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A rectangular picture is surrounded by a boarder, as shown [#permalink]

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17 Aug 2010, 18:21

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C

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27% (03:05) wrong based on 183 sessions

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Attachment:

Rectangular.jpg [ 11.23 KiB | Viewed 6404 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?

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: A rectangular picture is surrounded by a boarder, as shown [#permalink]

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26 Jul 2013, 23:37

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 Regards,

Re: A rectangular picture is surrounded by a boarder, as shown [#permalink]

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01 Jun 2015, 23:00

Hello from the GMAT Club BumpBot!

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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.

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
_________________

Re: A rectangular picture is surrounded by a boarder, as shown [#permalink]

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07 Jun 2015, 03:47

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.
_________________

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 [ 332.5 KiB | Viewed 3864 times ]

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

Re: A rectangular picture is surrounded by a boarder, as shown [#permalink]

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23 Jun 2015, 06:10

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This post received KUDOS

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

Re: A rectangular picture is surrounded by a boarder, as shown [#permalink]

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16 Aug 2016, 20:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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