The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface : GMAT Problem Solving (PS)
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# The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface

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The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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07 Oct 2009, 19:53
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56% (02:15) correct 44% (01:49) wrong based on 45 sessions

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The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface area is 4*pi*r^2. If a spherical balloon has a volume of 972 pi cubic centimeters, what is hte surface area of the balloon in square centimeters?

a. 40
b. 100
c. 400
d. 1,000
e. 10,000
[Reveal] Spoiler: OA
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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08 Oct 2009, 00:48
amitgovin wrote:
the volume of a sphere with radius r is (4/3)*pi*r^3 and the surface area is 4*pi*r^3. If a sperical balloon has a volume of 972 pi cubic centimeters, what is hte surface area of the balloon in square centimeters?

a. 40
b. 100
c. 400
d. 1,000
e. 10,000

The surface area is 4.pi.r^2 (its area remember not volume)

as 4/3.pi.r^3=972pi

r=9

so area = 4.pi.r^2= 324.pi= 324 x 3.14 = 1000 (approx)
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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01 Nov 2010, 03:32
It's a very simple problem. But what if I don't know that 9^3 = 729? There must be some solution that can be used without that knowledge. Anyone? Thanks.
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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01 Nov 2010, 06:03
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nonameee wrote:
It's a very simple problem. But what if I don't know that 9^3 = 729? There must be some solution that can be used without that knowledge. Anyone? Thanks.

You can always factorize 729 to get 3^6 but to ensure that you do not waste time, it is essential to know the following powers:
i. $$2^2 = 4; 2^3 = 8; 2^4 = 16; 2^5 = 32; 2^6 = 64; 2^7 = 128; 2^8 = 256; 2^9 = 512; 2^10 = 1024$$
ii. $$3^2 = 9; 3^3 = 27; 3^4 = 81; 3^5 = 243; 3^6 = 729$$
iii. $$4^2 = 16; 4^3 = 64; 4^4 = 256, 4^5 = 1024$$(of course, you see the link with the powers of 2 here)
iv. $$5^2 = 25; 5^3 = 125; 5^4 = 625$$
v. $$6^2 = 36; 6^3 = 216$$
vi. $$7^2 = 49; 7^3 = 343$$
vii. $$8^2 = 64; 8^3 = 512$$
viii. $$9^2 = 81; 9^3 = 729$$
and it is good to know the squares of the next 10 numbers.

You should be so comfortable with them that if someone shakes you in your sleep and asks you, "What is 243?" You should say, "3^5"!!!
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 10 Sep 2010 Posts: 133 Followers: 2 Kudos [?]: 33 [0], given: 7 Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink] ### Show Tags 01 Nov 2010, 06:38 I see that this particular question provides a formula for volume of a sphere. But generaly speaking, are we suppose to know such fomulas by heart? Thanks. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2138 Kudos [?]: 13694 [0], given: 222 Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink] ### Show Tags 01 Nov 2010, 07:00 Fijisurf wrote: I see that this particular question provides a formula for volume of a sphere. But generaly speaking, are we suppose to know such fomulas by heart? Thanks. No! But, you should know the formula of calculating area of common shapes like square, circle, triangle, rectangle etc. Using this knowledge, you can calculate the area of 3D figures where one of these shapes is the base and the height is perpendicular to base. Volume for such figures = Area of Base x Height e.g. A cylinder - Circular base and height perpendicular to base Volume of cylinder = Area of base x Height = π.r^2 x h Volume of a rectangular solid = lb x h You may also wish to remember that volume of a pyramid/cone kind of 3D figures is 1/3 x Area of Base x Height e.g. Volume of cone = 1/3 x π.r^2 x h Volume of pyramid with rectangular base of sides l and b = 1/3 x lb x h _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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01 Nov 2010, 14:45
VeritasPrepKarishma wrote:
Fijisurf wrote:
I see that this particular question provides a formula for volume of a sphere.
But generaly speaking, are we suppose to know such fomulas by heart?
Thanks.

No! But, you should know the formula of calculating area of common shapes like square, circle, triangle, rectangle etc.
Using this knowledge, you can calculate the area of 3D figures where one of these shapes is the base and the height is perpendicular to base.
Volume for such figures = Area of Base x Height
e.g. A cylinder - Circular base and height perpendicular to base
Volume of cylinder = Area of base x Height = π.r^2 x h
Volume of a rectangular solid = lb x h

You may also wish to remember that volume of a pyramid/cone kind of 3D figures is 1/3 x Area of Base x Height
e.g. Volume of cone = 1/3 x π.r^2 x h
Volume of pyramid with rectangular base of sides l and b = 1/3 x lb x h

Thanks. One more related question (though, not geometry, but number properties):
I know the formula for sum of consequtive intergers: sum=(average)x(number of terms).
I can remeber it becuase it is easy and makes sense.
Much harder formular is for sum of squares of consequtive integeres: I do not really understand how it works:

(sum of squares)=n(n+1)(2n+1)/6

Have you ever had a need to use this formula on GMAT?

Thanks.
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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01 Nov 2010, 15:18
This formula for the sum of squares can be useful in some cases, but you really don't need to know how it is derived. Just remembering it is sufficient (although it is quite unlikely you'll need it anywhere)
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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01 Nov 2010, 17:19
Fijisurf wrote:
I can remeber it becuase it is easy and makes sense.
Much harder formular is for sum of squares of consequtive integeres: I do not really understand how it works:

(sum of squares)=n(n+1)(2n+1)/6

Have you ever had a need to use this formula on GMAT?

Thanks.

We need to know the sum of consecutive integers, but we do NOT NEED to know the sum of squares of consecutive integers. GMAT does not expect you to know it. Still, it is not a bad idea to remember it since knowing these basic formulas can make a tough question easier sometimes...
Its derivation is beyond our scope and is not a simple straight forward method. Though, it can easily be proved by induction. To see a derivation, check out this link
http://mathforum.org/library/drmath/view/56982.html
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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26 Oct 2016, 09:21
This question requires you to know the surface area of a sphere, which is 4pi(r^2)

So, using volume formula..

4/3(pi)(r^3) = 972(pi) --> you'll end up with r = cube rt(729) = 9

SA --> 4(9^2)(pi) = 324pi ~1000
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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27 Oct 2016, 02:06
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is it surface area 4pi r ^2 or 4pi r ^3.Please correct the question
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface [#permalink]

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27 Oct 2016, 02:59
sanaexam wrote:
is it surface area 4pi r ^2 or 4pi r ^3.Please correct the question

Edited. Thank you.
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Re: The volume of a sphere with radius r is (4/3)*pi*r^3 and the surface   [#permalink] 27 Oct 2016, 02:59
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