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A border of uniform width is placed around a rectangular

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Re: A border of uniform width is placed around a rectangular [#permalink]

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New post 07 Feb 2016, 12:30
Suryanshu wrote:
Given Info: We are given a rectangular photograph of size 8X10 inches. The photograph is surrounded by a uniform width border whose area is 144 square inches.

Interpreting the Problem: In order to find the width of the border, we need to first assume the width of the border in terms of x, and then use the relation of the difference of the areas of 'Photograph+Border' and 'Photograph' to calculate the width of the border.

Solution: Let the width of the border be a.

Then the width of the larger rectangle formed as indicated in the figure will be 8+2a (Uniform width on both sides)
The length of the larger rectangle formed as indicated in the figure will be 10+2a (Unifrom width on both sides)

This is shown in the figure
Attachment:
6.png

Now calculating areas

Area of the larger rectangle = (8+2a)*(10+2a) = \(4a^2+36a+80\)
Area of the photograph = 8*10 = 80

Now the difference of the areas of larger rectangle and photograph is given to us as 144 square inches.

Forming an equation based on above information

Area of larger rectangle- Area of the photograph = \(4a^2+36a+80\)-\(80\)=144
\(4a^2+36a\)=\(144\)
\(4a^2+36a-144=0\)
\(a^2+9a-36=0\)
\(a^2+12a-3a-36=0\)
\((a-3)(a+12)=0\)

Only solution valid for this quadratic equation is a=3 (Because the length of the border of photograph cannot be negative)

So the width of the border will be equal to 3 inches.

Hence, option A is correct.


I still don't get why the width is 8+2a and the length is 10+2a. If I look at the diagram, if the total area is supposed to be 224, shouldn't the length be 10+a and 8+a?

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A border of uniform width is placed around a rectangular [#permalink]

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New post 07 Feb 2016, 13:37
oa7 wrote:
Suryanshu wrote:
Given Info: We are given a rectangular photograph of size 8X10 inches. The photograph is surrounded by a uniform width border whose area is 144 square inches.

Interpreting the Problem: In order to find the width of the border, we need to first assume the width of the border in terms of x, and then use the relation of the difference of the areas of 'Photograph+Border' and 'Photograph' to calculate the width of the border.

Solution: Let the width of the border be a.

Then the width of the larger rectangle formed as indicated in the figure will be 8+2a (Uniform width on both sides)
The length of the larger rectangle formed as indicated in the figure will be 10+2a (Unifrom width on both sides)

This is shown in the figure
Attachment:
6.png

Now calculating areas

Area of the larger rectangle = (8+2a)*(10+2a) = \(4a^2+36a+80\)
Area of the photograph = 8*10 = 80

Now the difference of the areas of larger rectangle and photograph is given to us as 144 square inches.

Forming an equation based on above information

Area of larger rectangle- Area of the photograph = \(4a^2+36a+80\)-\(80\)=144
\(4a^2+36a\)=\(144\)
\(4a^2+36a-144=0\)
\(a^2+9a-36=0\)
\(a^2+12a-3a-36=0\)
\((a-3)(a+12)=0\)

Only solution valid for this quadratic equation is a=3 (Because the length of the border of photograph cannot be negative)

So the width of the border will be equal to 3 inches.

Hence, option A is correct.


I still don't get why the width is 8+2a and the length is 10+2a. If I look at the diagram, if the total area is supposed to be 224, shouldn't the length be 10+a and 8+a?


You are not looking at it correctly. For length, the width of the border needs to be added twice to 10 to make it equal to the length of the photo and the frame. Thus you get 10+x+x = 10+2x (x for the width on the right hand side and another x for the width on the left hand side).

Hope this helps.
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Re: A border of uniform width is placed around a rectangular [#permalink]

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New post 23 Jun 2016, 09:35
Walkabout wrote:
A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches. If the area of the border is 144 square inches, what is the width of the border, in inches?

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9


To solve this problem it’s easier to refer to a diagram of what is provided in the problem stem. In creating the diagram we can also fill in any variables we need to solve the problem.

Image

Notice that we labeled the width of the border as x. We are given that the area of the border is 144 square inches. Knowing this, we can set up the following equation:
Area of Entire Shape – Area of Picture = Area of Border

(8 + 2x)(10 + 2x) – (8) (10) = 144

80 + 16x + 20x + 4x^2 – 80 = 144

4x^2 + 36x – 144 = 0

x^2 + 9x – 36 = 0

(x + 12)(x – 3) = 0

x = -12 or x = 3

Since we cannot have a negative length, x = 3. Thus, the width of the border is 3 inches.

Answer is A.
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Re: A border of uniform width is placed around a rectangular [#permalink]

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New post 30 Jun 2016, 10:50
GhostA wrote:
Length of Picture L1 = 10
Width of Picture W1 = 8

Area of Picture = L1 X W1 = 80



Total Area including frame = L2 X W2 = 144

So L2 and W2 should be multiples/factors of 144.

Prime factorization of 144 = 2 X 7 X 11
Only multiples that make any sense are 14 and 11. (2 x 77 is impossible and so is 22 x 7; total length/width can't be less than length/width of the photograph).

So length (L2) should be 14 and Width (W2) should be 11.

So Width of border = (W2-W1) = 11-8 = 3

Answer choice A.


How do you get that highlighted part????

\(2*11*7\) = 154

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A border of uniform width is placed around a rectangular [#permalink]

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New post 18 Aug 2016, 10:17
I solved this in 1:03 by testing answers, including drawing a picture to make sense of the question.

Answers are always in ascending order. For questions like this, you need only test (B) and (D):

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9

Plugging in B, we have the Border area =(10+(4*2))x(8+(4+2))-80=(10+8)x(8+8)-80=18*16-80=288-80=208.

Notice that 208 is > 144, the given area. Eliminate (B).

Notice that (C), (D), and (E) all give you an even larger border. Eliminate them as well.

The only remaining answer is (A). You're done!

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A border of uniform width is placed around a rectangular [#permalink]

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New post 05 Jul 2017, 21:59
Plugging in values might be faster here.

Starting with B since it is easier. The area of the border comes out to be more than the given value of 144. This leaves us with only one option A
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Re: A border of uniform width is placed around a rectangular [#permalink]

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New post 06 Jul 2017, 07:56
Say length of outer rectangle are 'a' and 'b' respectively.

Then

Area of Border=Area of outer rectangle-Area of inner rectangle

144=ab-80
ab=324

Now substitute options to get area as 324.

only option A:3 satisfies.

ab=(10+3+3)*(8+3+3)
ab=16*14
ab=324

Cheers!

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Re: A border of uniform width is placed around a rectangular [#permalink]

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New post 20 Sep 2017, 09:00
sorry my concern with the problem is still not solved. Can someone please assist me with this .So 1) If 144 is the area of the border then does that include the area of the photograph ?? I am really not understanding that point.

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Re: A border of uniform width is placed around a rectangular   [#permalink] 20 Sep 2017, 09:00

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