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Square divided into rectangles

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by gmattesttaker2 » Thu Mar 20, 2014 9:56 pm
Hello,

Can you please tell me how to solve this:

The figure above is a square divided into rectangles. The measures of some of the
sides of some of the rectangles are shown in the figure. What is the value of
x if the shaded area equals the unshaded area?

(A) 0.5
(B) 1
(C) 1.5
(D) 3
(E) 4.5

OA: C

Thanks a lot,
Sri
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by [email protected] » Thu Mar 20, 2014 11:26 pm
Hi Sri,

The key to solving this question is that big shape is a SQUARE.

From the drawing, we know that we're dealing with a 9x9 square (the right "side" of the square is 3 3's, so we have a length of 9).

The first "column" of rectangles has a length of X
The third "column" of rectangles has a length of 3
The second "column" of rectangles is unlabeled, so let's say is has a length of Y

We're told that the total area of the shaded region = the total area of the unshaded region.

The area of the 5 shaded regions = 3X + 9 + 3Y + 3X + 9

The area of the 4 non-shaded regions = 3Y + 3X + 9 + 3Y

Now, we set them equal to one another...

6X + 3Y + 18 = 3X + 6Y + 9

Combine like terms...

3X + 9 = 3Y

X + 3 = Y

Since the square is a 9x9 square, we know that X+Y+3 = 9; this means....

X + Y = 6

Now we have a system of equations:
X+3 = Y
X+Y = 6

X=1.5 and Y=4.5

We're asked for the value of X:

Final Answer: C

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by theCodeToGMAT » Thu Mar 20, 2014 11:31 pm
{C}
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by GMATGuruNY » Fri Mar 21, 2014 1:13 am
Image

Since the height of each row is 3:
All 3 A regions have the same area (3x).
All 3 B regions have the same area (3y).
All 3 C regions have the same area (3*3 = 9).

Sum of the shaded regions = 2A + B + 2C.
Sum of the unshaded regions = A + 2B + C.
Since the sums are equal, we get:
2A + B + 2C = A + 2B + C
A + C = B.

Since A=3x, C=9, and B=3y, we get:
3x + 9 = 3y
x+3 = y.
x-y = -3.

Since the garden is square, each side = 9.
Thus, x+y=6, as shown in the figure above.
Adding the two equations, we get:
(x-y) + (x+y) = -3+6
2x = 3
x = 1.5.

The correct answer is C.
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by gmattesttaker2 » Mon Mar 24, 2014 10:31 pm
Hello Rich, Rahul and Mitch,

Thank you all for your excellent explanations.

Best Regards,
Sri
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