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Manager  Joined: 16 Feb 2012
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The shaded region in the figure above represents a rectangular frame  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 62% (03:06) correct 38% (03:23) wrong based on 675 sessions

### HideShow timer Statistics 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. $$\frac {9}{2}$$

PS35461.01

Attachment: Frame.png [ 2.69 KiB | Viewed 42780 times ]

Originally posted by Stiv on 29 Jun 2012, 13:14.
Last edited by Bunuel on 21 Sep 2019, 05:40, edited 6 times in total.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Stiv wrote:
Attachment: Frame.png [ 2.69 KiB | Viewed 35616 times ]
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 $$xy=9*15$$ --> $$x*(\frac{5}{6}x)=9*15$$ --> $$x^2=2*81$$ --> $$x=9\sqrt{2}$$.

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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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The total area is 15*18=270, the area of the picture is half of the whole area = 135. the ration of the width and length of the picture is the same as the frames 15/18 or 5/6. We need to find the length of the picture 5x*6x=135, 30x^2=135, x^2=135/30, x=3/sqrt2, so the length = 6*3/sqrt2=9sqrt2
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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l-length of the pic and w-width

18*15-LW - Sof the frame
LW - S of the pic

18/15=L/W --> W= 15L/18

according to the statement 18*15-LW=LW --> 18*15-L*15L/18=L*15L/18 --> L=9\sqrt{2}

A
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Why do I get a different answer if I just left at l/w=18/15 than simplifying to l/w=6/5 ?? Aren't the ratios the same?
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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b00gigi wrote:
Why do I get a different answer if I just left at l/w=18/15 than simplifying to l/w=6/5 ?? Aren't the ratios the same?

To point out why you are getting a different answer you have to show your work. Anyway, yes, the ratio is the same but we need to find the length of the picture, which can be done as explained here: the-shaded-region-in-the-figure-above-represents-a-135095.html#p1100419

Hope this helps.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Total area of the given figure= 18*15 = 270
Area of frame = Area of the picture => We need to divide the total area into two parts, 270/2 = 135. The frame and picture have 135 inch^2 area each.
l(pic) l(frame)
----- = ---------- = 6/5 ==> Area of picture = 135= 6k * 5k ==> 30k^2=135 ==> k =3/sqrt(2). So, l(pic)= 6* 3/sqrt(2) = 9*sqrt(2)
w(pic) w(frame)
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Hi Math Experts,

I am unable to understand the reason to divide by 2 as stated in the reply, can somebody help?

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
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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joepc wrote:
Hi Math Experts,

I am unable to understand the reason to divide by 2 as stated in the reply, can somebody help?

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 combined area of black and white is 18*15 (black + white = 18*15). The area of black = the area of white, so black + black = 18*15 --> black =18*15/2.

Hope it's clear.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Stiv wrote:
Attachment:
Frame.png
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. $$\frac {9}{2}$$

We see that the total area of the frame and the picture is 18 x 15 = 270. Since we know that the length and width of the picture have the same ratio as the length and width of the frame, let’s denote the length of the picture by 18k and the width of the picture by 15k, where k is some positive constant.

Then, the area of the picture is (18k)(15k) = 270k^2

The area of the frame can be found by subtracting the area of the picture from the total area of the frame and the picture: 270 - 270k^2

Since the area of the frame is equal to the area of the picture, we have:

270 - 270k^2 = 270k^2

270(1 - k^2) = 270k^2

1 - k^2 = k^2

2k^2 = 1

k^2 = 1/2

k = 1/√2

Since the length of the picture was represented by 18k, the length is 18(1/√2) = 18/√2 = (18/√2)*√2/√2= 18√2/2 = 9√2.

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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Confounding language - from where to more such difficult language questions...?
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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DAakash7 wrote:
Confounding language - from where to more such difficult language questions...?

Check:
Our Questions' Banks
Shaded Region Problems from our Special Questions Directory.

