The shaded region in the figure above represents a rectangular frame

Say the length and the width of the picture are \(x\) and \(y\) respectively. Since they have the same ratio as the length 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 \(xy=9*15\) --> \(x*(\frac{5}{6}x)=9*15\) --> \(x^2=2*81\) --> \(x=9\sqrt{2}\).

Answer: A.
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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

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

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)

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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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.

Answer: A
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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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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.

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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 my 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

Answer: A

RELATED VIDEO

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Say length of the picture is x and width of the picture is y.

Info 1) length = 18, width = 15
Info 2) (length * width) — (xy) = (xy)
Info 3) 18:15 = x:y

So we can calculate this.


Posted from my mobile device

Hi guys, I have a concerning question about this. The questions asks for the length, which corresponds to y. I therefore did x = 6/5y and then used y for the rest of the problem. I did not get any of the answers because of this. I understand how x can get A, but x corresponds to the width, right? Y corresponds to the length.

Bunuel

Is there an error in my understanding (likely) or in the question stem?

My work:

x = 6/5y
9*15 = xy
9*15 = (6/5y)y
from here I go awry and get the wrong answer.

Why are you guys using x instead when that doesn't relate to the length?

You can denote by x and y anything you want. It's not necessary that x is width and y is length. If you denote y to be length and x to be width then solution is as follows:

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

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 \(xy=9*15\) --> \(y*(\frac{5}{6}y)=9*15\) --> \(y^2=2*81\) --> \(y=9\sqrt{2}\).

Answer: A.
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Simple approximation question

Since they have the same ratio as the length and width of the frame, the length and breadth of the pic can only be 6 and 5 or 12 and 10

Only option A is closest to 12

WHY are we doing math on this question?!?!

Just LOOK at the figure. The length of the white rectangle surely can't be less than 9. B, C, D, and E are ridiculous!!

Answer choice A.


Note that this is an Official geometry question rated on this forum as 700-level. ZERO math skills required and it shouldn't take longer than 20 seconds. LOOK at geometry questions before you go head-first into mathy mcmath math land.

Edited to add: I just noticed that the average time spent is over 3 minutes. Yikes! But what makes things really bad is that 34% of people are missing it, with C and D each accounting for 11%. If you spend 3 minutes doing a bunch of work to arrive at an answer choice, you should at least take the extra second at the end to make sure that it makes sense. Look back at the figure. Is the length of that white rectangle really 6.4 (answer choice C) or 4.3 (answer choice D)? Not even close. If you're going to spend 3 minutes (even when you don't have to ), you might as well make sure you're not completely wasting those three minutes!


ThatDudeKnowsBallparking
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