Ball-shaped oranges of uniform size are packed in a cubical box: in a row one behind another, rows of equal length abreast, similar layers one on the other. The oranges of a layer are directly above the ones in the previous layer, not in the dimples. The size of the box is chosen so that the oranges are not able to move. Ball-shaped tangerines of uniform size are packed in the same way in an identical box. The diameter of a tangerine is a half of the diameter of an orange. Find in percentages how much larger the number of the tangerines in the box is than the number of oranges. Which box has a larger empty space? How much larger in percentages?
+118723 consider a cube that is just big enouugh for one orange.
Let the radius of the orange be 2 units.
The volume of the orange is $$\frac{4}{3}\pi \times 2^3 = \dfrac{8*4\pi}{3}\;units^3$$
Now in that same box you can fit 8 tangarines. the radius of each one is 1unit
The volume of the tangarines is $$8\times \frac{4}{3}\pi \times 1^3 = \dfrac{8*4\pi}{3}\;units^3$$
The volume of the 8 tangarines is exactly the same as the volume of 1 orange so the ratio of the spaces in the boxes must be 1:1
The two boxes have exactly the same amount of empty space. 
+118723 2 times as many across
2 times as many back
2 times as many up
2x2x2=8
8 times more tangerines.
800% times as many tangerines as oranges.
Haven't thought about the rest yet.
It's hard to think on a phone. :)
+118723 consider a cube that is just big enouugh for one orange.
Let the radius of the orange be 2 units.
The volume of the orange is $$\frac{4}{3}\pi \times 2^3 = \dfrac{8*4\pi}{3}\;units^3$$
Now in that same box you can fit 8 tangarines. the radius of each one is 1unit
The volume of the tangarines is $$8\times \frac{4}{3}\pi \times 1^3 = \dfrac{8*4\pi}{3}\;units^3$$
The volume of the 8 tangarines is exactly the same as the volume of 1 orange so the ratio of the spaces in the boxes must be 1:1
The two boxes have exactly the same amount of empty space.
