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Geometry

by rintoo22 » Wed Mar 20, 2013 5:14 am
The figure attached represents a square garden that is divided into 9 rectangular regions with indicated dimensions in meters. The shaded regions are planted with peas, and the unshaded regions are planted with tomatoes. If the sum of the areas of the regions planted with peas is equal to the sum of the areas of the regions planted with tomatoes, what is the value of x?

A. 0.5
B. 1
C. 1.5
D. 2
E. 2.5
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by Anju@Gurome » Wed Mar 20, 2013 5:24 am
rintoo22 wrote:The figure attached represents a square garden that is divided into 9 rectangular regions with indicated dimensions in meters. The shaded regions are planted with peas, and the unshaded regions are planted with tomatoes. If the sum of the areas of the regions planted with peas is equal to the sum of the areas of the regions planted with tomatoes, what is the value of x?
Let us mark the regions as follows,

Image

Now, area of each of the region A, D, and G = 3x
And, area of each of the region B, E, and H = 3(9 - (3 + x)) = 3(6 - x)
Now, area of each of the region C, F, and I = 3*3 = 9

We can see that in the first two rows, in each column one of the shaded region has exactly equal area as of the unshaded region in the same column, i.e. A = D, B = E, and C = F

Hence, for the total shaded area to be equal to the total unshaded area, area of (G + I) must be equal to the area of H.

So, (3x + 9) = 3(6 - x)
--> 6x = 9
--> x = 9/6 = 1.5
Last edited by Anju@Gurome on Wed Mar 20, 2013 5:30 am, edited 1 time in total.
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by srcc25anu » Wed Mar 20, 2013 5:26 am
area of all pea region = 9 + 9 + 6x + 18-3x = 36 + 3x
area of all tomato regions = 9 + 36-6x + 3x = 45 - 3x
since the two area are equal: 36 + 3x = 45 - 3x => 6x = 9 or x = 1.5
hence C
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by Brent@GMATPrepNow » Wed Mar 20, 2013 5:33 am
First recognize that the middle length must be 6-x.
Then we can find the areas of all 9 rectangles.
Image

The sum of the areas of the regions planted with peas is equal to the sum of the areas of the regions planted with tomatoes.

3x + 3x + 3(6-x) + 9 + 9 = 3x + 3(6-x) + 3(6-x) + 9
3x + 36 = -3x + 45
6x = 9
x = 9/6 = [spoiler]3/2 = C[/spoiler]

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by GMATGuruNY » Wed Mar 20, 2013 8:23 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 rairavig » Thu Mar 21, 2013 12:13 am
Image

Since it is clear from the solutions that shaded and unshaded areas in first two coloums are equal and will cancel each other, we will consider the area present in last coloum, as per question the shaded area is equal to unshaded area:-
3x + 9 = 3 * (6-x)
x + 3 = 6-x
2x = 3
x = 1.5
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