How many liters of a 20% alcohol solution and a 50% alcohol solution must be mixed to make 20 liters of a 35% alcohol solution?

Step-by-step solution

1. 1

   Name the two unknowns. Let $x$ be the number of liters taken from the $20\%$ solution and $y$ the number of liters taken from the $50\%$ solution. Two unknowns means two independent equations are needed — one for total volume, one for the amount of pure alcohol.

2. 2

   Equation 1: total volume. The two amounts must add up to the required $20$ liters:

   $x+y=20.$

3. 3

   Equation 2: pure alcohol is conserved. This is the key idea in every mixture problem: the alcohol in the final mix equals the alcohol carried in by each ingredient. The target holds $0.35\times 20=7$ liters of pure alcohol, while the ingredients contribute $0.20x$ and $0.50y$:

   $0.20x+0.50y=7.$

4. 4

   Solve the system by substitution. From the first equation $x=20-y$. Substituting into the second:

   $0.20(20-y)+0.50y=7\ \Longrightarrow\ 4+0.30y=7\ \Longrightarrow\ 0.30y=3\ \Longrightarrow\ y=10.$

   Then $x=20-10=10$.

5. 5

   Check the result against the original wording. The $20\%$ solution supplies $0.20\times 10=2$ L of alcohol and the $50\%$ solution supplies $0.50\times 10=5$ L, for $7$ L in $20$ L of mixture, i.e. $\frac{7}{20}=35\%$. Correct.

6. 6

   Sanity check with the midpoint shortcut. $35\%$ sits exactly halfway between $20\%$ and $50\%$, so equal parts of the two solutions had to be the answer. When the target concentration is not the midpoint, the mix is weighted toward whichever stock solution is closer to it — a useful way to catch a sign or arithmetic slip before trusting the algebra.