Thread: How many basketballs will fit into a room? (Dimensions and amounts included)

1. How many basketballs will fit into a room? (Dimensions and amounts included)

So there's two questions that I have to solve. I started solving the first one, but I didn't get how to finish it. Here's how it goes:
A basketball has a circumference of 29.5 inches. Because of this shape, we decide to put the basketballs into “cubic” boxes. How many of these boxes can we fit into a room with the dimensions of 27 ft x 27 ft 6 in x 9 ft 2 in ? Show all setups, work and round
to the nearest hundredth. (Hint: Do not take room volume divided by cube volume).

So, basically I found out the radius of the basketball (4.7in) by using the circumference formula, and then found the volume of the basketball (434.89in^3). After this I found the volume of the box (830.58in) and the volume of the room (11,761,200in^3). I am now confused how to find out how many boxes will fit into the room since you cannot take room volume divided by cube volume? Please help me Dx

The second question is almost the same but this time, there are no boxes enclosing the basketballs and it just asks how many basketballs can fit into the same room (with the same dimensions I mentioned above), and answers rounded to the nearest hundredth. This time, it says not to take room volume divided by ball volume. Please help me solve this one too! Thanks!!

2. 1) Cubes is not a good plan. That's definitely a lower bound. Put four balls together, their centers making a lovely tetrahedron, and see how much space is between them. This might be a good start on increasing the number. There are other problems at the walls, however.

2) You'll love this. If you number the balls, you might get a different outcome. I explain:

a1) Throw in 10 balls. b1) Find #1 and remove it.
a2) Throw in 10 balls. b2) Find #2 and remove it.
a3) Throw in 10 balls. b3) Find #3 and remove it.
...
an) Throw in 10 balls. bn) Find #n and remove it.
...

Continue this process. This leads to no balls at all. Every ball eventually will be removed. :-)

3. You seem to be "missing the point" in question#1; WHY are you calculating volumes?
Let's take a simple case: you have a 7by7by7 square box and a 4by4by4 square box:
how many 4by4by4 boxes can you fit in the 7by7by7 box? ONLY 1, right?
But 7by7by7 box volume = 343 and 4by4by4 box volume = 64 ; 64 * 5 = 320:
so that "seems" to mean that 5 boxes can be fit in...but ONLY 1 can...GET IT?

4. Could you elaborate on the throwing the balls please, I'm not grasping the concept.
You put 10 balls into the room, and take out the ball labelled #1?
Why?

5. Who cares why? Follow the logic.