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Intern
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Shape PS [#permalink] New post 27 Aug 2006, 12:51
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Please explain this...
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GMAT Instructor
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 [#permalink] New post 27 Aug 2006, 15:10
You can see that the right side of the square is y+2/3y, so the area is 25/9*y^2.

Find the area of the triangles in terms of y: For example, the one at the bottom right is (4/3y )*(y )/2
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 [#permalink] New post 27 Aug 2006, 15:44
B 14/25

Area of the square = (w + x)^2 = (1/4x + x)^2 = (5/4x)^2 = 25/16 x^2

Area of the shaded part = Area of the square - sum of the unshaded parts
unshaded parts are right angled triangles

unshaded parts = 1/2xw + 1/2yx + 1/2*2*w^2 + 1/2wx
= wx + w^2 + 1/2yx
=1/4x^2 + 1/16x^2 + 3/8x^2 = 11/16x^2

area of the shaded part = 7/8 x^2

What fraction of the square is the shaded part? 7/8 / 25/16 = 14/25

OA??
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 [#permalink] New post 28 Aug 2006, 12:05
Answer B

Solution
Basic Funda : convert all in terms of W
from the equations
1)x=4w
2)y=3w

Area of the NON shaded region [wx+wx+2w^2+xy]/2

substitute the values of X & Y we get
[wx+wx+2w^2+xy]/2
=11w^2


Area of square
(w+x)^2
again substitute
we get
(w+x)^2 =25 w^2


thus Area of shaded region=Area of square-Area of the NON shaded
=14 w^2

Thus the ratio 14/25
VP
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 [#permalink] New post 28 Aug 2006, 16:00
12w=3x=4y

rewrite this as

12w = 12x/4 =12y/3

you get w =x/4 =y/3 and fight it out
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 [#permalink] New post 31 Aug 2006, 21:43
D is the trap answer for those who forget to sutract 11 from 25.

Intuitively, pressed for time, one could "eyeball" this problem.

(B)
  [#permalink] 31 Aug 2006, 21:43
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