Kane’s bakery makes four flavours of pie. the weight of each flavour pie is follows: apple pie- 700 grams, banoffee pie- 725 grams, custard pie- 600 grams , rhubarb pie - 675 grams. robert makes an order for 4 apple pies, 8 banoffee pies, 3 custard pies and 5 rhubarb pies to sell at a fundraiser. kane packs the order into boxes that each hold 4 kilograms . what is the fewest number of boxes he can pack the order into?

It is a very difficult problem if the box sizes are smaller or if there are more pies to pack.

For this particular problem, we first calculate the total weight of pies:
4 apple pies at 700 g = 2800 g
8 banoffee pies at 725 g =5800 g
3 custaard pies at 600 g = 1800 g
5 rhubarb pies at 675 g = 3375 g
for a total of 13775 g
The minimum number of boxes needed will be 13775/4000=3.44, or
at least 4 boxes will be required.

We also note that 4 boxes will hold 16000 g, and we only need to pack 13775g, so there is a lot of margin.

We can proceed to do the mechanical way, fill in order of apple, banoffee, custard, and finally rhubarb pies.
Following table shows how we fill, naively:
      box content               Remaining to pack
Box A B C R     Weight    A B C R   
0                                      4  8  3 5     (original order)
1     4  1  0 0     3525       0  7  3 5
2     0  5  0 0     3625       0  2  3 5
3     0  2  3 0     3250       0  0  0 5
4     0  0  0 5     3375       0  0  0 0
Total weight     13775 g

So that does the job, 4 boxes (minimum) with none exceeding 4000 g to pack a total of 13775 g of pies.