Who packed all the pies? kane’s bakery makes four flavour pies. the weight of each flavour pie is follows: apple pie-700g banoffee pie-725g custard pie-600g rhubarb pie-675g 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 can each hold 4kg. what is the fewest number of boxes he can pack the order into? show your solution

This is an identical problem answered by myself a few days ago.  The answer is reproduced below for your convenience.

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.
 
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The answer would be 4x+3