A clerk worked out the mean of 20 readings and got 35. It later came to light that he had typed the reading 46 twice and had left out the reading 28 altogether, so his list still held 20 entries. What is the correct mean?

His total was $20\times 35=700$, and it carries one 46 too many and no 28 at all. The correct total is 700 − 46 + 28 = 682, so the correct mean is 682 ÷ 20 = 34.1. Reversing the two corrections gives $(700+46-28)÷20=35.9$; deleting the spare 46 but forgetting to insert the 28 gives 680 ÷ 20 = 34. The count stays 20 throughout, because one entry simply held the wrong number — no extra reading was ever added.