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31 December, 09:49

A tank 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.

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  1. 31 December, 13:40
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    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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