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655-705 (Hard)|   Geometry|                              
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sakshamchhabra
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A border of uniform width is placed around a rectangular 03 Apr 2019, 00:33
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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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A border of uniform width is placed around a rectangular 19 Apr 2021, 09:21
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Bunuel
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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:



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.

Show Spoiler
Attachment:
photograph.png
Show more

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?
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A border of uniform width is placed around a rectangular 19 Apr 2021, 09:42
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AnkitSingh2020
Bunuel
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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:



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.

Show Spoiler
Attachment:
photograph.png

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?
Show more

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?
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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.

Show Spoiler
Attachment:
photograph.png
[/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.
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A border of uniform width is placed around a rectangular 19 Apr 2021, 10:05
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AnkitSingh2020


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.
Show more

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.
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A border of uniform width is placed around a rectangular 19 Apr 2021, 10:14
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Bunuel
AnkitSingh2020


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.

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.
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Cristal clear now! Thank so much :)
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A border of uniform width is placed around a rectangular 19 Apr 2021, 13:26
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Bunuel
AnkitSingh2020


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.

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.
Show more


Cannot find the question link
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A border of uniform width is placed around a rectangular 19 Apr 2021, 13:31
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ManyataM
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AnkitSingh2020


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.

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.


Cannot find the question link
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Here is the question: https://gmatclub.com/forum/a-border-of- ... l#p1158911
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A border of uniform width is placed around a rectangular 08 Jun 2021, 15:49
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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?
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A border of uniform width is placed around a rectangular 08 Jun 2021, 15:56
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Raki12
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?
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It's given in the stem: "A border of uniform width ..."
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A border of uniform width is placed around a rectangular 01 Nov 2021, 09:21
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Walkabout
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
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quick way to solve

Area of small region Sa= 10X8=80, area of border Sb=144
Area of big rect. including frame = 80+144 = 244 this also equals to (10+2x)*(8+2x) = 244
scan the answer choice lets start with C, 10+12 * 8+12 = 22*20= bigger than 244 so D & E also no. only 16X14 will result in ending 4. so ans is A
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A border of uniform width is placed around a rectangular 09 Mar 2022, 11:23
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Walkabout
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
Show more

Total area = (photograph area) + (border area) = (8*10) + 144 = 224.

Since the border SURROUNDS the 8 by 10 photograph, the total area can be represented as follows:
(border width + 8 + border width)(border width + 10 + border width) = 224

We can PLUG IN THE ANSWERS, which represent the border width.
When the correct answer is plugged in, total area = 224.

B: 4
In this case, (border width + 8 + border width)(border width + 10 + border width) = (4+8+4)(4+10+4) = 16*18 = 288.
Since the total area is TOO BIG, a smaller answer choice is needed.

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A
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A border of uniform width is placed around a rectangular 09 Mar 2022, 23:23
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Walkabout
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
Show more


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.
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A border of uniform width is placed around a rectangular 01 Apr 2022, 04:57
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Walkabout
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
Show more

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.
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A border of uniform width is placed around a rectangular 01 Apr 2022, 06:08
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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! :cool:

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. :) :thumbsup:

Lets try to use this strategy instead of conventional equations! :cool:
:idea: 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

:idea: 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

:idea: 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.

:idea: 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
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A border of uniform width is placed around a rectangular 03 Oct 2022, 03:01
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JeffTargetTestPrep
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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.



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.
Show more

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.
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A border of uniform width is placed around a rectangular 12 Dec 2022, 17:22
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Video solution from Quant Reasoning:
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A border of uniform width is placed around a rectangular 12 Dec 2022, 23:14
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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?

Attachment:
Screenshot 2022-12-13 at 11.41.16 AM.png
Screenshot 2022-12-13 at 11.41.16 AM.png [ 15.59 KiB | Viewed 2549 times ]

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
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A border of uniform width is placed around a rectangular 13 Feb 2023, 04:04
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Kimberly77
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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


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.
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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!


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Aditi Gupta
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A border of uniform width is placed around a rectangular 03 Jun 2023, 09:10
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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.
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