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GMAT OG 11 Ed. Section 5.3 Problem 238 Page 265

Expert replies
by paradox » Sun Aug 05, 2007 11:26 am
The shaded region in the figure represents a rectangular frame with length 18 and width 15. 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?

The official answer is as follows:

Let l and w be the length and width of the picture, so:
1) lw = 270 - lw
2) l/w = 18/15

From these two equations, we can solve to get the answer of 9(2)^1/2

My question is this - for equation #1, I wrote it as:
1) lw = (18-l)(15-w)
The reasoning here is that the area of the picture (lw) is equal to the area of the shaded region or the frame which should be (18-l)(15-w).
I then use the same equation 2 to solve for l.

I know there is a flaw in my reasoning here, but could someone please help point out what it is?...I can't seem to get my head around what I'm doing wrong.

Thanks!


[/img]
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Source: — Problem Solving |

by keepsmilinyaar » Sat Aug 11, 2007 12:39 am
I get the flaw in ur reasoning.

lw = (18-l)(15-w)

Have a look at the equation here: (18-l)(15-w). This means that the length of the outer frame is (18-l) and the width is (15-w). However this is incorrect as the length of the two outer frames along the length would now be.. ((15-w)/2) * 2 (l). The width of the two outer frames along the width would be ((18-l)/2) * 2(15).

Adding the above two equations will give us the area of the picture frame:

[((15-w)/2) * 2 (l) ] + [ ((18-l)/2) * 2(15)] = (15-w)l + (18-l)15.

This would reduce to ur equation 1.

I am sorry i wouldnt be able to show u the image from which I got the equations. But picture the entire frame as being divided into 4 sections. Sumthing like what is shown in the image below
Attachments
untitled.JPG
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explanation

by resilient » Wed Mar 26, 2008 11:45 am
is there a better explanation for the solution in the back of the book?

I am trying to find the reasoning for this!
Appetite for 700 and I scraped my plate!
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by chidcguy » Sun Jun 15, 2008 12:04 pm
It is very tempting to write down the equation as

lw= (18-l) (15-w)

How ever, we have to always keep in mind that the

area of shaded region = area of the whole region - area of unshaded region.

If we go by the above rule in mind, Area of the whole region = 18 X 15

Area of the unshaded region = l X W

Area of shaded region = 270 - lw

That means the problem is saying lw=270-lw
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by Stuart@KaplanGMAT » Sun Jun 15, 2008 2:46 pm
Let's not forget a key piece of info:

The area of the picture is the same as the area of the frame.

We know that total area is 18*15 = 270

Therefore, the area of the picture is 135 and the area of the frame is 135.

At this point the easiest way to solve would be to backsolve, but you haven't provided the choices, so it's hard to describe. We know that l * w (if those are the dimensions of the picture) = 135 and we need to find an answer (value of l) that will make everything work out nicely.

If you provide the choices, I'll do the actual backsolving.
Image

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