Q. 417 - Intermediate Algebra

Question

A pharmacist needs 20 liters

of a 2% saline solution. He has a 1%

and a 5% solution available. How

many liters of the 1% and how

many liters of the 5% solutions

should she mix to make the 2%

solution?

Step-by-Step Solution

Answer

For $1\%$ solution required is $15 litres$ and for $5\%$ solution is $5 litres$.

Step 1. Given information.

Total quantity of solution $=20 litres$.

Available % of solution present $= 1\% and 5\%$

Total % of solution required $=2\%$

Step 2. Write the concept.

Let $x$ litres be from $1\%$ of solution and $y$ litres be from $5\%$ of solution.

such that:

$$\begin{gathered} 1\% of x litres + 5\% of y litres = 2\% of 20 litres \\ \frac{x}{100}+\frac{5y}{100}=\frac{2\times 20}{100} .........\left(1\right) \\ Total solution: \\ x+y=20 ........\left(2\right) \end{gathered}$$

Step 3. Calculation.

Given equations:

$$\begin{gathered} x+5y=40 .......\left(1\right) \\ x+y=20 .........\left(2\right) \end{gathered}$$

subtracting equation (2) from equation (1)

Solve for $y$:

$$\begin{gathered} 4y=20 \\ y=5 \end{gathered}$$

put the values of $y$ in equation (1)

Solve for $x$:

$$\begin{gathered} x+25=40 \\ x=15 \end{gathered}$$

Step 4. Check the values.

Substituting values of $x and y$ in equation (1) we get:

$$\begin{gathered} x+5y=40 \\ 15+5\left(5\right)=40 \\ 40=40 \end{gathered}$$

which holds true.