How much of 20% alcohol solution and 50% alcohol solution must be mixted to get 9 liters of 30% alcohol solution?
Jake, don't just state an answer, which is wrong anyway
amount of the 20% stuff --- x
amount of the 50% stuff --- 9-x
.2x + .5(9-x) = .3(9)
times 10
2x + 5(9-x) = 27
2x + 45 - 5x = 27
-3x = -18
x = 6
So he needs 6 L of the 20% solution, and
3 L of the 50% solution
I think it is 3.86 liters you should double check.
To determine the amount of each solution needed to create a 30% alcohol solution, we can use a mathematical approach known as the mixture problem.
Let's represent the amount of the 20% alcohol solution as x liters and the amount of the 50% alcohol solution as y liters.
To solve the problem, we need to consider two aspects: the total amount of solution and the amount of alcohol in the mixture.
First, let's focus on the total amount of solution:
x + y = 9 --- (Equation 1)
This equation states that the sum of the amounts of the 20% and 50% alcohol solutions should equal 9 liters, as we want a total mixture of 9 liters.
Now, let's consider the amount of alcohol in the mixture:
0.20x + 0.50y = 0.30(9) --- (Equation 2)
In this equation, 0.20x represents the amount of alcohol from the 20% solution, and 0.50y represents the amount of alcohol from the 50% solution. The right side of the equation represents the amount of alcohol in the 30% solution obtained from mixing the two solutions. Multiplying the total volume (9 liters) by the desired alcohol concentration (30%) gives us 0.30(9) or 2.7 liters of alcohol.
Now, let's solve the system of equations (Equations 1 and 2) to find the values of x and y:
From Equation 1, we can rearrange it to obtain:
y = 9 - x --- (Equation 3)
Now, substitute Equation 3 into Equation 2:
0.20x + 0.50(9 - x) = 2.7
Simplifying further, we have:
0.20x + 4.5 - 0.50x = 2.7
Combining like terms, we get:
-0.30x + 4.5 = 2.7
Next, subtract 4.5 from both sides of the equation:
-0.30x = -1.8
Finally, divide both sides by -0.30 to solve for x:
x = -1.8 / -0.30
x = 6
Now that we have obtained the value of x (which represents the amount of 20% alcohol solution, in this case), we can substitute it back into Equation 3 to find y:
y = 9 - x
y = 9 - 6
y = 3
Therefore, to get 9 liters of a 30% alcohol solution, you need to mix 6 liters of a 20% alcohol solution with 3 liters of a 50% alcohol solution.