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Updated on: 14 Jun 2021, 20:07
Originally posted by AlexTheTrainer on 12 Jun 2021, 07:02.
Last edited by chetan2u on 14 Jun 2021, 20:07, edited 2 times in total.
Last edited by chetan2u on 14 Jun 2021, 20:07, edited 2 times in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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61% (02:48) correct
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Yvonne invited x people for dinner and plans to buy a pizza with a diameter of 14 inches. When she discovers that x + z people will be coming to dinner, how much greater in diameter does the pizza need to be, in terms of x and z, so that each person receives the same amount of pizza that Yvonne had originally planned?
(A) \( 14*(\frac{\sqrt{x+z}}{\sqrt{x}}-1)\)
(B) \(\frac{14\sqrt{xz}}{x}\)
(C) \(\frac{49𝜋z}{x}\)
(D) \(\frac{7\sqrt{xz}}{z}\)
(E) \(\frac{z}{7\pi x}\)
Note from a GMAT Club newbie (not new to the GMAT - wrote this about 10 years ago):
(B) and (D): radical sign intended to exactly cover xz in the numerator.
Posted from my mobile device
(A) \( 14*(\frac{\sqrt{x+z}}{\sqrt{x}}-1)\)
(B) \(\frac{14\sqrt{xz}}{x}\)
(C) \(\frac{49𝜋z}{x}\)
(D) \(\frac{7\sqrt{xz}}{z}\)
(E) \(\frac{z}{7\pi x}\)
Note from a GMAT Club newbie (not new to the GMAT - wrote this about 10 years ago):
(B) and (D): radical sign intended to exactly cover xz in the numerator.
Posted from my mobile device
12 Jun 2021, 21:12
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AlexTheTrainer
The distribution would always be on the area.
So, area =\(\pi r^2=49\pi\), and each of x would get \(\frac{49\pi }{x}\).
If z join in to make the new strength x+z, the requirement of additional quantity of pizza such that each gets \(\frac{49\pi }{x}\) = \(\frac{49z\pi }{x}\)
Thus the total area required for x+z people=\(\pi (r_n)^2=\) \(\frac{49(x+z)\pi }{x}\)
\((r_n)^2=\frac{49(x+z)}{x}......r_n=7*\frac{\sqrt{x+z}}{\sqrt{x}}\)
The diameter or 2r would be \( 2*7*\frac{\sqrt{x+z}}{\sqrt{x}}\).
Since we are looking at the increase, we require to subtract the initial diameter that is 14.
Additional diameter = \( 2*7*\frac{\sqrt{x+z}}{\sqrt{x}}-14\)
\( 14*(\frac{\sqrt{x+z}}{\sqrt{x}}-1)\)
12 Jun 2021, 21:46
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AlexTheTrainer
We could also substitute some values for x and z to get to our answer.
Say initially there was only one person and one more joins in, that is x=1 and z is also 1.
Initially the one person gets \(\pi *7^2=49\pi\)
So, for two persons to get \(49\pi\) each, the area should be \(98\pi\)
The new radius => \(\pi r^2=98\pi........r=7\sqrt{2}.......d=2r=14\sqrt{2}\)
The additional diameter = \(14\sqrt{2}-14=14(\sqrt{2}-1)\)
The correct answer that is missing from the options is \( 14*(\frac{\sqrt{x+z}}{\sqrt{x}}-1)\)
\( 14*(\frac{\sqrt{1+1}}{\sqrt{1}}-1)=14(\sqrt{2}-1)\)
