GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Nov 2018, 08:45

# LBS is Calling R1 Admits - Join Chat Room to Catch the Latest Action

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Key Strategies to Master GMAT SC

November 24, 2018

November 24, 2018

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

# A border of uniform width is placed around a rectangular

Author Message
TAGS:

### Hide Tags

CEO
Joined: 20 Mar 2014
Posts: 2635
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: 2830
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: 5
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!
Manager
Joined: 22 Nov 2016
Posts: 213
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: 20
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: 205
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: 50730
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: 9
Location: United States (NY)
GMAT 1: 690 Q44 V40
GMAT 2: 710 Q44 V43
GPA: 3.25
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.
VP
Joined: 09 Mar 2016
Posts: 1112
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:)
VP
Joined: 07 Dec 2014
Posts: 1118
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
Intern
Joined: 15 Sep 2018
Posts: 30
A border of uniform width is placed around a rectangular  [#permalink]

### Show Tags

03 Oct 2018, 03:47
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 [ 16.12 KiB | Viewed 196 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.

Intern
Joined: 15 Sep 2018
Posts: 30
Re: A border of uniform width is placed around a rectangular  [#permalink]

### Show Tags

16 Nov 2018, 00:13
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 $$x$$:

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

Re: A border of uniform width is placed around a rectangular &nbs [#permalink] 16 Nov 2018, 00:13

Go to page   Previous    1   2   [ 32 posts ]

Display posts from previous: Sort by

# A border of uniform width is placed around a rectangular

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.