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A right cylinder and a cube have the same surface area. If the height

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A right cylinder and a cube have the same surface area. If the height  [#permalink]

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New post 18 Jun 2019, 00:35
1
1
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

43% (02:42) correct 57% (03:28) wrong based on 37 sessions

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Re: A right cylinder and a cube have the same surface area. If the height  [#permalink]

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New post 18 Jun 2019, 11:34
2
Bunuel wrote:
A right cylinder and a cube have the same surface area. If the height of the cylinder is equal to its diameter, then the volume of the cylinder is approximately what percent greater than the volume of the cube?

A. 6%
B. 13%
C. 26%
D. 32%
E. 35%


Given, h = 2r
Let r = 1, h = 2

Surface area of cylinder = 2πr(h + r)
= 2πr(2r + r) = 6π

Surface area of cube = 6a^2

A right cylinder and a cube have the same surface area
—> 6π = 6a^2
—> a^2 = π
—> a = √π

% greater = (volume of cylinder - volume of cube)/volume of cube*100
= (πr^2h - a^3)/a^3*100
= (2π - π√π)/π√π*100
= (2/√π - 1)*100
= (2/√3.14 - 1)*100
= 13%

IMO Option B

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Re: A right cylinder and a cube have the same surface area. If the height  [#permalink]

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New post 23 Jun 2019, 19:33
1
Bunuel wrote:
A right cylinder and a cube have the same surface area. If the height of the cylinder is equal to its diameter, then the volume of the cylinder is approximately what percent greater than the volume of the cube?

A. 6%
B. 13%
C. 26%
D. 32%
E. 35%


We can let d = the diameter (or height) of the cylinder. Since the surface area of a cylinder is 2πrh + 2πr^2, the surface area of this cylinder is 2π(d/2)(d) + 2π(d/2)^2 = πd^2 + πd^2/2 = 3πd^2/2.

Since the cube has the same surface area and the surface area of a cube is 6s^2, we have:

6s^2 = 3πd^2/2

s^2 = πd^2/4

s = (√π)d/2

Since the volume of a cylinder is πr^2h, the volume of this cylinder is π(d/2)^2*d = πd^3/4. Since the volume of a cube is s^3, the volume of this cube is [(√π)d/2]^3 = (π√π)d^3/8. Therefore, the volume of the cylinder is:

(πd^3/4)/[(π√π)d^3/8] = (1/4)/(√π/8) = 2/√π ≈ 1.13 times the volume of the cube.

In other words, the volume of the cylinder is approximately 13% greater than the volume of the cube.

Answer: B
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A right cylinder and a cube have the same surface area. If the height  [#permalink]

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New post 04 Aug 2019, 13:45
Dillesh4096 wrote:
Bunuel wrote:
A right cylinder and a cube have the same surface area. If the height of the cylinder is equal to its diameter, then the volume of the cylinder is approximately what percent greater than the volume of the cube?

A. 6%
B. 13%
C. 26%
D. 32%
E. 35%


Given, h = 2r
Let r = 1, h = 2

Surface area of cylinder = 2πr(h + r)
= 2πr(2r + r) = 6π

Surface area of cube = 6a^2

A right cylinder and a cube have the same surface area
—> 6π = 6a^2
—> a^2 = π
—> a = √π

% greater = (volume of cylinder - volume of cube)/volume of cube*100
= (πr^2h - a^3)/a^3*100
= (2π - π√π)/π√π*100
= (2/√π - 1)*100
= (2/√3.14 - 1)*100
= 13%


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is there any simple way to calculate the last step here?
i.e 2/√\(\pi\) - 1 = 0.13 ??

This seems too tedious and difficult to solve.

Please help
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Re: A right cylinder and a cube have the same surface area. If the height  [#permalink]

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New post 04 Aug 2019, 23:32
DarkHorse2019 wrote:
is there any simple way to calculate the last step here?
i.e 2/√\(\pi\) - 1 = 0.13 ??

This seems too tedious and difficult to solve.

Please help


Hi DarkHorse2019,

So √3.14 is close to 1.8 [ Since 18^2 = 3.24]

Now, let’s see value of 2/1.8 - 1 = 0.2/1.8 = 2/18 = 1/9 in fraction
In % = 1/9*100 = 100/9 = (11-12)%

So, 13% after assumptions

Also note that other options are quite distant.

Hope it helps!

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Re: A right cylinder and a cube have the same surface area. If the height   [#permalink] 04 Aug 2019, 23:32
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