A rectangular picture is surrounded by a boarder, as shown in the

Ans : C

Assume the length of inner rectangle as L and the width as w then L = 2W
so the length of outer rectange will be L+4 and width as W+4

The area of border is
area of outer rectangle - area of inner rectangle
[(L+4)*(W+4)]- [ L * W] =196

4L +4W +16 = 196
substituting L = 2W we get
12W = 180
W= 15
L = 2W = 30

In border area there are 4 square , two simillar rectangle and two another similar rectangle
L= length of inner rectangle, l= width of inner rectangle
Given L= 2l ,
Border area 196= 4 square ( 2*2) + 2 rectangle (2*l)+ 2 rectangle (2*L)
196= 16+ 2*L+ 2*L + L/2*2+L/2*2
180=6L
L= 30.
Hope the approach is correct
Regards,

Hi All,

Many Test Takers would use an algebraic approach to solving this problem (which is fine). Since the prompt asks for the length of the picture and the answer are NUMBERS, we can TEST THE ANSWERS.

We're given some facts to work with:
1) The length of the picture is TWICE its width
2) The area of the BORDER is 196 sq. inches
3) The border "adds 4" to the length and width of the picture

We're asked for the LENGTH of the picture.

Since the Length is TWICE the width, chances are that we're looking for an answer that is TWICE another answer (which means either 30 & 15 or 60 & 30).

Let's TEST Answer C: 30

IF.....
Length = 30
Width = 15
Area of picture = 450

Total Length = 34
Total Width = 19
Total area = 646

Total area - picture area = 646 - 450 = 196. This is a MATCH for what we were told, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Here is a 20 second estimation method:
Say the width of the picture is b.

Area of the frame can be approximated to 2*b + 2*b + 2*2b + 2*2b = 12b. Some area is left out here. Ignore it.
12b is a bit less than 196 so b would be more than 10 but less than 20. So length is more than 20 but less than 40.
The only option is (C)

Alternatively, if you want to find the exact value, the area of the left out region is 4 squares of 2*2 i.e. 16 square inches.
So 12b = 196 - 16 = 180
b = 15
So length = 30

C
(l+4) (b+4) - lb = 196
l+b = 45
if l=2b then b=15 and l=30

Hi karishma,

I did not understand how you can approximate area of the frame as 2*b + 2*b + 2*2b + 2*2b = 12b.
Can you please explain this in detail.

The shaded region is approximately 2b + 2b + 4b + 4b. The four little squares (with question marks) is the only area unaccounted for.

Hey,

I started with calculating the area of the frame.

The small corners (red rectangles): We know that the frame is 2 wide from every side. So each of the red rectangles will have an area of 2^2 = 4. We have four of those, so 4*4 = 16.

The green rectangles: 2*2w = 4w. We have two of those, so 2*4w = 8w.

The black (remaining two rectangles): 2*w = 2w. We have two of those, so 2*2w = 4w.

Adding the different areas gives as the total frame area: 16 + 12w

From the stem we know that 16 + 12w = 196.

So, w = 15 and the length is 30.

Let \(x\) be the width and \(2x\) the length of the photo.
Therefore, the width of the frame is \(x+4\) and the length is \(2x+4\)

Area of Frame & Photo - Area of Photo = Area of Frame
\((2x+4)*(x+4)-2x^2=196\)
\(2x(x+4)+4(x+4)-2x^2=196\)
\(2x^2+8x+4x+16-2x^2=196\)
\(12x=196-16\)
\(12x=180\)
\(x=\frac{180}{12}\)
\(x=15\)

Length of the photo \(= 2x = 30 inches\)

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