Golden Spheres

Geometry Level 2

A solid golden cube with side 10 cm is melted into solid golden spheres of radius 2 cm. What is the maximum number of golden spheres that can be made?


The answer is 29.

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4 solutions

let n \large n be the maximum number of golden spheres that can be made

then, we have

n = v o l u m e o f c u b e v o l u m e o f s p h e r e = x 3 4 3 π ( r 3 ) = 1 0 3 4 3 π ( 2 3 ) = 1000 33.51032 \large n=\dfrac{volume~of~cube}{volume~of~sphere}=\dfrac{x^3}{\frac{4}{3}\pi (r^3)}=\dfrac{10^3}{\frac{4}{3}\pi (2^3)}=\dfrac{1000}{33.51032} \approx 29 \color{#D61F06}\boxed{\large 29}

Pranshu Gaba
Feb 12, 2016

The volume of the cube is C = 1 0 3 = 1000 cm 3 C = 10^{3} = 1000 \text{ cm}^{3} .
The volume of one sphere is S = S = 4 3 π r 3 = \dfrac{ 4 } { 3 } \pi r^{3} = 4 3 π × ( 2 3 ) = \dfrac{ 4 } {3} \pi \times (2^{3} ) = 32 π 3 cm 3 \dfrac{ 32 \pi}{3 } \text{ cm}^{3} .

The maximum number of spheres that can be made from the cube are C S = \left \lfloor \dfrac{ C} {S} \right \rfloor = 1000 32 π 3 = \left \lfloor \dfrac{ 1000 } { \frac{32 \pi}{3} } \right \rfloor = 3000 32 π = \left \lfloor \dfrac{ 3000 } { 32 \pi } \right \rfloor = 29.84 = 29 \lfloor 29.84 \ldots \rfloor = \boxed{ 29 } _\square

29.8 ~30 not 29 how come the answer is 29 u asked maximum

Debasis Rath - 5 years, 4 months ago

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In these kind of problems, we cannot round off our answer to the nearest integer. Take a look at this example.

Suppose one pencil costs $1.00 and you have $29.84 . How many pencils will you able to buy with this money?

We cannot buy 30 pencils because we fall short of $0.16 . So even though the nearest integer to 29.84 is 30, we cannot buy 30 pencils. We can only buy 29 pencils. We must consider the greatest integer less than equal to 29.84 instead, which is 29.

Pranshu Gaba - 5 years, 4 months ago

Number of complete spheres = floor(29.84...)=29.

Prasit Sarapee - 5 years, 3 months ago

Why is it that you multiplied the volume of the cube by 3 in your final step? I may be missing something obvious.

Alexander Caines - 5 years, 4 months ago

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.

1000 32 π 3 = 1000 × 3 32 π = 3000 32 π \frac {1000}{\frac{32π}{3}} = \frac {1000\times 3}{32π} = \frac {3000}{32π}

Gabriel Gomes - 5 years, 4 months ago

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Oh my... Thank you.

Alexander Caines - 5 years, 4 months ago

Looking for more easier for no use of cal.

One could be 4/3 = 1.33 pi = 3.14 so 4/3 x pi = 4.2 roughly multiply by 2 cube (8) 33.6 vplume of the sphere. 1000/33 = 30.30303030..... (those who know ryles for 1/3, 1/33, 1/33 shd know how) as well 1000/33.3 will be 30.03003003.... so 1000/33.4 will be close to 30 or 29.xxxxx

Rakibul Raihan - 5 years, 4 months ago

The volume of a cube is given by v = a 3 v=a^3 where a a is the edge length of the cube. The volume of a sphere is given by v = 4 3 π r 3 v=\dfrac{4}{3}\pi r^3 where r r is the radius of the sphere.

number of spheres = volume of the cube volume of one sphere = 1 0 3 4 3 π ( 2 3 ) = 29 \color{#3D99F6}\text{number of spheres}=\dfrac{\text{volume of the cube}}{\text{volume of one sphere}}=\dfrac{10^3}{\frac{4}{3}\pi (2^3)} = \color{#D61F06}\boxed{29}

Thank you! This helped me a lot.

Viyah Zarisse - 9 months, 1 week ago
Prasit Sarapee
Feb 19, 2016

The maximum number of spheres = f l o o r ( 1000 / ( 4 3 π × 2 3 ) ) = f l o o r ( 29.84.... ) = 29. =floor(1000/( \dfrac {4} {3} \pi\times 2^{3}))=floor(29.84....)=29.

If the side of the cube was 12, we could insert only three roughs of spheres, each formed by 3 spheres. So you have a layer oft spheres, and could insert three layers, 27 spheres at all. If th side is shorter, you can't introduce 29 spheres, where is the mistake?

STELVIO ANDREATTA - 4 years, 12 months ago

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