January 26, 2019 January 26, 2019 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. January 27, 2019 January 27, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 177

A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
20 Dec 2012, 04:33
Question Stats:
62% (02:17) correct 38% (02:07) wrong based on 2717 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) 3 (B) 4 (C) 6 (D) 8 (E) 9
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 52428

A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
20 Dec 2012, 04:42




SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1823
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
27 Aug 2014, 01:27
Refer diagram below: Attachment:
photograph.png [ 3.89 KiB  Viewed 106998 times ]
Area of multicoloured shaded region is given = 144 Area of each square (Yellow) = \(x^2\) ; No. of squares = 4 Area of rectangle (pink) = 10x ; No. of rectangle (pink) = 2 Area of rectangle (blue) = 8x ; No. of rectangle (pink) = 2 Setting up the equation \(4x^2 + 20x + 16x = 144\) \(x^2 + 9x  36 = 0\) x = 3 Answer = A
_________________
Kindly press "+1 Kudos" to appreciate




Manager
Joined: 18 Aug 2006
Posts: 87
Location: United States
WE: Consulting (Telecommunications)

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
25 May 2013, 04:42
Let the width of the uniform border be X
Area of photo 80 (given)
Area of Photo + border = (10+2X)*(8+2X) (deduced)
Difference in areas, (10+2X)*(8+2X)  80 = 144 (given)
(10+2X)*(8+2X) =224 (10+2X)*(8+2X) =16*14 (10+2X)*(8+2X) =(10+6)(8+6)
=> 2X= 6 =>X=3



Intern
Joined: 09 Sep 2013
Posts: 16

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
18 Oct 2013, 12:24
Why does this not work....
(2x+10)(2x+8)=224
Divide above by 2 to get...
(x+5)(x+4)=112



Math Expert
Joined: 02 Sep 2009
Posts: 52428

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
18 Oct 2013, 12:33



Intern
Joined: 04 Feb 2014
Posts: 12

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
11 Feb 2014, 13:54
why are we adding both sides by 2x



Math Expert
Joined: 02 Sep 2009
Posts: 52428

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
12 Feb 2014, 00:21



Manager
Joined: 25 Sep 2012
Posts: 245
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
25 Aug 2014, 21:19
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
Answer A



Director
Joined: 23 Jan 2013
Posts: 560

A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
Updated on: 12 Mar 2015, 19:42
Backsolving: we are looking for area 144+80=224 C. 22*20=440, too much B. 18*16=288, too much
must be A
Originally posted by Temurkhon on 26 Aug 2014, 21:29.
Last edited by Temurkhon on 12 Mar 2015, 19:42, edited 1 time in total.



Manager
Joined: 25 May 2014
Posts: 66

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
02 Jan 2015, 17:03
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.
So Width of border = (W2W1) = 118 = 3
Answer choice A.



Intern
Joined: 11 Apr 2015
Posts: 32
Location: Germany
Concentration: General Management, Entrepreneurship
GPA: 3.1
WE: Project Management (Energy and Utilities)

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
24 Apr 2015, 11:40
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(102x). 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



Manager
Joined: 26 Feb 2015
Posts: 115

A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
18 May 2015, 02:22
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: \((x+y)(xy)\), which in this case is \((15+1)(151)\). 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



Manager
Joined: 11 Oct 2013
Posts: 106
Concentration: Marketing, General Management

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
18 Nov 2015, 23:52
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!
_________________
Its not over..



Intern
Joined: 18 Aug 2012
Posts: 10

A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
29 Nov 2015, 06:39
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 91942 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 \(4a^2+36a\)=\(144\) \(4a^2+36a144=0\) \(a^2+9a36=0\) \(a^2+12a3a36=0\) \((a3)(a+12)=0\) 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.



Intern
Joined: 14 Jan 2016
Posts: 7

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
21 Jan 2016, 14:48
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 largerthanphotograph size?



CEO
Joined: 20 Mar 2014
Posts: 2636
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
21 Jan 2016, 17:09
vignuol wrote: 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 largerthanphotograph 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 aborderofuniformwidthisplacedaroundarectangular144434.html#p1158913 with dimensions of the photograph = 8 x 10 while those of the entire thing (photograph+border) = 14 x 16.



Intern
Joined: 14 Jan 2016
Posts: 7

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
22 Jan 2016, 00:13
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...



CEO
Joined: 20 Mar 2014
Posts: 2636
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
22 Jan 2016, 02:26
vignuol wrote: 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. Hope this helps.



Intern
Joined: 18 May 2015
Posts: 12

Re: A border of uniform width is placed around a rectangular
[#permalink]
Show Tags
07 Feb 2016, 11:30
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 \(4a^2+36a\)=\(144\) \(4a^2+36a144=0\) \(a^2+9a36=0\) \(a^2+12a3a36=0\) \((a3)(a+12)=0\) 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?




Re: A border of uniform width is placed around a rectangular &nbs
[#permalink]
07 Feb 2016, 11:30



Go to page
1 2
Next
[ 32 posts ]



