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How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to obtain 3 liters of 40 percent alcohol solution?
Wiki User
∙ 6y ago
Let there be x litres of 50% solution and y% of 20 % solution, then you have two equations by considering the amount of alcohol and the total amount of liquid:
- 50% x + 20% y = 40% of 3 litres
- x + y = 3 litres
These are two simultaneous equations involving 2 unknowns which can be solved:
Double {1} and subtract from {2} to give:
x - x + y - 40%y = 3 - 80% of 3
→ 60% y = 20% of 3
→ y = 20%/60% of 3 = 1/3 of 3 = 1
Substitute for y in {2} to get:
x + 1 = 3
x = 2
Therefore you need 2 litres of 50% solution and 1 litre of 20% solution.
Wiki User
∙ 6y ago
Wiki User
∙ 6y ago2 litres of the 50% solution and 1 L of the 20%.
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How many liters of a 20 percent solution of acid should be added to 10 liters of a 30 percent solution of acid to obtain a 25 percent solution?
10 liters.
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A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
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A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
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How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to get obtain 18 liters of 30 percent alcohol solution?
The total volume of solution equals 18 L. So if we let X be the 50% solution & Y be the 20% solution we can solve a set of two equations with 2 unknowns: X + Y = 18 (total volume of solution is 18 L) 0.5X + 0.2Y = 18*(0.30) (the total amount of alcohol in the new 18L solution) Solving simultaneously, multiply the 1st equation by 0.5 then substract to solve for Y. X + Y = 18 {*0.5} => 0.5X + 0.5Y = 9 0.5X + 0.2Y = 5.4 => 0.5X + 0.2Y = 5.4 ------------------------------------------------------- 0.3Y = 3.6 Y = 12 Then substitute for Y back into the 1st equation & solve to get X=6. So you need 12 L of the 20% solution & 6 L of the 50% solutions to mix together to get 18 L of a 30% alcohol solution.
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