BTGmoderatorDC 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. \(\frac {9}{2}\)

OA A

Source: Official Guide

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

Thus, we have xy = 18*15 - xy => 2xy = 18*15 => xy = 9*15

Also, given that x/y = 18/15 => y = 5x/6

Plugin-in the value of y in xy = 9*15, we get x*5x/6 = 9*15 => x = 9√2 in

The correct answer: A

Hope this helps!

-Jay
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BTGmoderatorDC 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. \(\frac {9}{2}\)

ALTERNATE SOLUTION:

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

Answer: A

BTGmoderatorDC 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. \(\frac {9}{2}\)

OA A

Source: Official Guide

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