A tank contains a solution of salt in water. initially a tank contains 1000L of water in 10 Kg of salt disolved in it. the mixture is poured off at a rate of 20L/min, and simultaneously pure water is added at a rate of 20L/min. All the time the tank is stirred to keep the mixture uniform. Find the mass of salt in the tank after 5min. The tank must be topped up by adding more salt when the mass of salt in the tank falls to 5kg; after how many minutes will it need topping up?

Ans: 9.05kg, 37.4 minutes

in Calculus Answers by Level 3 User (4.0k points)

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1 Answer

Initially, there is 10kg of salt in 1000L of water. 20L of solution will contain 1000/20*10=10/50=0.2kg of salt. That leaves 9.8kg in the original solution. When the lost water is topped up, there will be 9.8kg in solution after a minute. The next minute will see 9.8/50=0.196kg of salt lost, leaving 9.8-0.196=9.604kg; the third minute, 9.604(1-0.02) and so on. The amount of t minutes is given by 10(1-0.02)^t, so when t=5, the amount of salt is 9.0392kg approx.

We need to find t when 10*0.98^t=5, 0.98^t=0.5, tlog0.98=log0.5, t=34.31mins.

After 34.31 minutes the amount of salt will have dropped to 5kg, and the tank will need to be topped up with 5kg to restore to the initial level. Merry Christmas and a Happy New Year!

 

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