GMAT question

[shamrocked]

I am about to take the GMAT in a couple of weeks and when going over an old test I ran into this problem. I would appreciate any help in explaining how you ended up getting to the answer. Thanks!

I can't attach the document for some reason so I will try to write out the problem.

It shows a rectangle with a width of 15 and a length of 18 with a shaded area inside (supposed to look like a picture frame).

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 square root of 2
(b) 3/2
(c) 9/ sqaure root of 2
(d) 15 (1-1/square of 2)
(e) 9/2

 

[Denis]

HINT:
picture length = x, picture width = y ; then x / y = 18 / 15 ; y = 5x / 6

x(5x / 6) = 18 * 15 / 2 ; solve for x.

If you don't follow, only your teacher can help you.

 

[soroban]

Hello, shamrocked!

It shows a rectangle with a width of 15 and a length of 18 with a shaded area inside.
(Supposed to look like a picture frame.)

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?

. . \(\displaystyle (a)\;9\sqrt{2} \qquad (b)\;\frac{3}{2} \qquad (c)\;\frac{9}{\sqrt{2}} \qquad (d)\;\frac{15}{1-\frac{1}{\\\sqrt{2}}} \qquad(e)\;\frac{9}{2}\)

    - * - - - - - - - *
    : |               |
    : |   * - - - *   |
    : |   |       |   |
   18 |  x|       |   |
    : |   |   y   |   |
    : |   * - - - *   |
    : |               |
    - * - - - - - - - *
      : - - -15 - - - :

The length of the picture is \(\displaystyle x\), the width of the picture is \(\displaystyle y.\)

\(\displaystyle \text{We are told that: }\; \frac{x}{y} \:=\:\frac{18}{15} \quad\Rightarrow\quad y \:=\:\frac{5}{6}x\;\;[1]\)


The area of the picture is: .\(\displaystyle xy\text{ in}^2\)

The area of the frame is: .\(\displaystyle 18\!\cdot\!15 - xy\text{ in}^2\)

These areas are equal: .\(\displaystyle xy \;=\;270 - xy \quad\Rightarrow\quad 2xy \:=\:270 \quad\Rightarrow\quad xy \:=\:135\;\;[2]\)


\(\displaystyle \text{Substitute [1] into [2]: }\;x\left(\frac{5}{6}x\right) \:=\:135 \quad\Rightarrow\quad \frac{5}{6}x^2\:=\:135 \quad\Rightarrow\quad x^2 \:=\:162\)


Therefore: .\(\displaystyle x \;=\;\sqrt{162} \;=\;9\sqrt{2}\) . . . answer (a).

 

[shamrocked]

Thank you so much! Great explanations... huge help!