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A rectangular picture is surrounded by a boarder, as shown

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A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 17 Aug 2010, 17:21
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Attachment:
Rectangular.jpg
Rectangular.jpg [ 11.23 KiB | Viewed 2970 times ]
A rectangular picture is surrounded by a boarder, as shown in the figure above. Without the boarder the length of the picture is twice its width. If the area of the boarder is 196 square inches, what is the length, in inches, of the picture, excluding the boarder?

A. 10
B. 15
C. 30
D. 40
E. 60
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Jul 2013, 01:59, edited 1 time in total.
Renamed the topic, edited the question, added the OA and the image.
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Re: Determining Area [#permalink] New post 17 Aug 2010, 17:54
kwhitejr wrote:
FIGURE: Imagine a rectangle within a rectangle. Every side of the interior rectangle is 2 inches from the congruous side of the larger rectangle.

A rectangular picture is surrounded by a border, as shown in the figure above. Without the border the length of the picture is twice the width. If the area of the border is 196 square inches, what is the length, in inches, of the picture, excluding the border?

A. 10
B. 15
C. 30
D. 40
E. 60


Answer : 30.

Length of the outer rectangle - L, Breadth of the outer rectangle - B.
Length of the inner rectangle - l, Breadth of the inner rectangle - b.

Now given that - L*B = 196 sq. unit and l = 2b.

Also l = L - 4 and b = B - 4

Hence area of the outer rectangle (border area) = 196 - Area of the inner rectangle

(L * B) = 196 - (l*b)

((l+4) * (b+4)) = 196 - (l*b)

Simplifying => (lb + 4l + 4b + 16) = (196 - lb)

(4l + 4b) = 180
(4(2b) + 4b) = 180
12b = 180

b = 15, l = 30.

Hence the length of the inner rectangle or picture is 30 units. Answer C.
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Re: Determining Area [#permalink] New post 17 Aug 2010, 18:15
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
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Re: Determining Area [#permalink] New post 25 Jul 2013, 01:55
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Re: Determining Area [#permalink] New post 25 Jul 2013, 02:00
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 26 Jul 2013, 22:37
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
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 29 Oct 2013, 02:02
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 01 Jun 2015, 22:00
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 03 Jun 2015, 14:48
Expert's post
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:
[Reveal] Spoiler:
C


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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 03 Jun 2015, 20:26
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kwhitejr wrote:
Attachment:
Rectangular.jpg
A rectangular picture is surrounded by a boarder, as shown in the figure above. Without the boarder the length of the picture is twice its width. If the area of the boarder is 196 square inches, what is the length, in inches, of the picture, excluding the boarder?

A. 10
B. 15
C. 30
D. 40
E. 60


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
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 03 Jun 2015, 23:37
C
(l+4) (b+4) - lb = 196
l+b = 45
if l=2b then b=15 and l=30
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 07 Jun 2015, 02:47
VeritasPrepKarishma wrote:
kwhitejr wrote:
Attachment:
Rectangular.jpg
A rectangular picture is surrounded by a boarder, as shown in the figure above. Without the boarder the length of the picture is twice its width. If the area of the boarder is 196 square inches, what is the length, in inches, of the picture, excluding the boarder?

A. 10
B. 15
C. 30
D. 40
E. 60


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


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.
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 08 Jun 2015, 06:56
Expert's post
akhil911 wrote:
VeritasPrepKarishma wrote:
kwhitejr wrote:
Attachment:
The attachment Rectangular.jpg is no longer available
A rectangular picture is surrounded by a boarder, as shown in the figure above. Without the boarder the length of the picture is twice its width. If the area of the boarder is 196 square inches, what is the length, in inches, of the picture, excluding the boarder?

A. 10
B. 15
C. 30
D. 40
E. 60


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


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.


Attachment:
Approx Area.jpg
Approx Area.jpg [ 332.5 KiB | Viewed 445 times ]

The shaded region is approximately 2b + 2b + 4b + 4b. The four little squares (with question marks) is the only area unaccounted for.
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Re: A rectangular picture is surrounded by a boarder, as shown [#permalink] New post 23 Jun 2015, 05:10
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.
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Re: A rectangular picture is surrounded by a boarder, as shown   [#permalink] 23 Jun 2015, 05:10
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