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A rectangular picture is surrounded by a boarder, as shown in the
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Updated on: 11 Feb 2018, 22:23
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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 Attachment:
Rectangular.jpg [ 11.23 KiB  Viewed 10060 times ]
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Originally posted by kwhitejr on 17 Aug 2010, 18:21.
Last edited by Bunuel on 11 Feb 2018, 22:23, edited 2 times in total.
Renamed the topic, edited the question, added the OA and the image.



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Re: A rectangular picture is surrounded by a boarder, as shown in the
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17 Aug 2010, 18:54
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: A rectangular picture is surrounded by a boarder, as shown in the
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17 Aug 2010, 19:15
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: A rectangular picture is surrounded by a boarder, as shown in the
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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,



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Re: A rectangular picture is surrounded by a boarder, as shown in the
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29 Oct 2013, 03:02
Please refer fig below:
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Rectangular.jpg [ 21.25 KiB  Viewed 8989 times ]
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Re: A rectangular picture is surrounded by a boarder, as shown in the
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03 Jun 2015, 15:48
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
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Re: A rectangular picture is surrounded by a boarder, as shown in the
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03 Jun 2015, 21:26
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 in the
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04 Jun 2015, 00:37
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 in the
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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.
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Re: A rectangular picture is surrounded by a boarder, as shown in the
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08 Jun 2015, 07:56
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 [ 332.5 KiB  Viewed 6951 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 in the
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23 Jun 2015, 06:10
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 in the
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06 Oct 2018, 13:10
kwhitejr wrote: 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 Attachment: Rectangular.jpg 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 \((2x+4)*(x+4)2x^2=196\) \(2x(x+4)+4(x+4)2x^2=196\) \(2x^2+8x+4x+162x^2=196\) \(12x=19616\) \(12x=180\) \(x=\frac{180}{12}\) \(x=15\) Length of the photo \(= 2x = 30 inches\)
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