It is currently 23 Jan 2018, 10:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A border of uniform width is placed around a rectangular

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 20 Mar 2014
Posts: 2685
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

07 Feb 2016, 12:37
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.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1835
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

23 Jun 2016, 08:35
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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Joined: 12 Apr 2015
Posts: 6
A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

18 Aug 2016, 09:17
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!
Senior Manager
Joined: 22 Nov 2016
Posts: 251
Location: United States
GPA: 3.4
A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

05 Jul 2017, 20:59
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
_________________

Kudosity killed the cat but your kudos can save it.

Intern
Joined: 30 Jun 2017
Posts: 25
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

06 Jul 2017, 06:56
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!
Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 186
Location: India
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

20 Sep 2017, 08:00
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.
Math Expert
Joined: 02 Sep 2009
Posts: 43381
A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

07 Dec 2017, 19:52
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.

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 $$(4x^2+36x+80)-80=144$$ --> $$4x^2+36x-144=0$$ --> $$x^2+9x-36=0$$ --> $$(x-3)(x+12)=0$$ --> $$x=3$$.

_________________
Intern
Joined: 24 Oct 2017
Posts: 1
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

17 Dec 2017, 16:02
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.
Manager
Joined: 09 Mar 2016
Posts: 157
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

30 Dec 2017, 08:52
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?

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:)
Director
Joined: 07 Dec 2014
Posts: 885
Re: A border of uniform width is placed around a rectangular [#permalink]

### Show Tags

30 Dec 2017, 18:40
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
Re: A border of uniform width is placed around a rectangular   [#permalink] 30 Dec 2017, 18:40

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by