A border of uniform width is placed around a rectangular

Let the width of the uniform border be X

Area of photo 80 (given)

Area of Photo + border = (10+2X)*(8+2X) (deduced)

Difference in areas, (10+2X)*(8+2X) - 80 = 144 (given)

(10+2X)*(8+2X) =224
(10+2X)*(8+2X) =16*14
(10+2X)*(8+2X) =(10+6)(8+6)

=> 2X= 6
=>X=3

Backsolving:
we are looking for area 144+80=224
C. 22*20=440, too much
B. 18*16=288, too much

must be A

Once we get the idea that the width (w) is to be added twice to 10 and 8 we can find by substitution
This how I did the problem
Area of frame = 80 Area of photograph = 144
Total Area = 224

Now we know the new area will be (8+2w)*(10+2w)
Substitute option A for w
(8+6)*(10+6) = 14*16 = 224

Answer A

There is actually quite a nifty little way of doing this. For those who struggle with the whole quadratic solution.

You can do the following:

Since a uniform width is placed around it. We need to find a number where A remains 2 bigger than B. (Since just as much is added to the length as to the width. A will always be 2 more than B)

Since area is 224. We need to find a number for A that is 2 greater than B. You might ask yourself well how am I supposed to do that?

If you can realize that 224 is 1 less than a perfect square. You can use the following: \((x+y)(x-y)\), which in this case is \((15+1)(15-1)\). So we got a = 16 and b = 14.

Now, since the inner border is 8. and we add 2x we got 8 + 2x = 14. x = 3.

This might seem like a weird approach, but if you can quickly realize the relation here, you can solve it in less than a minute

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.



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.

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.

I found substitution as an easier solution to this problem.

Given that photograph dimensions are 8 X 10. Its area is 80.
Total area(frame + photo) - area of photo = 144
(8+2x) (10+2x) - 80 = 144 --> Here x is the border width. Adding the border on both sides, extra padding is 2x.
(8+2x) (10+2x) = 224

Solving this will give a quadratic equation, which will consume some time to get to. Easier would be to substitute values.

Substitute smallest option, i.e A. Fits right in!!
Answer Choice A is correct!

Check the diagram:The length of the longer side is 10 + x + x = 10 + 2x and the length of the shorter side is 8 + x + x = 8 + 2x.

Hope it's clear.

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

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.

why are we adding both sides by 2x

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.

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!

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.



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 .

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

I see what you are doing there. The area of the photograph WITH the border is \(4x^2+36x+80\) square inches. We don't have an equation there, we have an expression, so you cannot reduce \(4x^2+36x+80\) by 4 to get \(x^2+9x+20\) (it's just \(4x^2+36x+80\) not \(4x^2+36x+80=0\)).

Hope it's clear.

It's doesn't indicate whether the figure was drawn to scale or not. So, how can you assume that each side between the rectangular photograph and the uniform is x? i.e: Couldn't the left side be bigger/smaller than the right side?