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?
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The volume of the cube is
C
=
1
0
3
=
1
0
0
0
cm
3
.
The volume of one sphere is
S
=
3
4
π
r
3
=
3
4
π
×
(
2
3
)
=
3
3
2
π
cm
3
.
The maximum number of spheres that can be made from the cube are ⌊ S C ⌋ = ⌊ 3 3 2 π 1 0 0 0 ⌋ = ⌊ 3 2 π 3 0 0 0 ⌋ = ⌊ 2 9 . 8 4 … ⌋ = 2 9 □
29.8 ~30 not 29 how come the answer is 29 u asked maximum
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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.
Why is it that you multiplied the volume of the cube by 3 in your final step? I may be missing something obvious.
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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
The volume of a cube is given by v = a 3 where a is the edge length of the cube. The volume of a sphere is given by v = 3 4 π r 3 where r is the radius of the sphere.
number of spheres = volume of one sphere volume of the cube = 3 4 π ( 2 3 ) 1 0 3 = 2 9
Thank you! This helped me a lot.
The maximum number of spheres = f l o o r ( 1 0 0 0 / ( 3 4 π × 2 3 ) ) = f l o o r ( 2 9 . 8 4 . . . . ) = 2 9 .
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?
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let n be the maximum number of golden spheres that can be made
then, we have
n = v o l u m e o f s p h e r e v o l u m e o f c u b e = 3 4 π ( r 3 ) x 3 = 3 4 π ( 2 3 ) 1 0 3 = 3 3 . 5 1 0 3 2 1 0 0 0 ≈ 2 9