How many liters each of 30% and 10% acid should be mixed to obtain 20 liters of 11% acid?
Kind of Solution
Liters of Solution
Amount of Pure Acid
0.3
x
0.1
y
0.11
20
Let x be the amount of 30% acid and y be the amount of 10% acid.
The total volume of the solution is x + y = 20 liters.
The amount of pure acid in the 30% solution is 0.3x liters.
The amount of pure acid in the 10% solution is 0.1y liters.
The total amount of pure acid in the solution is 0.3x + 0.1y liters.
Since the total volume of the solution is 20 liters, we can write the equation:
0.3x + 0.1y = 0.11 * 20
0.3x + 0.1y = 2.2
Since x + y = 20, we can write y = 20 - x.
Substituting this into the equation above, we get:
0.3x + 0.1(20 - x) = 2.2
0.3x + 2 - 0.1x = 2.2
0.2x = 0.2
x = 1
Substituting this into the equation y = 20 - x, we get:
y = 20 - 1
y = 19
So, 1 liter of 30% acid and 19 liters of 10% acid should be mixed to obtain 20 liters of 11% acid.
To solve this problem, we can set up a system of equations. Let's call the volume of the 30% acid solution x liters and the volume of the 10% acid solution y liters.
The first equation represents the total volume: x + y = 20.
The second equation represents the amount of pure acid in the final mixture: (0.3x + 0.1y)/(x + y) = 0.11.
We can rewrite the first equation as y = 20 - x.
Substituting this into the second equation, we get (0.3x + 0.1(20 - x))/20 = 0.11.
Now we can solve for x:
(0.3x + 2 - 0.1x)/20 = 0.11.
Combining like terms, we have 0.2x + 2 = 2.2.
Subtracting 2 from both sides, we get 0.2x = 0.2.
Dividing both sides by 0.2, we find x = 1.
Now substitute this value back into y = 20 - x:
y = 20 - 1.
y = 19.
Therefore, you should mix 1 liter of the 30% acid solution and 19 liters of the 10% acid solution to obtain 20 liters of an 11% acid solution.
To find the number of liters of each solution needed to obtain a 20% acid solution, we can use the concept of mixtures.
Let's assign variables to represent the liters of each solution. Let x be the number of liters of the 30% acid solution, and y be the number of liters of the 10% acid solution.
Now, we need to set up an equation based on the amount of pure acid in each solution:
Amount of pure acid from the 30% solution + Amount of pure acid from the 10% solution = Amount of pure acid in the final 11% solution.
0.3x + 0.1y = 0.11 * 20
Simplifying the equation:
0.3x + 0.1y = 2.2
Now, we have a system of equations:
0.3x + 0.1y = 2.2
x + y = 20
We can solve this system of equations using various methods such as substitution or elimination to find the values of x and y.