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A project requires a rectangular sheet of cardboard satisfying the fol
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19 Sep 2017, 12:50
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35% (02:46) correct 65% (02:29) wrong based on 275 sessions
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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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Re: A project requires a rectangular sheet of cardboard satisfying the fol
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19 Apr 2018, 14:42
bkpolymers1617 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 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 GMAT, everyone should know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2 So, we know that x/y ≈ 1.4In 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 satisfying the fol
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19 Sep 2017, 13:09
Algebraic Solution Say the original sheet measures L by W, with L > W, so that the cut yields two rectangular halves measuring L/2 by W each. Per the project requirements, the ratio of these new dimensions must be the same as the ratio of L to W. Consider both possible ways in which the ratio might be set up. If the ratio of original to new is set up as L : W = L/2 : W, then it reduces to 1 = 1/2, an impossible condition. (Or, if you prefer to reason theoretically, it is impossible to reduce L by half but leave W unchanged and have the ratio stay the same.) Therefore, the width (W) of the original rectangle, must become the longer side for the smaller rectangles. The ratio must be set up as L : W = W : L/2. Set up the ratio: \(L/W = 2L/W\) L/W= 1.4 ( \(Root 2 = 1.4\)) The condition thus requires original dimensions that come as close as possible to satisfying , or, equivalently, . The one closest to 1.4 is the correct answer. (A) 10/7 = approximately 1.4 (B) 14/8 = 7/4 = 1.75 (C) 13/10 = 1.3 (D) 5/3 = = approximately 1.7 (E) 8/5 = = 1.6 Choice A comes closest to the requirement. The correct answer is (A).
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A project requires a rectangular sheet of cardboard satisfying the fol
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19 Sep 2017, 13:02
The first and the foremost thing to realize is that one side of the figure would be cut by half. Now when you go through the options, please make sure on thing. There is no need to half the smaller side in the option, as it would be the total waste of time. For example, if you have a rectangular figure with a ratio of 1:4, then halfing the smaller side would bring the ration to .5 to 4, which means that the ratio will decrease further rather than remaining the same. So now let's go to the option choices: a) Sheet measuring 7*10 : Lets half the larger side"10", we get 7:5, now is 5/7 equal to 7/10. I would say pretty close lets keep. b) A sheet measuring 8 inches by 14 inches: Half the larger side we get, 7/8 vs 8/14: No chance of being close eliminate c) A sheet measuring 10 inches by 13 inches: Half the larger side we get, 6.5/10 vs 10/13: Not close eliminate d) A sheet measuring 3 feet by 5 feet: 2.5/3 vs 3/5 : No chance e) A sheet measuring 5 feet by 8 feet: 4/5 vs 5/8: Not close Your answer is A
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A project requires a rectangular sheet of cardboard satisfying the fol
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24 Jun 2018, 23:01
GMATPrepNow wrote: bkpolymers1617 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 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 GMAT, everyone should know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2 So, we know that x/y ≈ 1.4In 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 Hi Brent a lil basic question : While trying to maintain the ratio of the new triangle to the Original Triangle why did you reverse the side . SHould the Sides of both the triangles not corelate that is height / width vs Height / width ie x/y ( Old ratio ) = (x/2 )/ y ( new ratio ) ? in the solution provided by you you have done the following " In other words, we want x/y = y/(x/2) Cross multiply to get: x²/2 = y² '' PLease explain



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Re: A project requires a rectangular sheet of cardboard satisfying the fol
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18 Sep 2019, 02:40
Do the math. OR print thousands of documents in letter (7x11) and tabloid (11x14) sizes and KNOW that when you fold a tabloid in half you get 2 letter sizes. Closest to these dimensions is 7x10 HENCE A.
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Re: A project requires a rectangular sheet of cardboard satisfying the fol
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30 Sep 2019, 07:08
bkpolymers1617 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 An easy method to do this would be  a) 7/10, 5/7, now multiply the denominator with the numerator and see which numbers are closest. so, in this case, 49 , 50 b) 4/7, 7/8 =>32, 49 c) 10/13, 13/20 . 200, 169 d) 3/5, 5/6. 18, 15 e) 5/8, 4/5. 25, 32 As you can see that A is closest, so you dont have to approximate the values in this.



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A project requires a rectangular sheet of cardboard satisfying the fol
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19 Oct 2019, 10:26
mimajit wrote: GMATPrepNow wrote: bkpolymers1617 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 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 GMAT, everyone should know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2 So, we know that x/y ≈ 1.4In 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 Hi Brent a lil basic question : While trying to maintain the ratio of the new triangle to the Original Triangle why did you reverse the side . SHould the Sides of both the triangles not corelate that is height / width vs Height / width ie x/y ( Old ratio ) = (x/2 )/ y ( new ratio ) ? in the solution provided by you you have done the following " In other words, we want x/y = y/(x/2) Cross multiply to get: x²/2 = y² '' PLease explain Hi mimajit, The question you have asked is actually the key to getting the question correct. If you observe the diagram from Brent's post then x is the longer side and y is the shorter side of the rectangle. And we know that in a rectangle the length is the longer side and the width is the shorter side. When we slice the rectangle into two halves x divides into two \(\frac{x}{2}\)'s and now \(\frac{x}{2}\) no longer remains the longer side and it is y that becomes the length now. Hence, when we take the ratio of length by width y becomes the length and \(\frac{x}{2}\) becomes the width. As a side note, if you continue to keep \(\frac{x}{2}\) as the length then you will end up with a tautology  \(\frac{x}{y} = \frac{x}{2y}\) where \(\frac{x}{y}\) cancel out on both sides. Warm Regards, Pritish
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