Arctic antifreeze is 18% alcohol and frost no more is 10% alcohol. How many liters of each should be mixed to get 20 liters of a mixture that is 15% alcohol?

Question

Arctic antifreeze is 18% alcohol and frost no more is 10% alcohol. How many liters of each should be mixed to get 20 liters of a mixture that is 15% alcohol?

Answer 1

This problem deserves an organized table.

 

                               Liters solution       % Alcohol         Total Liters Alcohol

Arctic Antifreeze    x                           0.18                0.18x

Frost No More       y                           0.10                0.10y

Mixture                x+y=20                 0.15                3

 

The rows are multiplied across. The last row is the mixture we're after, a total of 20 L with 15% alcohol. Multiplied it equals 3 L. Since x+y=20 then x=20-y (substitution method to eliminate a variable). Let's rewrite the table.

 

                               Liters solution        % Alcohol        Total Liters Alcohol
Arctic Antifreeze    20-y                     0.18                0.18(20-y)
Frost No More        y                         0.10                0.10y
Mixture                 x+y=20               0.15                3

 

Now we use the last column to find y.

0.18(20-y) + 0.10y = 3

3.6 - 0.18y + 0.10y = 3

-0.08y = -0.6

y = 7.5 L

Remember, this is only y. Plug back in for x.

We need 7.5 L of Frost No More and (x=20-7.5) 12.5 L of Arctic Antifreeze.