For more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Bunuel wrote:
Stiv wrote:
Attachment:
Frame.png
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 $$xy=9*15$$ --> $$x*(\frac{5}{6}x)=9*15$$ --> $$x^2=2*81$$ --> $$x=9\sqrt{2}$$.

bunuel why do place x in place of y? x∗(5/6x)
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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SandhyAvinash wrote:
Bunuel wrote:
Stiv wrote:
Attachment:
Frame.png
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 $$xy=9*15$$ --> $$x*(\frac{5}{6}x)=9*15$$ --> $$x^2=2*81$$ --> $$x=9\sqrt{2}$$.

bunuel why do place x in place of y? x∗(5/6x)

In $$xy=9*15$$, we substitute y in terms of x, which we found above (check the highlighted part) to get $$x*(\frac{5}{6}x)=9*15$$. This allows us to get an equation with only one variable x, and solve it.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Stiv 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 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. $$\frac {9}{2}$$

From the question stem ("...the length and width of the picture have the same ratio as the length and width of the frame") we know:

The frame+picture ("big" rectangle) and the picture ("small" rectangle) are two SIMILAR rectangles. (*)

(*) From above we have proportionality on the corresponding sides. The necessary additional condition - equality in the corresponding internal angles - is guaranteed: they are all 90 degrees!

Again from the question stem we know what the examiner defines as "length" and "width" (by the dimensions associated to these words), so that our FOCUS is:

$$? = x\,\,\,\,\left[ {{\text{inches}}} \right]\,\,\,\,\,\,\left( {{\text{See}}\,\,{\text{figure}}\,\,{\text{below}}} \right)$$

From "The frame encloses a rectangular picture that has the same area as the frame itself." we know that the "big" (rectangle) has TWICE the area of the "small" (rectangle).

To avoid using the second dimension of the picture, as it was done in previous (correct) solutions, let´s remember an important geometric property:

In any two similar polygons, the ratio of their areas is equal to the square of the ratio of similarity of the polygons!

Therefore:

$$2 = \frac{{{S_{\,{\text{big}}}}}}{{{S_{\,{\text{small}}}}}} = {\left( {\frac{{18}}{x}} \right)^2}\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,\,\sqrt 2 = \frac{{18}}{x}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x\sqrt 2 = 18$$

$$x\sqrt 2 = 18\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x\sqrt 2 \cdot \sqrt 2 = 18\sqrt 2 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = 9\sqrt 2$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Attachments 20Set18_8h.gif [ 6.86 KiB | Viewed 12204 times ]

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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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Stiv wrote:
Attachment:
Frame.png
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. $$\frac {9}{2}$$

IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.
So, we can use this fact to solve the question by simply "eyeballing" the diagram.

See our video below on this topic as well as other assumptions we can make about diagrams on the GMAT

If you had to ESTIMATE the length of the picture, what would you say it is?
12? 13? 14? 15?

As long as you're in this range, you should be able to solve this one.

ASIDE: On test day, you should have memorized the following approximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2

Now let's check the answer choices....

A. 9√2 ≈ (9)(1.4) ≈ 13. This is within our estimated range. KEEP

B. 3/2 = 1.5. This is WAYYYY outside our estimated range. ELIMINATE

C. 9/√2 ≈ 9/1.4 ≈ 6. This is WAYYYY outside our estimated range. ELIMINATE

D. 15(1 - 1/√2) ≈ 15(1 - 0.7) ≈ (15)(0.3) ≈ 4.5. This is WAYYYY outside our estimated range. ELIMINATE

E. 9/2 = 4.5. This is WAYYYY outside our estimated range. ELIMINATE

RELATED VIDEO FROM OUR COURSE

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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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+1 for checking the answers method:

Since we know that the area of the small rectangle is half the big area, the small length should be a bit more than half of 18, so we can estimate the answers using √2 ≈ 1.4 (that L has to be >9).

Only A) fits this criteria.
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Re: The shaded region in the figure above represents a rectangular frame  [#permalink]

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18*15 = 270

We are told that the frame and photo have equal areas... we can express this as a ratio for conceptual understanding:
Ratio of
Frame: Photo: Total
1x: 1x: 2x
2x = 270
Thus x=135

SO we know the area of each object = 135

Using ratios again we see that both objects are proportionate (share the same ratio):
length:width: total
18: 15: 33
6: 5 : 11

length/ width = 6/5
since we want to solve for length we should determine what Width is in terms of length:
5*length = 6*width
width= 5*length/6

Area= Length * width
substitute in:
Area = l*((5*l)/6)
we know area= 135
135 = (5l^2)/6
135*6=5*L^2
27*6=L^2
162=L^2
L= root (81) * root(2)
L= 9root(2)
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Here's how I went from 430 to 710, and how you can do it yourself: Re: The shaded region in the figure above represents a rectangular frame   [#permalink] 18 Nov 2019, 17:25
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