How many litres of 30percent alcohol solution must be mixed with 30litre of a90percent solution to get a 80percent solution?
.3 x + .9 (30) = .8 (x+30)
.30x + .90*30 = .80(x+30)
Now just find x.
To solve this problem, we need to find out how many liters of a 30% alcohol solution should be mixed with 30 liters of a 90% alcohol solution to obtain a 80% alcohol solution.
Let's use the following variables:
x: the number of liters of the 30% alcohol solution
0.3x: the amount of alcohol in the 30% solution (since it is 30% alcohol)
30: the amount of alcohol in the 90% solution (since it is 90% alcohol)
0.8(30 + x): the amount of alcohol in the 80% solution (since it is 80% alcohol)
Since we want to mix the two solutions, the total amount of alcohol in the final solution should be equal to the sum of the amounts of alcohol in the individual solutions.
0.3x + 30 = 0.8(30 + x)
Now we can solve this equation to find the value of x:
0.3x + 30 = 0.8(30 + x)
0.3x + 30 = 24 + 0.8x
0.3x - 0.8x = 24 - 30
-0.5x = -6
x = -6 / -0.5
x = 12
Therefore, you would need to mix 12 liters of the 30% alcohol solution with the 30 liters of the 90% alcohol solution to obtain a 80% alcohol solution.