How do you do this problem ?

Question

Solve.
A chemist needs 70 milliliters of a 69% solution but has only 60% and 81% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

40 ml of 60%; 30 ml of 81%
30 ml of 60%; 40 ml of 81%
50 ml of 60%; 20 ml of 81%
20 ml of 60%; 50 ml of 81%

Answer 1

Let x = ml of 60% solution.
Let y = ml of 81% solution.

You've two variables, so you need two equations to solve them. The first is easy... you need a total of 70 ml of the 69% solution:
x + y = 70

The second is trickier and deals with the amount of the chemical in the solution from each bottle. Only 60% of x is really the potent stuff within the solution, and only 81% of y is what you're looking for within that solution. When you combine them, the total amount of liquid (70 ml) is made of of 69% of the chemical. Adding these gives your other equation:
.60x + .81y = .69(70). Multiplying through by 100 gets rid of the decimals. You now have this system of equations:
x + y = 70
60x + 81y = 4830

If you subtract y from both sides on the top equation, you can substitute the expression for x into the second:
x + y = 70. Adding -y to both sides,
x = 70 - y. Substituting into the other equation,
60(70 - y) + 81y = 4830. Distributing,
4200 - 60y + 81y = 4830. Combining like terms and subtracting 4200,
21y = 630. Dividing by 21,
y = 30.

Since x = 70 - y,
x = 70 - (30) = 40.

The answer is (a), 40ml of 60% and 30ml of 81%.

Answer 2

call x(ml 60%) and cally(ml81%)
x+y=70
and
(0.6x+0.81y)/(x+y)=0.69
so
0.6x+0.81y= 48.3
and x+y=70
solving the siystem as y=70-x
you get x=40ml and y =30ml