A chemist needs 10L of 40% solution of alcohol. He plans to
mix a 60% solution with a 20% solution. How many liters of
each solution must he mix to produce the required solution?
Actually in this problem ONLY, since 40% just happens to be
exactly half-way between 60% and 20%, we know immediately that
it would take exactly the same amount of each. So we know that
to get 10 liters, we must use 5 liters of each. However, below
is the way we would solve such a problem in ALL cases, whether
the desired percentage is half-way between the given percentages
or not.
Let x = the number of liters of the 40% solution, and
Let y = the number of liters of the 60% solution.
Then we make this chart:
number percent number
of of pure of liters
liters alcohol of pure
of in each alcohol
each solution in each
solution as a decimal solution
--------------------------------------------------------
First solution x 0.60 0.60x
Second solution y 0.20 0.20y
-------------------------------------------------------
Final solution 10 0.40 0.40(10)
The first equation comes from the number of liters of
solution in each solution:
x + y = 10
The second equation comes from the the number of liters
of pure alcohol in each solution:
0.60x + 0.20y = .40(10)
We remove the decimals by multiplying each term by 100,
which involves moving the decimals two place right.
60x + 20y = 40(10)
60x + 20y = 400
We can simplify that further by dividing through by 20
3x + y = 20
Now we have the system of equations
We solve that either by substitution or elimination. I'll
choose substitution. Solve the first equation for y:
y = 10-x
Substitute in the second equation:
3x+(10-x) = 20
3x+10-x = 20
2x+10 = 20
2x = 10
x = 5 liters of 60%
Substitute x=5 in y = 10-x
y = 10-5
y = 5 liters of 20%
Edwin
