A lab worker needs $5$ litres of $20\%$ alcohol solution.

Question

A lab worker needs $5$ litres of $20\%$ alcohol solution. A $10\%$ solution is mixed with a $60\%$ solution. How many litres of the $60\%$ solution will be required?

Answer 1

Call the amount of the 10% solution x and the amount of 60% solution y. Then you have clearly:

x + y = 5

Also, you want

0.1*x + 0.6*y = 0.2*5

as you want the total amount of alcohol to equal 1 litre.

Now you have a system of equations that you can solve.

Answer 2

Solve the system given by the two equations

$$0.1x+0.6y=0.2*5=1$$

$$x+y=5$$

to get that $x=4$ and $y=1$, so the answer is 1 liter.

Answer 3

Alcohol percentages are usually measured by volume.

Suppose you pour $a$ litres of 10% solution together with $b$ litres of 60% solution. Then the resulting mixture contains $\frac{10}{100}a+\frac{60}{100}b$ litres of alcohol. If you're aiming for 5 litres of 20% solution, that has to contain 1 liter of pure alcohol, so we have the equation $$ \frac{10}{100}a + \frac{60}{100}b = 1 $$

Now you're probably supposed to assume that the volume of the mixture is $a+b$. (Note: This is not true in reality!) So we have the additional equation $$ a+b=5$$

Now you have two equations with two unknowns. Solve them.