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# The shaded region in the figure above represents a rectangul

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The shaded region in the figure above represents a rectangul [#permalink]  17 Oct 2010, 12:02
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Difficulty:

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Question Stats:

71% (02:57) correct 29% (04:22) wrong based on 30 sessions

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 lenght and width of the frame, what is the length of the picture, in inches?

A. $$9\sqrt2$$
B. $$\frac {3}{2}$$
C. $$\frac {9}{\sqrt2}$$
D.$$15 ( 1 - \frac {1}{\sqrt2})$$
E. $$\frac {9}{2}$$

Source: Paper Test
Test Code 28
Section 5
# 15

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-shaded-region-in-the-figure-above-represents-a-135095.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jan 2014, 03:22, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Picture Frame question [#permalink]  17 Oct 2010, 14:11
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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, 15:52
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, 19:26
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, 13:23
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, 17:28
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, 22:05
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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, 22:23
agreed, small calc mistake there ... willl correct
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Re: Picture Frame question [#permalink]  15 Apr 2011, 01:06
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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 6207 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]  06 Jan 2014, 03:15
Is this one really supposed to be easy?
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Re: Picture Frame question [#permalink]  06 Jan 2014, 03:23
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FirstScore350 wrote:
Is this one really supposed to be easy?

It's ~650 level question.

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 lenght and width of the frame, what is the length of the picture, in inches?

A. $$9\sqrt2$$
B. $$\frac {3}{2}$$
C. $$\frac {9}{\sqrt2}$$
D. $$15 ( 1 - \frac {1}{\sqrt2})$$
E. $$\frac {9}{2}$$

Say the length and the width of the picture are $$x$$ and $$y$$ respectively. Since they have the same ratio as the lenght and width of the frame, then $$\frac{x}{y}=\frac{18}{15}$$ --> $$y=\frac{5}{6}x$$.

Next, since the frame encloses a rectangular picture that has the same area as the frame itself and the whole area is $$18*15$$, then the areas of the frame (shaded region) and the picture (inner region) are $$\frac{18*15}{2}=9*15$$ each.

The area of the picture is $$xy=9*15$$ --> $$x*(\frac{5}{6}x)=9*15$$ --> $$x^2=2*81$$ --> $$x=9\sqrt{2}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-shaded-region-in-the-figure-above-represents-a-135095.html
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Re: Picture Frame question   [#permalink] 06 Jan 2014, 03:23
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