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The formula for the surface of a sphere is S=4πr^2. Two lead balls
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petocities
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The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  24 Oct 2015, 11:52
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84% (01:37) correct 16% (01:47) wrong based on 263 sessions

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The formula for the surface of a sphere is \(S=4πr^2\). Two lead balls have radii 3 inches and 6 inches, respectively. A third ball has surface area equal to the sum of the surface areas of the two smaller balls. What is the radius, in inches, of the third ball?

    (A) \(3\sqrt{10}\)
    (B) \(3\sqrt{5}\)
    (C) \(36π\)
    (D) \(45π\)
    (E) \(180π\)

It's not really a hard question, but I got it wrong because I didn't take enough time to read carefully the question. So i'll just leave it here for others to avoid my mistake.
Good luck ya'll

PS: Difficulty is probably under 600, but couldn't edit the tags
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erratic15
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  24 Oct 2015, 12:28
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petocities
The correct answer is B.
The surface area of the bigger sphere is = 4π (3^2 +6^2) = 4πr^2
4π (9 + 36) = 4πr^2
45 = r^2
r = \sqrt{45}
= 3 \sqrt{5}
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petocities
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GMAT 1: 620 Q38 V37
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  25 Oct 2015, 09:45
erratic15 wrote:
petocities
The correct answer is B.
The surface area of the bigger sphere is = 4π (3^2 +6^2) = 4πr^2
4π (9 + 36) = 4πr^2
45 = r^2
r = \sqrt{45}
= 3 \sqrt{5}


Yeah, upon revising my calculations I got it right, but on my first time I stopped at 4π(9+36) = 180π. Since that's (E), I just selected it. That's the silly mistake that one has to avoid!
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rasti12
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  27 Feb 2017, 10:51
Can please somebody explain how the translation from /sqrt(45) --> 3*/sqrt(5) works?
I got this one right but only because I canceled the 3 obvious wrong ones and then estimated what value r approximately has to be
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Bunuel
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  03 Mar 2017, 01:04
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rasti12 wrote:
Can please somebody explain how the translation from /sqrt(45) --> 3*/sqrt(5) works?
I got this one right but only because I canceled the 3 obvious wrong ones and then estimated what value r approximately has to be


\(\sqrt{45}=\sqrt{9*5}=3\sqrt{5}\)

Hope it's clear.
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gmat831
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The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  07 Oct 2018, 07:18
Can someone please explain why the SA of the third lead ball is not 180π? I calculated the first SA as 36π and the second as 144π then added them together. What am I missing to get to 45π as the third ball's SA?
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Aaditya96
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The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  20 Jul 2019, 06:21
First we have to determine the sum of the surface areas of the two smaller balls.
1. 4∏3^2 = 4*∏*9 = 36∏
2. 4∏6^2 = 4*∏*36=144∏
144∏+36∏= 180∏
Now we can determine the radius
180∏/4= 45∏

√45 = 5*3*3 = 3√5
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink] New post  26 Jul 2020, 05:44
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Re: The formula for the surface of a sphere is S=4πr^2. Two lead balls [#permalink]
26 Jul 2020, 05:44
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