The diagram shows a cubic block of marble ($1 \times 1 \times 1\ \mathrm {m^3}$) having a planar fracture. What is the maximum number of slabs sized $20 \times 20 \times 5\ \mathrm {cm^3}$ that can be cut from this block avoiding the fracture?
How to solve this aptitude question? Please help me in this regard.
Thank you very much.
It is rather difficult to demonstrate what the optimal strategy is for packing problems like this and it often results in a more trial and error approach.
Without the fracture we could of course slice the whole block in 25 smaller blocks of dimension $20 \times 20 \times 100$ cm, each containing 20 pieces and since there is no unused part the maximum number of possible plates would be $25 \times 20 = 500$.
If we align the smaller blocks along the fracture (assumed to be perfectly flat and thin), the fracture would intersect only 5 blocks on the diagonal, leaving 20 perfect smaller blocks and hence $20\times20=400$ plates.
So can we do better? This is where trial and error comes in. An "obvious" other approach would be to make big rectangular slices parallel to the fracture both thickness 5 cm. The bottom of the first slice has dimensions $100 \sqrt{2} \times 100 \times 5$ but the top is a bit smaller,i.e. $(100 \sqrt{2}-10) \times 100 \times 5$, which is big enough to fit $6 \times 5 = 30$ plates. The next slice is a bit smaller, but can also fit 30 plates. If you do the math you will find that the successive slices can contain $30,30,25,25,20,20,15,15,10,10,5,5$ plates, which gives you a total of 210 plates above the fracture and likewise 210 plates below. Together you therefore get 420 plates and hence this is better than before. (Better also from a practical point of view, because there is a bit more space available, whereas for the other arrangement cuts would have to be perfect.)
I might be wrong of course, but I would expect that this is the best result possible. Proving this is a whole different matter.