A merchant can place 8 large boxes or 10 small boxes into a carton for shipping. In one shipment, he sent a total of 96 boxes. If there are more large boxes than small boxes, how many cartons did he ship?

Why this works

This riddle involves a bit of logical reasoning and arithmetic. The merchant has two types of boxes that he can ship: large boxes (8 per carton) and small boxes (10 per carton). Given the total of 96 boxes and the condition that there are more large boxes than small boxes, we can set up equations to find the right combination. First, let’s denote the number of cartons for large boxes as \(L\) and for small boxes as \(S\). The total number of boxes can be represented as \(8L + 10S = 96\). The riddle also states that \(L > S\). By testing different combinations, we find that if \(L = 7\) (which gives 56 boxes) and \(S = 4\) (which gives 40 boxes), we satisfy both the total box requirement and the condition of having more large boxes. Adding these cartons together gives us \(L + S = 7 + 4 = 11\), which is the final answer!