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A student cuts 80 rectangles from construction paper, all of which are

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A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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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
[Reveal] Spoiler: OA
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Re: A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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l = 10l > 10Total
w = 1022830
w > 10183250
Total404080


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

Answer: D
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Re: A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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New post 27 Feb 2017, 04:33
Vyshak wrote:



l = 10l > 10Total
w = 1022830
w > 10183250
Total404080


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

Answer: D


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.
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Re: A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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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.
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Re: A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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grichagupta wrote:
Vyshak wrote:



l = 10l > 10Total
w = 1022830
w > 10183250
Total404080


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

Answer: D


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.
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Re: A student cuts 80 rectangles from construction paper, all of which are [#permalink]

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New post 08 Nov 2017, 23:35
Vyshak wrote:
grichagupta wrote:
Vyshak wrote:



l = 10l > 10Total
w = 1022830
w > 10183250
Total404080


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

Answer: D


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"
Re: A student cuts 80 rectangles from construction paper, all of which are   [#permalink] 08 Nov 2017, 23:35
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