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by J N » Fri Apr 19, 2013 9:18 am
Not official GMAT - just using to get math skills refreshed.



The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?


(A) 9SQrt2
(B) 3/2
(C) 9/sqrt2
(D) 15(1-(1/sqrt2)
(E) 9/2
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by GMATGuruNY » Fri Apr 19, 2013 9:24 am
The shaded region in the figure below represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the lenght and width of the frame, what is the length of the picture, in inches?


(A) 9*sqrt(2)
(B) 3/2
(C) 9/sqrt(2)
(D) 15(1-(1/2))
(E) 9/2
Total area = 18*15 = 270.
Since the frame and the picture have equal areas, the area of the picture = (1/2)270 = 135.
In the frame, L:W = 18:15 = 6:5.
The dimensions of the picture are in the same ratio:
Since 6:5 is close to 1:1, the length of the picture and the width of the picture are almost equal.

We can plug in the answers, which represent the length of the picture.
Since 12*12 = 144, the length of the picture must be around 12 inches.
Only A works:
9√2 ≈ 9(1.4) = 12.6.

The correct answer is A.

Algebraic solution:
Since L:W = 6:5, let L = 6x and the W = 5x.
Since L*W = 135, we get:
(6x)(5x) = 135
x² = 9/2
x = 3/√2.

The length of the picture = 6x = 6(3/√2) = 18/√2 * √2/√2 = (18√2)/2 = 9√2.

The correct answer is A.
Last edited by GMATGuruNY on Fri Apr 19, 2013 10:44 am, edited 2 times in total.
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by Anju@Gurome » Fri Apr 19, 2013 9:28 am
J N wrote:The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
Let us assume the length of the picture is L.
Hence, width of the picture = (L/18)*15 = 5L/6

So, area of the picture = 5L²/6
--> Area of the frame = (15*18 - 5L²/6)

So, 5L²/6 = (15*18 - 5L²/6)
--> 2*(5L²/6) = 15*18
--> L² = 15*18*3/5 = 3*3*18 = 9*9*2
--> L = 9√2

The correct answer is A.
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by Anju@Gurome » Fri Apr 19, 2013 9:42 am
Another less calculation intensive but conceptual solution is as follows...

As the area of the frame and area of the picture are equal, area of the picture = (total area)/2 = (15*18)/2

Now, the length and width of the picture are in the same ratio as the length and width of the frame. Therefore, we need to reduce the length and width of the frame (the length and width of the picture will be less than those of the frame) by the same factor (as we need to maintain the ratio) so that their product is (15*18)/2

This can be done in only one way : (15/√2 )*(18/√2 )

So, length of the picture = 18/√2 = 9√2

The correct answer is A.
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by sonalibhangay » Sat Apr 20, 2013 10:22 am
Area of Picture and Frame together = 18*15 = 270

For dimensions of picture:
If we say Length = L and Breadth = B then L*B = Area = 15 *18 / 2 = 135. Therefore B = 135 /L.

But we also know that Ratio of the L and B = 18 / 15 (Same as that of the frame)

So substituting L/ B = 6/5 ---> L / 135/L = 6 / 5 ---> L^2 = 6 *135 /5 -->

L^2 = 6*27 therefore L=sqrt (6*27) --> 9sqrt2
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