StringArgs wrote:
Hi
Bunuel,
KarishmaB for concept clarity, say we have to find the maximum number that can fit. Am I wrong to think that that would require all 9 cases to be calculated or is there a smart way to select the l*b pair for area covered and h for layers made to maximize the fit?
3 of the 9 cases: (where we assume the smaller boxes are kept 12*6 with height 4)
Bigger box: 30*60 with height 20
Number of boxes per layer = (30*60)/(12*6)= 25
Number of layers = 20/4=5
Total fit = 25*5 = 125
Bigger box: 20*60 with height 30
Number of boxes per layer = (20*60)/(12*6)= 100/6 (fraction so some space is wasted, nearest lower integer = 16)
Number of layers = 30/4
Total fit = 16*30/4 =120
Bigger box: 20*30 with height 60
Number of boxes per layer = (20*30)/(12*6)= 25/3 (fraction so some space is wasted, nearest lower integer = 8)
Number of layers = 60/4=15
Total fit = 8*15 =120
Now here the maximum fit 125 but can the other orientation of smaller box (say 6*4 with height 12) give us a better fit?. For example in a similar question(
https://gmatclub.com/forum/what-is-the-maximum-number-of-cubic-blocks-of-wood-that-will-fit-into-247028.html#p1907074) i see that all 3 cases gave maximum fit, in that case was there a way to foresee that by looking at the dimensions itself and avoiding the calculation of all 3 but doing just 1 .
Thank you for your time.
Whether we will need to calculate the values in various orientations depends on the sizes given . We need to maximise the volume used of Box X. So I will look for orientations in which the rectangular blocks fit perfectly. Dimensions of the box are 60, 30 and 20 and those of the blocks are 12, 6 and 4. Since 30 is divisible by 6 only (out of 12, 6 and 4), side of 6 should correspond to side of 30 to ensure that no volume is leftover. Also 20 is divisible by 4 so side of 4 should correspond to 20. Since 60 is divisible by all 12, 6 and 4, we have no problem there.
This is how we will place the blocks inside the box such that all volume of the box is utilised.
Consider the length of box X as 60 so the rectangular block can be placed inside such that its length side is 12 which means 5 of these will be placed next to each other. The width of box X is 30 cm so 6 cm should be the width of the rectangular blocks. So 5 of these will be placed width-wise too. Hence, we will have 25 boxes in the bottom layer. Now, height of box X is 20 cm and height of each rectangular block is 4 cm so 5 layers can be stacked one of top of another.
Since all the volume is utilised, there is no way we can place anymore blocks inside the box. Had the dimensions not matched perfectly, we would need to consider a few orientations.
Say if the box dimensions were 60, 30 and 18, how many maximum blocks can be fit in?
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