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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A border of uniform width is placed around a rectangular [#permalink]

Show Tags

20 Dec 2012, 05:33

2

This post received KUDOS

45

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

57% (03:22) correct
43% (02:18) wrong based on 1106 sessions

HideShow timer Statistics

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 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:

Attachment:

photograph.png [ 5.15 KiB | Viewed 34563 times ]

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

Re: A border of uniform width is placed around a rectangular [#permalink]

Show Tags

02 Jan 2015, 18:03

1

This post received KUDOS

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.

Re: A border of uniform width is placed around a rectangular [#permalink]

Show Tags

19 Nov 2015, 00:52

1

This post received KUDOS

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! _________________

Re: A border of uniform width is placed around a rectangular [#permalink]

Show Tags

25 Aug 2014, 22:19

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

Re: A border of uniform width is placed around a rectangular [#permalink]

Show Tags

24 Apr 2015, 12:40

I didn't find this question difficult, however I still managed to make a mistake. I basically thought that the outer size of the board matches with the edges of the picture. This would give the following equation: 144= 2(8x) +2x(10-2x). However , the inner sides of the board match with the picture. I think that question did not state it clear enough. _________________

"I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times." Bruce Lee

"I hated every minute of training, but I said, "Don’t quit. Suffer now and live the rest of your life as a champion."" Muhammad Ali

A border of uniform width is placed around a rectangular [#permalink]

Show Tags

18 May 2015, 03:22

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

A border of uniform width is placed around a rectangular [#permalink]

Show Tags

29 Nov 2015, 07:39

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 [ 8.39 KiB | Viewed 12783 times ]

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.

I'd like to ask you the main question I have: if the photograph has sides 8 x 10, the border should not have a larger-than-photograph size?

You mean the total or final dimensions of picture with border? If that is the case then yes, you are correct and that is what is mentioned in a-border-of-uniform-width-is-placed-around-a-rectangular-144434.html#p1158913 with dimensions of the photograph = 8 x 10 while those of the entire thing (photograph+border) = 14 x 16. _________________

Re: A border of uniform width is placed around a rectangular [#permalink]

Show Tags

22 Jan 2016, 01:13

I mean the size of the border ALONE: the question says that the PHOTOGRAPH is 8x10, so it's logical to think that the border has a width bigger than 8 or 10. You said 14x16,but the answer is 3...I don't understand...

I mean the size of the border ALONE: the question says that the PHOTOGRAPH is 8x10, so it's logical to think that the border has a width bigger than 8 or 10. You said 14x16,but the answer is 3...I don't understand...

No, you are confusing width of the border with the final dimensions of photograph + border.

Yes, x=3 is the width but the final dimensions of photo + border are greater than those of photo alone.

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...