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Intern  B
Joined: 09 Sep 2016
Posts: 35
Location: Georgia
GPA: 3.75
WE: Analyst (Investment Banking)
A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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4
22 00:00

Difficulty:   95% (hard)

Question Stats: 54% (02:56) correct 46% (03:23) wrong based on 213 sessions

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A student cuts 80 rectangles from construction paper, all of which are at least 10 inches in length and in width, and 20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide. if 40 of the rectangles have a length of exactly 10 inches and 50 of the rectangles are greater than 10 inches wide. How many rectangles have a perimeter of greater than 40 inches? (Note: assume that width and length are interchangeable; in other words, width does not have to be shorter than length)

(A) 18
(B) 22
(C) 32
(D) 58
(E) 66
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1682
Location: India
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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1
 l = 10 l > 10 Total w = 10 22 8 30 w > 10 18 32 50 Total 40 40 80

Condition: Perimeter > 40
2(l + w) > 40
l + w > 20
Number of rectangles satisfying this condition = 8 + 18 + 32 = 58

##### General Discussion
Intern  S
Joined: 13 Dec 2016
Posts: 41
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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Vyshak wrote:
 l = 10 l > 10 Total w = 10 22 8 30 w > 10 18 32 50 Total 40 40 80

Condition: Perimeter > 40
2(l + w) > 40
l + w > 20
Number of rectangles satisfying this condition = 8 + 18 + 32 = 58

Dear Vyshak, Request you to explain as to how you arrived at the break up of 22 / 8 / 18 and 32. I understood rest of the figures and calculation.
Intern  Joined: 26 Feb 2017
Posts: 3
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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1
grichagupta

"20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide"
l>10 w=10 : (20%*40) = 8, then you just complete the rest of the table.
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1682
Location: India
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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2
grichagupta wrote:
Vyshak wrote:
 l = 10 l > 10 Total w = 10 22 8 30 w > 10 18 32 50 Total 40 40 80

Condition: Perimeter > 40
2(l + w) > 40
l + w > 20
Number of rectangles satisfying this condition = 8 + 18 + 32 = 58

Dear Vyshak, Request you to explain as to how you arrived at the break up of 22 / 8 / 18 and 32. I understood rest of the figures and calculation.

Length and width are atleast 10 inches.
Total number of rectangles = 80

Number of rectangles with length of 10 inches = 40 --> Number of rectangles with length greater than 10 inches = 40
Number of rectangles with width greater than 10 inches = 50 --> Number of rectangles with width of 10 inches = 30

20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide --> (1/5)*40 = 8 rectangles with l > 10 and w = 10.

You will be able to find the other blank cells once you know the above values.
Intern  B
Joined: 05 Jun 2014
Posts: 9
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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Vyshak wrote:
grichagupta wrote:
Vyshak wrote:
 l = 10 l > 10 Total w = 10 22 8 30 w > 10 18 32 50 Total 40 40 80

Condition: Perimeter > 40
2(l + w) > 40
l + w > 20
Number of rectangles satisfying this condition = 8 + 18 + 32 = 58

Request you to explain as to how you arrived at the break up of 22 / 8 / 18 and 32. I understood rest of the figures and calculation.

Length and width are atleast 10 inches.
Total number of rectangles = 80

Number of rectangles with length of 10 inches = 40 --> Number of rectangles with length greater than 10 inches = 40
Number of rectangles with width greater than 10 inches = 50 --> Number of rectangles with width of 10 inches = 30

20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide --> (1/5)*40 = 8 rectangles with l > 10 and w = 10.

You will be able to find the other blank cells once you know the above values.

Thank you, this helped.
I was confused with the wording "20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide"
Duke & Cornell Moderator S
Joined: 29 May 2018
Posts: 100
Location: India
GMAT 1: 700 Q48 V38 GMAT 2: 720 Q49 V39 GPA: 4
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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How can the width of any rectangle be greater than the length of that rectangle. I was able to solve the question using double matrix method but I had a hard time digesting the question stem. Wasted a lot of time re-reading the question to make sure my understanding was correct.

Bunuel can you help?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9866
Location: Pune, India
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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giobas wrote:
A student cuts 80 rectangles from construction paper, all of which are at least 10 inches in length and in width, and 20 percent of rectangles that are greater than 10 inches long are exactly 10 inches wide. if 40 of the rectangles have a length of exactly 10 inches and 50 of the rectangles are greater than 10 inches wide. How many rectangles have a perimeter of greater than 40 inches? (Note: assume that width and length are interchangeable; in other words, width does not have to be shorter than length)

(A) 18
(B) 22
(C) 32
(D) 58
(E) 66

Given by the question:

.................Exactly 10 ................ > 10......... Total

Length ........40 ...............................................80
Width .......... .................................50..............80

We can infer the:

.................Exactly 10 ................ > 10 ......... Total

Length ........40 .............................40 .............80
Width ...........30.............................50 .............80

20% of 40 rectangles (>10 length) are 8 rectangles. They are a part of 30 that have exactly 10 width. So other 22 rectangles must have exactly 10 length too.
So only 22 rectangles have exactly 10 length and exactly 10 width which gives exactly 40 perimeter.

Rest 80 - 22 = 58 rectangles must have perimeter greater than 40.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  S
Joined: 10 Jun 2019
Posts: 117
Re: A student cuts 80 rectangles from construction paper, all of which are  [#permalink]

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This is a very nasty question and a bit difficult to spell out but this is how i approached it. 40 have L=10.meaning 40 have L>10. Now 50 have W>10 meaning 30 have w=10. 20 % of those with L>10 have W= 10. So rectangles wth L>10 & W=10 are 8 in total. Meaning L>10 and W>10 are 32 in total. Now we know that W>10 are 50 in number . We have already accounted for 32 of them which have dimensions L>10 and W> 10.The rest of the 18 with W>10 will have L=10. Now we know L=10 are 40 and we have accounted for 18 of them.Meaning the rest of the L=10 have W=10.Hence the number of L=10 and W=10 are 22. Looking at these the total number of rectangles with perimeter greater than 40 is 50+ 8=58 if you want to go by the 50 -30 grouping or just as well 18+32+8 if you want to go by the 40-40 grouping Re: A student cuts 80 rectangles from construction paper, all of which are   [#permalink] 08 Jul 2019, 09:14
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# A student cuts 80 rectangles from construction paper, all of which are  