( 1+20%)a *( 1- 1/6)b = ab
So there is no change.
Question 2: Because there are always an integer number of good peaches and bad peaches.
So the number of peaches assigned to Class A is a multiple of 9.
The number of peaches assigned to Class B is a multiple of 16.
(1) If Class B is given to peach 16* 1= 16, then there is 95- 16=79, but 79 is not a multiple of 9.
Therefore, the peaches assigned to Class B are not 16.
(2) If the peach allocated by Class B is 16*2=32, then the remaining 63 peaches are multiples of 9.
So Class A got 63 and Class B got 32.
Therefore, Class A is given to the bad peach 63*(2/9)= 14, and Class B is given to the bad peach 32*(3/ 16)=6.
So Class A and Class B allocate 95- 14-6=75 good peaches.