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For a party, three solid cheese balls with diameters of 2 inches, 4 in
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20 Oct 2015, 02:57
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For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.) (A) 12 (B) 16 (C) \(\sqrt[3]{16}\) (D) \(3\sqrt[3]{8}\) (E) \(2\sqrt[3]{36}\) Kudos for a correct solution.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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20 Oct 2015, 05:27
Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. diameters of 2 inches, 4 inches, and 6 inches i.e. Radie of 1 inches, 2 inches, and 3 inches Sum of their volumes = \(\frac{4}{3}\pi (1^3+2^3+3^3)\) = \(\frac{4}{3}\pi (36)\) Volume of New Ball = \(\frac{4}{3}\pi (R^3)\) = \(\frac{4}{3}\pi (36)\) i.e. \(R^3 = 36\) i.e. \(R = \sqrt[3]{36}\) i.e. \(Diameter = 2\sqrt[3]{36}\) Answer: Option E
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For a party, three solid cheese balls with diameters of 2 inches, 4 in
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Updated on: 21 Oct 2015, 09:22
On combining the 3 cheese balls, we get a combined volume of (4/3) π (1+8+27)= 48π.
Equating 48π with the formula for vol.of sphere 48π=4/3πr3 r3=36 So diameter= 236−−√3
Option E



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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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20 Oct 2015, 09:34
HarrishGowtham wrote: On combining the 3 cheese balls, we get a combined volume of (4/3) π (1+8+27)= 36π.
Equating 36π with the formula for vol.of sphere 36π=4/3πr3 r3=36 So diameter= 236−−√3
Option E The answer that you have calculated is correct however the Highlighted calculation seems to be flawed. (4/3) π (1+8+27) is NOT equal to 36π. and 36π is NOT equal to 4/3πr3 I am sure the idea that you used is correct and it's just some typo.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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20 Oct 2015, 09:38
Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. The volume of the combined cheese ball is \(\frac{4}{3}\pi*36\). This is based on the three single \(r^3\) from each cheese ball with radius 1, 2 and 3. Now the new diameter will be 2*r = \(2\sqrt[3]{36}\), where \(r^3 = 36\) Answer E.
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For a party, three solid cheese balls with diameters of 2 inches, 4 in
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21 Oct 2015, 05:09
Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. \(\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3\) equating the volumes of cheese balls, we get \(\frac{1}{6}\pi d^3 = \frac{1}{6}\pi 2^3 + \frac{1}{6}\pi 4^3 + \frac{1}{6}\pi 6^3\) => \(d^3 = 2^3 + 4^3 + 6^3\) => \(d^3 = 2^3 ( 1 + 8 + 27)\) => \(d= 2\sqrt[3]{36}\) Answer choice EKudos, if you like the explanation



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For a party, three solid cheese balls with diameters of 2 inches, 4 in
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28 Oct 2015, 06:31
v12345 wrote: Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. \(\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3\) equating the volumes of cheese balls, we get \(\frac{1}{6}\pi d^3 = \frac{1}{6}\pi 2^3 + \frac{1}{6}\pi 4^3 + \frac{1}{6}\pi 6^3\) => \(d^3 = 2^3 + 4^3 + 6^3\) => \(d^3 = 2^3 ( 1 + 8 + 27)\) => \(d= 2\sqrt[3]{36}\) Answer choice EKudos, if you like the explanation How do you get \(\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3\) ?



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For a party, three solid cheese balls with diameters of 2 inches, 4 in
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29 Oct 2015, 02:51
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is 43πr3, where r is the radius.) (A) 12 (B) 16 (C) 3√16 (D) 33√8 (E) 23√36 Since the diameters are 2, 4, 6 the radii are 1, 2, 3. So the total volume is (4/3)*π*1^3 + (4/3)*π*2^3 + (4/3)*π*3^3 =(4/3)*π*(1^3 + 2^3 +3^3 )=(4/3)*π*36. Let the radius of the new cheese ball be r. Then the volume of the new cheese ball is (4/3)*π*r^3. So (4/3)*π*r^3 should be (4/3)*π*36. That means r= 3√36. So diameter is 2 3√36. So the answer is (E).
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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01 Nov 2015, 05:47
Let R be the radius of the single combined cheese ball Sum of volumes of the 3 solid cheese balls = Volume of the single combined cheese ball => 4/3 * pi *( 1^(3) + 2^(3) + 3^(3) ) = 4/3 * pi * R^(3) => R^(3) = 36 Diameter = 2R = 2(36^(1/3)) Answer E
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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28 Feb 2016, 02:15
My mistake was to calculate (1+2+3)³ instead of (1³+2³+3³)  can somebody explicate on why exactly it is wrong? I see the mistake but I just can't find a sound explanation..



