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A border of uniform width is placed around a rectangular

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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
1
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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A border of uniform width is placed around a rectangular  [#permalink]

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New post 18 Aug 2016, 10:17
1
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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A border of uniform width is placed around a rectangular  [#permalink]

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New post 07 Dec 2017, 20:52
longhaul123 wrote:
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.


144 square inches is the area of the border without the photograph.
The area of the photograph is 8*10=80 square inches
Total area = 80 + 144.

Check the solution below: hope it's explained there clearly.

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


Consider the diagram below:

Image

The area of just the photograph is 8*10=80 square inches.

The area of the photograph with the border is \((10+2x)(8+2x)=4x^2+36x+80\) square inches.

The difference is 144 square inches, thus \((4x^2+36x+80)-80=144\) --> \(4x^2+36x-144=0\) --> \(x^2+9x-36=0\) --> \((x-3)(x+12)=0\) --> \(x=3\).

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

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New post 17 Dec 2017, 17:02
I took a bit of a different route:
The area of the rectangle (photo and border) is 144+80=224. I took the prime factorization of this to get 2^5 and 7. Therefore the large rectangle must have edges of 14 and 16, as this is these are the only multiples that are longer than the photograph length and width; (14-8)/2 =3 and (16-10)/2 = 3; E.

I'm not sure if this was longer or shorter than the OG solution, but I do prefer the OG solution.
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Re: A border of uniform width is placed around a rectangular  [#permalink]

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New post 30 Dec 2017, 09:52
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?


hi oa7 , to answer your question:
we write 2a because rectangle has two sides of equal width "a" which is a+a=2a and two sides of equal lengths a+a = 2a

i hope it helps to understand the solution:)
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Re: A border of uniform width is placed around a rectangular  [#permalink]

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New post 30 Dec 2017, 19:40
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


let x=width of border
(10+2x)(8+2x)-80=144➡
x^2+9x-36=0
x=3 inches
A
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A border of uniform width is placed around a rectangular  [#permalink]

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New post 03 Oct 2018, 04:47
Let \(x\) be the uniform width placed around the photograph. It’s best to draw the figure first so we can better understand the problem.

Attachment:
figure.JPG
figure.JPG [ 16.12 KiB | Viewed 2146 times ]


From the figure we can see that \(2x\) is added on both sides if we account for the border. Hence, the dimensions would be \((8+2x)\) and \((10+2x)\) if we include the border.

To find the area of the border alone, we subtract the smaller area of the photograph from the bigger rectangle formed.

We have \(8(10)=80 \ in^2\) for the photograph and \((8+2x)(10+2x)\) \(in^2\) for the bigger rectangle. Their difference must the border’s area which is given to be \(144 \ in^2\). From this, we can form the equation below.

\((8+2x)(10+2x)-80=144\)
\(80+16x+20x+4x^2-80=144\)
\(80+16x+20x+4x^2-80-144=0\)
\(4x^2+36x-144=0\)

Dividing both sides of the equation by \(4\), we have

\(x^2+9x-36=0\)
\((x+12)(x-3)=0\)
\(x= -12 \ and \ x=3\)

Since we’re looking for a positive value, this means that the only possible width of the border is \(3\) inches.

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

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New post 16 Nov 2018, 01:13
We are given a rectangle with sides of 8 inches and 10 inches. Thus, the area of the photograph is:

\(Area = (8)(10) = 80 \ square \ inches\)


We know that area of the border is 144 square inches. Thus, the total area of the border and the photograph is:

\(Total \ area = 80 + 144 = 224 \ square \ inches\)


Since the border is a uniform length, we can call that length \(x\).

Since there is a border on the left and right sides and the top and bottom sides, the two sides of the border can be referred to as:

\((8 + 2x)\) and \((10 + 2x)\)


We can use this to find the value of \(x\):

\((8 + 2x)(10 + 2x) = 224 \\
80 + 16x + 20x + 4x^2 = 224 \\
4x^2 + 36x = 144 \\
4x^2 + 36x –144 = 0 \\
x^2 + 9x -36 = 0 \\
(x + 12)(x - 3)=0 \\
x = 3\)


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

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New post 03 Apr 2019, 00:33
Although I could figure out the answer in less than 1 min but still ended up marking the wrong choice!

This question is great learning that one should always try to draw or imagine the figure, you might be doing all the right calculations but still end up choosing the wrong option.

Now, If you know 2 simple things you're good till the end

1. If 15*15= 225 then 14*16 = 224 (1 less than the perfect square) [i.e (15+1)(15-1)]

This works for all: 20*20=400, while 19*21=399
30*30=900 while 29*31=899

2. Area of frome+picture will be : (8+2x)(10+2x)=144+80
=(8+2x)(10+2x)=224 (there is a reason why GMAT has given this value)


PS. It is really important to take the extra dimensions frame as 2x and not x because the width is uniformly spread on both sides. (refer to diagram)


Now using the above information we can conclude that dimensions are 14 and 16

original dimensions of the picture: 8 and 10

so, 2x=6 and x=3 ( I could figure out the extra dimensions are 6 but did not consider that extra dimension of the frame is evenly spread out on both sides)

So (A)

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 19 Oct 2019, 10:29
Here is how I did it...

So I knew that our rectangle had an area of 80...
and our border had an area of 144...
So I knew that the area of the rectangle and its border was 80+144= 224
So then I filled in the equation for the rectangle and its border (8+2w)(10+2w)=224

For questions like this I find the fastest way is to not bother with the quadratic at all but rather start testing the answer choices!

So immediately I tested B (I like to test B first because if the answer is too high we can test A next and be pretty sure it will be correct)

B gave me (8+8)(10+8) = 288 Too High

So now I know there is only one way to go!

I tested A and got (8+6)(10+6)= (14)(16)= 224

That was quick!!!
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Re: A border of uniform width is placed around a rectangular   [#permalink] 19 Oct 2019, 10:29

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