Question

A pharmacist has a 10 \[\%\] alcohol solution and a 25 \[\%\] alcohol solution. How many milliliters of each solution will she need to mix together in order to have 200 mL of a 20 \[\%\] alcohol solution?

Answer 1

Hint: To solve the given problem, we have to assume the two variables for the amount in ml of solutions. Then, we have two variables to find we have to form two equations in two variables. This system of equations can be found using the relevant information given about the solution. We will use the substitution method to solve the system of equations.

Complete step-by-step solution:
To solve the given problem, we first have to assume the amount of each solution that is required to make the desired solution. Let’s assume that we take x ml of 10 \[\%\] alcohol solution and y ml of 20 \[\%\] alcohol solution. As the final solution is of 200 ml, we can form the first equation using these variables as \[x+y=200\].
The final solution is a 20\[\%\] alcohol solution, this alcohol content must be equal to the alcohol content of the original solution. Using this, we can form a second equation as
10 \[\%\] of x \[+\] 25 \[\%\] of y \[=\] 20 \[\%\] of 200
We can write this equation algebraically as \[0.1x+0.25y=0.2\times 200\]. Simplifying this equation, we get \[0.1x+0.25y=40\].
Now, we have a set of linear equations in two variables \[x+y=200\] and \[0.1x+0.25y=40\]. We can find the values of x and y by solving this system. We will use the substitution method to solve this system. Substituting \[y=200-x\] in the second equation, we get
\[\Rightarrow 0.1x+0.25\left( 200-x \right)=40\]
Solving the above one variable equation, we get \[x=66.67\]. Using this value, we can find the value of y as \[y=200-66.67=133.33\].
Thus, we need 66.67 ml of 10 \[\%\] solution and 133.33 ml of 25 \[\%\] solution.

Note: The important part while solving these types of equations is forming equations. We can form the equation by using the given information. Calculation mistakes while solving the system of equations should be avoided.