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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, 20:53

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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?

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, 01:48

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

please explain. thanks.

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)
_________________

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, 04:32

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.

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"!!!
_________________

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, 07:38

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.

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
_________________

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, 15:45

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?

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, 16:18

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)
_________________

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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