Eight machines running 6 hours a day take 15 days to complete an order. How many days will 12 machines running 5 hours a day take to complete an order one and a half times as large, all machines working at the same rate?

The first order costs $8\times 6\times 15=720$ machine-hours, so an order half as large again costs 1,080 machine-hours. The new set-up supplies $12\times 5=60$ machine-hours a day, giving 1,080 ÷ 60 = 18 days. 12 days is 720 ÷ 60 and forgets that the order is larger. 27 days is 1,080 ÷ 40 and keeps the original 8 machines. 15 days assumes the two changes cancel out, but 12 machines at 5 hours is only 60 machine-hours a day against 48 before, a rise of a quarter, while the order rises by a half.