above given figure and try to explain as much as possible-

Lets call Angle DOC = x = 45 Degree ( as Point C is 5,5 ) and angle OBC = y.

Now lets imagine BC is equal to OC. And hence <X=<Y = 45 Deg.

Therefore Triangle OBC is a right angled triangle.

Length of OC = BC = 5 sqt 2 ( \sqrt{5^2 + 5^2})

Therefore BO= \sqrt{OC^2 + BC^2}= 10 ( exactly)------

1But since BC > CO , therefore

angle y <

angle x = 45 Deg . Therefor

Angle BCO >90 Deg ( In a triangle Angle OBC= 180 - x- y) ------

2Equation

1 and 2 tells that BO>10.

That ladder thing, I can explain if you require.