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A project requires a rectangular sheet of cardboard satisfyi

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GreenlightTestPrep
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A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   30 Jan 2019, 09:28
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A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?

(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet
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A
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chetan2u
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   30 Jan 2019, 20:31
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GreenlightTestPrep wrote:
A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?

(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet



So, when you half a rectangle, one side will get halved and other will remain the same.
Thus the ratio of B:L should be same as L/2:B or L:2B..
This means B:L=L:2B or \(L^2=2B^2\)..
Let us check the options with this.


(A) A sheet measuring 7 inches by 10 inches
B=7 and L=10, so \(L^2=2B^2...10^2=2*7^2....100=98\)..YES almost equal

(B) A sheet measuring 8 inches by 14 inches
B=8 and L=14, so \(L^2=2B^2...14^2=2*8^2....196=128\)..NO

(C) A sheet measuring 10 inches by 13 inches
B=10 and L=13, so \(L^2=2B^2...13^2=2*10^2....169=200\)..NO

(D) A sheet measuring 3 feet by 5 feet
B=3 and L=5, so \(L^2=2B^2...5^2=2*3^2....25=18\)..NO

(E) A sheet measuring 5 feet by 8 feet
B=5 and L=8, so \(L^2=2B^2...8^2=2*5^2....64=50\)..NO

A
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   31 Jan 2019, 10:46
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GreenlightTestPrep wrote:
A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?

(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet


Here's an algebraic solution:

Let x be length of the LONG side of the original rectangle
Let y be length of the SHORT side of the original rectangle
Then cut the rectangle into two pieces
Image

We want the resulting rectangles to have the same ratio of length to width as the original sheet.
In other words, we want x/y = y/(x/2)
Cross multiply to get: x²/2 = y²
Multiply both sides by 2 to get: x² = 2y²
Divide both sides by y² to get: x²/y² = 2
Take square root of both sides to get: x/y = √2

IMPORTANT: For the GRE, it's often useful to know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2

So, we know that x/y ≈ 1.4
In other words, the ratio (LONG side)/(SHORT side) ≈ 1.4

Now check the answer choices:

(A) 10/7 = 1 3/7 ≈ 1.4 LOOKS GOOD!
(B) 14/8 = 1 6/8 = 1.75 ELIMINATE
(C) 13/10 = 1.3 ELIMINATE
(D) 5/3 = 1 2/3 ≈ 1.66 ELIMINATE
(E) 8/5 = 1.6 ELIMINATE

Answer: A

Cheers,
Brent
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   02 Aug 2019, 16:33
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GreenlightTestPrep wrote:

In other words, we want x/y = y/(x/2)


Wouldn't this mean we are comparing the new length/width ratio to the old width/length ratio? Shouldn't we keep it as x/y=x/(y/2)??
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   03 Aug 2019, 06:08
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chillMike wrote:
GreenlightTestPrep wrote:

In other words, we want x/y = y/(x/2)


Wouldn't this mean we are comparing the new length/width ratio to the old width/length ratio? Shouldn't we keep it as x/y=x/(y/2)??


I believe the intent here is to compare the longer sides to the shorter sides.
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   03 Feb 2021, 09:49
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chillMike wrote:
GreenlightTestPrep wrote:

In other words, we want x/y = y/(x/2)


Wouldn't this mean we are comparing the new length/width ratio to the old width/length ratio? Shouldn't we keep it as x/y=x/(y/2)??



When you cut the 2 rectangles, notice the Width of original sheet is now the length of new sheet and vice-versa.
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Re: A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   30 Aug 2021, 14:07
chetan2u wrote:
GreenlightTestPrep wrote:

So, when you half a rectangle, one side will get halved and other will remain the same.
Thus the ratio of B:L should be same as L/2:B or L:2B..
This means B:L=L:2B or \(L^2=2B^2\)..
Let us check the options with this.


A

would you show me where in the question say that (B:L should be same as L/2:B or L:2B.)
what I understand (B:L should be same as B:L/2)
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A project requires a rectangular sheet of cardboard satisfyi [#permalink] New post   05 Sep 2022, 11:51
------X-----
|-----|-----|
|-----|-----| Y
|-----|-----|
------------

X/Y = Y/(.5X)
0.5X^2 = Y^2

So we're basically looking for something where the square of one number is about half the square of the other number. Looking at option A, we see 7^2 = 49, 10^2 = 100, and 49 is pretty close to half of 100.

Answer = A.
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