Question

Plz help with this word problem

Answer 1

He will mix some 20% solution and some 70% solution to make 60% solution.

Let x = number of liters of 20% solution.
Let y = number of liters of 70% solution.

He wants to make 50 liters of 60% solution, so

x + y = 50 First Equation

The amount of acid in x amount of 20% solution is 20% of x, or 0.2x
The amount of acid in y amount of 70% solution is 70% of y, or 0.7y
The amount of acid in 50 liters of 60% solution is 60% of 50 liters, or 0.6 * 50 = 30
Now we add the amounts of acid.
0.2x + 0.7y = 30 Second Equation

x + y = 50
0.2x + 0.7y = 30

-0.2x - 0.2y = -10
0.2x + 0.7y = 30
------------------------
0.5y = 20

y = 40

x + y = 50

x + 40 = 50

x = 10

Answer: He needs 10 liters of 20% solution and 40 liters of 70% solution.

Answer 2

Using my favorite X-diagram method of solving mixture problems, we find Gabe needs 10 liters of 20% solution and 40 liters of 60% solution to make 50 liters of 60% solution.

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The numbers on the right end of the diagaonals are the difference between the center number and the number on the left end of the diagonal. The numbers on the right indicate the proportion of the associated (top or bottom) contributor. That is, the proportion of 70% : 20% solution is 40 : 10. As it happens, 40 + 10 = 50, so these numbers are the numbers of liters of corresponding constituent needed for the final mix.