ps from gmat club

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ps from gmat club

by bblast » Thu Jan 13, 2011 7:01 am
If equation |x/2| + |y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?


* 20
* 50
* 100
* 200
* 400
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by anshumishra » Thu Jan 13, 2011 7:36 am
|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.
Thanks
Anshu

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by bblast » Thu Jan 13, 2011 8:12 am
anshumishra wrote:|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

is this some property ?

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.
if i plot these points the side of square comes 20 :(
Cheers !!

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by anshumishra » Thu Jan 13, 2011 8:32 am
bblast wrote:
anshumishra wrote:|x|+|y| = 10, represents a square with side length equal to 10*sqrt(2),

is this some property ?

So area = 100*2 = 200

Try to plot them using four points [(10,0),(0,10),(-10,0),(0,-10)] on a co-ordinate plane.
if i plot these points the side of square comes 20 :(
No, the area is bounded by these four lines :
x+y = 10
x-y = 10
-x+y = 10
-x-y=10

I think, you have calculated the diagonal. Can you redraw the diagram, and check? [Else, I may be making mistake....sneaking the time out of office a bit...]
Thanks
Anshu

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by fskilnik@GMATH » Thu Jan 13, 2011 8:41 am
Hi there!

Anshu´s solution is correct, but let me help you (bblast) exploring the problem a bit more...

Please note that |x| + |y| = 10 must (at first sight and not counting symmetries) be considered in 4 scenarios:

(1) x <=0 and y >=0 :: in this case we get -x + y = 10 and this is a line that passes through points (-10,0) and (0,10)

This line is indicated in red in the figure I created.

(2) x <= 0 and y <= 0

(3) x >= 0 and y >=0

(4) x >=0 and y <= 0

Please work cases (2) to (4) to realize that the region mentioned in the question stem is (Anshu´s comment) "a square with side length equal to 10*sqrt(2)" and, more than that, with an area that is four times the area of the red right isosceles triangle in my figure!

We get 4* [(10*10)/2] = 200 and we are done.

Regards,
Fabio.


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by bblast » Thu Jan 13, 2011 11:46 am
wow this was a tricky problem, did not realize the solution at first, thanks guys.
Cheers !!

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by fskilnik@GMATH » Thu Jan 13, 2011 12:54 pm
bblast wrote:wow this was a tricky problem, did not realize the solution at first, thanks guys.
Thanks for the "thanks", bblast. Beautiful problem, by the way. :)
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