balls in a carton

[KK]

I thought of this question while working on a OG problem. However this is not the OG problem itself.

How many balls with diameter of 8 cm can be filled in a rectangular carton with dimentions of length of 48 cm, width of 32 cm and hight of 15 cm.

Looking forward to your help!

thanks

[TakingGMATInSep]

As length is 48, 6 balls can be arranged lengthwise.
As windth is 32, 4 balls can be arranged width wise.
So total balls = 6*4=24
As height is 15, we cant place another layer of balls on the first layer bcoz then total height will be 16.
Hence 24 balls with diameter 8 cm can be filled

[KK]

Thanks for your prompt reply.

However i still have a question:

Is there no duplication of balls between length and width which i assume there to be. In case i am righr we will have to subtract 4 due to dublication of 4 balls on the 4 corners.

[RA]

There should be no duplication. Think of it as 4 rows of 6 balls each

Hope that helps.

[RonPurewal]

As length is 48, 6 balls can be arranged lengthwise.
As windth is 32, 4 balls can be arranged width wise.
So total balls = 6*4=24
As height is 15, we cant place another layer of balls on the first layer bcoz then total height will be 16.
Hence 24 balls with diameter 8 cm can be filled

well, they won't give you a problem like this, because the correct answer is actually more complicated than that.
you can actually get another 5 x 3 = 15 balls on top of the original 24 balls, by putting them into the little depressions formed between the original balls.
if you don't know which depressions i'm talking about, picture the bottom layer of balls - six balls long, four balls wide - and then imagine placing another ball on top of that layer and letting it roll and settle where it "wants" to.
obviously, this ball won't sit right on top of one of the other balls; it will settle into one of the little nooks/crannies between the balls.
since there are 7 extra inches of vertical space, an 8-inch-diameter ball will easily be able to fit into the box once it settles into that space*, so you can fit 15 more balls into the box, for a total of 24 + 15, or 39.

this is actually a very active area of mathematics.

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* calculating the exact additional height required here is VERY difficult, but your intuition should tell you that 7 inches is plenty good.