How many boxes?

Is it D - 300?

For maximum # of boxes, you should align them within the carton so that the dimension of the carton is divisible by the dimension of the box (ie, no extra room is left). Therefore, you would align the side of the box with length 5 with the side of the carton of length 25, the side of the box with length 6 with the side of the carton of length 60, and the side of the box with length 7 with the side of the carton of length 42. This would allow for 25/5 = 5 boxes along one dimension, 60/6 = 10 along another, and 42/7 = 6 along the other.

Therefore we can fit 5*10*6 = 300 boxes in the carton.

(Another way to look at is would be to find the volume of the carton (25*42*60 = 63000) and divide by the volume of a box (5*6*7 = 210). 63000/210 = 300. Then since we know the the boxes can be packed tightly without leaving extra room, we know 300 is the answer.

That's the explanation verbatim from princeton review!

I am having a hard time understanding how these boxes can be lined up. Usually, to find max in a situation like this, you just can't divide the volume of carton by volume of boxes that this explanation seems to be doing.

Just to add to above -

Total volume of a carton = total volume of soap boxes.
= no. of soap boxes x volume of each soap box.
( assumption here is that they are of the same size)

no of soap boxes = total volume of carton / volume of each soap box

= 25 * 42 *60 / 7 * 6 * 5
Instead of multiplying the numerator and denominator just cancel out the factors. (42/7 = 6 ; 6 gets cancelled out with 6 and 25/5 =5, leaving us with 5*60)

= 300 boxes

Do not consider how they can be arranged inside the cartoon. The question is "what is the MAX..". Mathematically, 300 is MAX.

Now picture this, you somehow arrange 300 boxes in cartoon. I take a fist of sand and put into the cartoon. The boxes will not pop out of cartoon. Implies, practical volume is more than mathematical volume. Hence, forget arranging the boxes unless the question hints at you.

Lalit