Atank originally contains 100 gal of fresh water. then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. after 10 min the process is stopped, and fresh water is poured into the tank at a rate of 8 gal/min, with the mixture again leaving at the same rate. find the amount of salt in the tank at the end of an additional 10 min.

1,98 lb of salt

Explanation:

Gal at the beginning= 100 gal fresh water.

Inside rate1 =2 gal/min x 10min x 1/2lb/gal =  10 lb at the end of ten minutes.

outside rate1 = 1 lb / 100 gal = 0,001 lb /gal x 10 min =0,01 lb  at the end of ten minutes.

Final quantity of salt at ten minutes=  10 - 0,1 = 9,9 lb.

Inside rate 2= 8 gal/min x 0 lb = 0lbatthe end of next 10 minutes, + 9,9 lb from the initial state = 9,9 lb.

Outside rate 2= 9,9 lb /100 gal = 0,099 lb/gal x8 gal/min x 10min = 7,92 lb.

Final quantity = 9,9 lb - 7,92 lb = 1,98 lb of salt.

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