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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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05 May 2016, 00:24
momomo wrote: My mistake was to calculate (1+2+3)³ instead of (1³+2³+3³)  can somebody explicate on why exactly it is wrong? I see the mistake but I just can't find a sound explanation.. Hi, the problem with your approach is you combined their length of radius to create one large sphere. But in reality we are combining the volumes of the spheres to create a larger sphere.So we have to take sum of their volumes and not take sum of their radius.



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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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05 May 2016, 04:52
Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. Solution: We first need to determine the volume of each individual cheese ball. We have 3 cheese balls of diameters of 2, 4, and 6 inches, respectively. Therefore, their radii are 1, 2 and 3 inches, respectively. Now let’s calculate the volume for each cheese ball. Volume for 2inch diameter cheese ball(4/3)π(1)^3 = (4/3)π Volume for 4inch diameter cheese ball(4/3)π(2)^3 = (4/3)π(8) = (32/3)π Volume for 6inch diameter cheese ball(4/3)π(3)^3 = (4/3)π(27) = (108/3)π Thus, the total volume of the large cheese ball is: (4/3)π + (32/3)π + (108/3)π = (144/3)π = 48π We can now use the volume formula to first determine the radius, and then the diameter, of the combined cheese ball. 48π = 4/3π(r)^3 48 x 3 = 4(r)^3 144 = 4(r)^3 36 = r^3 r = (cube root)√36 Thus, the diameter = 2*(cube root)√36 Answer: E
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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10 Jun 2016, 09:11
( 1^3 + 2^3 + 3^3) = r ^3 r = cube root (36) d= 2r
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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17 Nov 2016, 09:57
Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. combined volume of 3 solid cheese balls=(4/3)*πr^3 >(4/3)*π(1^3+2^3+3^3) > (4/3)*π(36) Let, the radius of new ball=R, So, the volume of new balls=(4/3)*πR^3 So, we can write, (4/3)*πR^3=(4/3)*π(36) > R^3=36 > R=∛36 So, diameter (2*R)=2∛36 So, the correct answer is E.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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17 Nov 2016, 10:17
Key equation: Sum of individual volumes = Total volume 4/3*pi*(r1^3+r2^3+r3^3) = 4/3*pi*(R^3) r1, r2 and r3 are known > Solve for R. Be careful, that the question is diameter do answer = 2*R.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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18 Jun 2017, 16:45
GMATinsight wrote: Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. diameters of 2 inches, 4 inches, and 6 inches i.e. Radie of 1 inches, 2 inches, and 3 inches Sum of their volumes = \(\frac{4}{3}\pi (1^3+2^3+3^3)\) = \(\frac{4}{3}\pi (36)\) Volume of New Ball = \(\frac{4}{3}\pi (R^3)\) = \(\frac{4}{3}\pi (36)\) i.e. \(R^3 = 36\) i.e. \(R = \sqrt[3]{36}\) i.e. \(Diameter = 2\sqrt[3]{36}\) Answer: Option E Simple question : why we calculate the sum of their volumes? In my opinion, total volume of NEW cheese ball must be greater than total combined volume from three small cheese balls.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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18 Jun 2017, 21:40
septwibowo wrote: GMATinsight wrote: Bunuel wrote: For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is \(\frac{4}{3}\pi r^3\), where r is the radius.)
(A) 12
(B) 16
(C) \(\sqrt[3]{16}\)
(D) \(3\sqrt[3]{8}\)
(E) \(2\sqrt[3]{36}\)
Kudos for a correct solution. diameters of 2 inches, 4 inches, and 6 inches i.e. Radie of 1 inches, 2 inches, and 3 inches Sum of their volumes = \(\frac{4}{3}\pi (1^3+2^3+3^3)\) = \(\frac{4}{3}\pi (36)\) Volume of New Ball = \(\frac{4}{3}\pi (R^3)\) = \(\frac{4}{3}\pi (36)\) i.e. \(R^3 = 36\) i.e. \(R = \sqrt[3]{36}\) i.e. \(Diameter = 2\sqrt[3]{36}\) Answer: Option E Simple question : why we calculate the sum of their volumes? In my opinion, total volume of NEW cheese ball must be greater than total combined volume from three small cheese balls. This does not make sense. How can you make a ball larger in volume than the combined volume of three balls you are making it from?
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in
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18 Aug 2017, 10:53
Tip: Volume remains same when a solid is transformed from one shape to another!
Sum of their volumes = 43π(13+23+33)43π(13+23+33) = 43π(36)43π(36)
Volume of New Ball = 43π(R3)43π(R3) = 43π(36)43π(36)
i.e. R3=36R3=36 i.e. R=36‾‾‾√3R=363 i.e. Diameter=236‾‾‾√3




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