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# Picture Frame question

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Picture Frame question [#permalink]  17 Oct 2010, 13:02
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This is a tough one. Can anyone help me out with this:

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) 9\sqrt{2}

(B) \frac{3}{2}

(C) \frac{9}{\sqrt{2}}

(D) 15(1-\frac{1}{\sqrt{2}})

(E) \frac{9}{2}

Source: Paper Test
Test Code 28
Section 5
# 15
[Reveal] Spoiler: OA
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Re: Picture Frame question [#permalink]  17 Oct 2010, 15:11
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niheil wrote:
This is a tough one. Can anyone help me out with this:

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) 9\sqrt{2}

(B) \frac{3}{2}

(C) \frac{9}{\sqrt{2}}

(D) 15(1-\frac{1}{\sqrt{2}})

(E) \frac{9}{2}

Source: Paper Test
Test Code 28
Section 5
# 15

Let the length of the picture be x. Since its length and width are in the same ratio as that of the frame, the width must be (5/6)x.

Area of frame = 18*15 - Area of picture.

But we know area of picture = area of frame, hence :

Area of picture = 18*15/2
x * (5/6)x = 9*15
x^2 = 27*6 = 81*2
Hence x = 9 * sqrt(2)

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Re: Picture Frame question [#permalink]  17 Oct 2010, 16:52
I'm so sorry for forgetting to include the diagram or mentioning that the question came with one. Thanks for the help shrouded1.
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Re: Picture Frame question [#permalink]  17 Oct 2010, 20:26
hi, if its not much of a problem can you please include the diagram. thanx
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Re: Picture Frame question [#permalink]  18 Oct 2010, 14:23
18*15-X*5/6*X=X*5/6*
X=3\sqrt{2}
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Re: Picture Frame question [#permalink]  14 Apr 2011, 18:28
The explanation is not proper. Can someone else explain please.

The answer is coming down to 3/\sqrt{2} and not the OA 9/\sqrt{2}.
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Re: Picture Frame question [#permalink]  14 Apr 2011, 23:05
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shrouded1 wrote:
niheil wrote:
This is a tough one. Can anyone help me out with this:

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) 9\sqrt{2}

(B) \frac{3}{2}

(C) \frac{9}{\sqrt{2}}

(D) 15(1-\frac{1}{\sqrt{2}})

(E) \frac{9}{2}

Source: Paper Test
Test Code 28
Section 5
# 15

Let the length of the picture be x. Since its length and width are in the same ratio as that of the frame, the width must be (5/6)x.

Area of frame = 18*15 - Area of picture.

But we know area of picture = area of frame, hence :

Area of picture = 18*15/2
x * (5/6)x = 9*15
x^2 = 27/6 = 9/2
Hence x = sqrt(9/2)

x^2 = 27*6=9*9*2

So, x = 9\sqrt{2}

The approach is right but there is a calculation mistake (in red above).

If you make correct calculations (in blue above), you will get right answer as A.
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Re: Picture Frame question [#permalink]  14 Apr 2011, 23:23
agreed, small calc mistake there ... willl correct
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Re: Picture Frame question [#permalink]  15 Apr 2011, 02:06
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niheil wrote:
This is a tough one. Can anyone help me out with this:

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) 9\sqrt{2}

(B) \frac{3}{2}

(C) \frac{9}{\sqrt{2}}

(D) 15(1-\frac{1}{\sqrt{2}})

(E) \frac{9}{2}

Source: Paper Test
Test Code 28
Section 5
# 15

Attachment:

rectangular_picture_frame.PNG [ 10.34 KiB | Viewed 1999 times ]

Given:
Length of the frame(outer side of the black portion) = 18 inches
Width of the frame(outer side of the black portion) = 15 inches

Total Area of the frame(black portion) and picture(orange portion) = length*width = 18*15
A_t=18*15

Let the length of the picture(orange portion) be "l", we need to find this.
Let the width of the picture(orange portion) be "w"
Area of the picture(orange portion) = l*w
A_p=l*w

Area of the frame(black portion) = Total Area of the frame(black) and picture(orange) - Area of the picture(orange)
A_f=A_t-A_p

"The frame encloses a rectangular picture that has the same area as the frame itself"
A_p=A_f

A_p=A_t-A_p
A_t=2A_p
2A_p=18*15

A_p=\frac{18*15}{2}

l*w=\frac{18*15}{2} -----------------------1

"length and width of the picture(orange) have the same ratio as the length and width of the frame(black)"
\frac{l}{w}=\frac{18}{15}

w=\frac{15}{18}*l --------------------2

Substituting "w" from 2 in 1:

l*\frac{15}{18}*l=\frac{18*15}{2}
l^2=\frac{(18)^2}{2}

Taking the square root on both sides:
l=\frac{18}{\sqrt{2}}

l=\frac{2*9}{\sqrt{2}}

l=\frac{\sqrt{2}*\sqrt{2}*9}{\sqrt{2}}

l=9\sqrt{2}

Ans: "A"
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Re: Picture Frame question   [#permalink] 15 Apr 2011, 02:06
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