Question 1161169: How many quarts of pure antifreeze must be added to 9 quarts of a 50% antifreeze solution to obtain a 60% antifreeze solution? Found 2 solutions by ikleyn, math_helper:Answer by ikleyn(52788) (Show Source):
Let Q be that unknown amount of the pure antifreeze to add, under the problem's question.
Then the total antifreeze in the mixture will be Q + 0.5*9 quarts,
while the total volume of the mixture will be Q+9 quarts.
Therefore, the concentration equation is
= 0.6 (which is 60% concentration).
From the equation,
Q + 0.5*9 = 0.6*(Q+9),
Q + 4.5 = 0.6Q + 5.4
Q - 0.6Q = 5.4 - 4.5
0.4Q = 0.9
Q = = 2.25.
ANSWER. 2.25 quarts of the pure antifreeze should be added.
CHECK. = 0.6. ! Precisely correct !
Solved.
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It is a standard and typical mixture word problem.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
Read them and become an expert in solution mixture word problems.
At the start you have:
9 qts of solution, of which 9/2 = 4.5 qts are antifreeze
You want to add x qts of antifreeze (9/2 + x is the final amount of
antifreeze) and have the ratio of that to the total solution (9 + x)
equal to 60% (0.60):
Solve for x = 9/4 = 2.25 qts
