A border of uniform width is placed around a rectangular

Question: 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

Hey there!
I can see that most of the responses have used a formulaic approach using an unknown variable.
I would NOT suggest the same.
GMAT is a test of logic and not usage of equations & formulas!
Lets try to use the logical STRATEGY of BACKSOLVING!

What is the Strategy of Back-solving in GMAT?
Its a powerful and effective strategy that can help you eliminate multiple incorrect answer choices and involves working backwards from the answer choices to determine which one is correct. It helps you to narrow down the options.
On GMAT, the answer choices are in ascending or descending order and hence it makes sense to start with option C and eliminate extreme options A,B or D,E depending on the constraints.

Lets try to use this strategy instead of conventional equations!
GMAT Track of thought 1

The area of the rectangular photograph that measures 8 inches by 10 inches = 8 * 10 =80 square inches
All the 5 answer choices represent the width of the border, in inches

GMAT Track of thought 2
Area of the border = 144 and hence the area of the photograph along with the border = 144 + 80 =224 square inches

GMAT Track of thought 3
Begin with option(C) which is 6
If the width was 6 units the length would be 10 + 6 + 6 =22 units and the breadth 8 +6 +6 =20 units and hence the area of the photograph along with the border = 22 * 20 = 440 square inches. Its way greater than 224 square inches which is the area of the photograph along with the border as calculated above.

GMAT Track of thought 4
All values greater than 6 will result in area values even greater than 440.
So eliminate C,D & E
Use the same strategy of testing on option B now that provides 4 inches as the width of the border.
Then the length would be 10 + 4 + 4 =18 inches and breadth would be 8 + 4 +4 = 16 inches and hence the area of the photograph along with the border = 18 * 16 =288 and this is greater than 224 square inches.

That leaves us with option(A) as the answer.
You can go ahead and mark A without further calculations but if you do check with calculation,you would observe that if the width was 3 as mentioned in option A, the length would be 10 + 3 +3 =16 inches and the breadth would be 8 + 3 + 3 = 14 inchess.
and hence the area of the photograph along with the border = 16 * 14 = 224 square inches!!
BINGO!

Hope you learnt something new here and would apply it ahead.

Devmitra Sen
GMAT Mentor

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.



Consider the diagram below:



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.

We are given a rectangle with sides of 8 inches and 10 inches. Thus, the area of the photograph is:



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



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:



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



The final answer is .

Bunuel,

When I simplify the equation \(4x^2+36x+80\) to ===> \(x^2+9x+20\), the end result changes completely.

I end up getting the equation: \(x^2+9x-204\), what did I miss?

I'm not sure I follow you...

The area of the photograph WITH the border is \(4x^2+36x+80\) square inches;
The area of the photograph ONLY is 8*10 = 80;

We are told that the difference between them is 144, so we have \((4x^2+36x+80)-80=144\).

So, the final equation we get is \((4x^2+36x+80)-80=144\), which simplifies into \(x^2+9x-36=0\).

Does this make sense?

Consider the diagram below:



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.

[/quote]

Bunuel,

When I simplify the equation \(4x^2+36x+80\) to ===> \(x^2+9x+20\), the end result changes completely.

I end up getting the equation: \(x^2+9x-204\), what did I miss?[/quote]

I'm not sure I follow you...

The area of the photograph WITH the border is \(4x^2+36x+80\) square inches;
The area of the photograph ONLY is 8*10 = 80;

We are told that the difference between them is 144, so we have \((4x^2+36x+80)-80=144\).

So, the final equation we get is \((4x^2+36x+80)-80=144\), which simplifies into \(x^2+9x-36=0\).

Does this make sense?[/quote]

Yes Bunuel, It does make sense. It understood earlier too. My question is the area of the photograph with border when simplified to \(x^2+9x+20\) and then moved further, as I explained in my previous query it changes the end equation. Am I able to explain this now?

Thanks for the fast reply though.

Cristal clear now! Thank so much

Cannot find the question link

Here is the question: https://gmatclub.com/forum/a-border-of- ... l#p1158911

Hi everyone.
Can anyone please help in explaining how does one solve the equation: a2+ 9a−36=0 that we get while solving this question??
An elaborate explanation of the approach would be really helpful.

just take the new dimensions including border as K and K+2 (cause diff will still remain 2 units)

now K(K+2) = 224 (80+144)

do Trial & Error
since the difference between the 2 factors is just 2 units, take square root of 224 as reference for trial & error . it is approx 15.

now take one less and one more value than 15.
you get 14 and 16
14*16 is 224, hence it is successful attempt
it means after border dimension is increased by 6 units, and since border gets added twice so 6 = 2*border , border =3 units.

Great explanation always JeffTargetTestPrep, thanks. Question ask what is the width of the border, in inches? Do we not need to substituting 3 into (10 + 2x) here for final answer. Though 16 is not in the multiple choice. Just want to clarify and not sure what did I miss? Thanks for your time in advanced.

Video solution from Quant Reasoning:
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Given: A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches.

Asked: If the area of the border is 144 square inches, what is the width of the border, in inches?



Let the width of the border be w.
Area of the borders = (8+2w)(10+2w) - 8*10 = 144
(8+2w)(10+2w) = 144 + 80 = 224 = 14*16
8 + 2w = 14
w = 3 inch

IMO A

Hi Kimberly77
Thanks for your query.


You are confusing the width of the border with the width of the big rectangle (border + photograph). These two are completely different.


Let me explain the difference between the two. For this let’s first redraw and label the given situation in a figure.



Here, ABCD represents the rectangular photograph, and EFGH represents the photograph along with the border.

In this figure, observe that (10 + 2x) represents the width EFGH (photograph PLUS the width of the border), while what we want is the width of the border only.


Let me take you back to the question to read the highlighted parts again, carefully:

- “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?” you will see that border is only that part that is placed AROUND the photograph.

So, correct translation will tell you that the WIDTH of the border that was mentioned at the start of the question is precisely the one that is asked at the end. Since we took the width of the border as x inches, it is this ‘x’ that we need to find.


And that’s it!


Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT

To find the width of the border around the rectangular photograph, we need to consider the original dimensions of the photograph and the area of the border.

The photograph measures 8 inches by 10 inches.

Let's denote the width of the border as w inches.

When a border is added around the photograph, the new dimensions become (8 + 2w) inches by (10 + 2w) inches.

The area of the border is given as 144 square inches.

To find the width of the border, we need to set up an equation based on the given information:

(8 + 2w)(10 + 2w) - 8 * 10 = 144.

80 + 16w + 20w + 4w^2 - 80 = 144.

4w^2 + 36w = 144.

w^2 + 9w = 36.

w^2 + 9w - 36 = 0.

(w + 12)(w - 3) = 0.

Setting each factor equal to zero, we find two possible solutions:

w + 12 = 0, which gives w = -12 (extraneous solution since width cannot be negative).

w - 3 = 0, which gives w = 3.

Therefore, the width of the border is (A) 3 inches.