Answer to Question #279377 in Discrete Mathematics for Jaishree

Question

Suppose that in a bushel of 100 apples there are 20 that have worms in them

and 15 that have bruises. Only those apples with neither worms nor bruises can

be sold. If there are 10 bruised apples that have worms in them, how many of

the 100 apples can be sold?

Answer 1

$$N(W\cup B)=N(W)+N(B)-N(W\cap B)$$

$$=20+15-10=25$$

$$100-25=75$$

75 apples can be sold.