A tank contains 500 liters of water in which 500 grams of salt is dissolved. Brine containing 5 grams of salt per liter is pumped into the tank at a rate of 2 liters per minute. The well-stirred solution is then pumped out at a rate of 3 liters per minute. Find the amount of grams of salt in the tank when the tank contains 200 liters of brine.
Answers
Answer:
638 grams
Step-by-step explanation:
Input of fluid per minute = 2 liters
Output of fluid per minute = 3 liters
Net output of fluid per minute = 3 liters - 2 liters = 1 liter
The net amount of fluid pumped out to leave 200 liters in the tank = 500-200 = 300 liters
The time taken for net output of 300 liters = 1×300 = 300 minutes
The amount of salt pumped into the tank by 300 minutes = 10×300 = 3000 grams
The amount of salt which was already there = 500 grams
The total amount of salt collected to the tank by 300 minutes = 3000+500 = 3500 grams
The amount of fluid pumped in by 300 minutes = 2×300 = 600 liters
The amount of fluid which was already there = 500 liters
Total amount of fluid collected to the tank by 300 minutes = 600+500 = 1100 liters
Salt concentration = 3500 grams/1100 liters = 3.18 grams per liter
The amount of fluid pumped out of the tank = 3×300 = 900 liters
The amount of salt pumped out of the tank = 3.18×900 = 2862 grams
The remaining amount of salt in the tank = 3500-2862 = 638 grams