THE COSTERMONGER'S PUZZLE.
A costermonger bought 120 oranges at two for a penny, and 120 more at three for a penny, and mixed the oranges all together in a basket. He sold them out, hoping to receive back his money again, at the rate of five for twopence; but on counting his money he found that he had sold the oranges for fourpence less than they had cost him. How this happened will be seen by following the accompanying figures. The first forty purchasers of the oranges would take 200 out of the 240 oranges, and taking it for granted that the fruit was equally mixed, would receive for their money 100 of the oranges originally bought at two a penny and 100 of those at three a penny, and would pay for them the sum of 6s. 8d. The forty remaining oranges would bring in, at the same rate, 1s. 4d. only, making 8s. in all. The cost of the fruit was, for the first 120, 5s., and for the second 120, 3s. 4d., or 8s. 4d. in all, making the loss of 4d. on the lot. To more fully explain the matter, we will suppose the oranges not mixed, but standing in separate baskets, from which, for each purchaser, the costermonger takes two of the two a penny oranges and three of the three a penny oranges, disposing of them in that way for twopence; it will then be clearly seen that the basket containing the three a penny oranges will be first exhausted, for the first forty purchasers, each having three oranges from one basket, will take all the 120 oranges purchased at three a penny, but will require only 80 oranges from the other basket, thus leaving 40 of the two a penny oranges to be sold at five for twopence, or a loss of fourpence on the last 40 sold.