I've been working on this problem for 2 hours now and can't figure it out. How much must be removed from a 5qt container containing 12% antifreeze and replaced by pure antifreeze to increase it to a 20% antifreeze solution? Can anyone help me please?
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Algebra mixture problem. I can't figure it out.
- Thread starter festey1
- Start date
I've been working on this problem for 2 hours
now and can't figure it out. How much must be removed from a
5qt container containing 12% antifreeze and replaced by pure
antifreeze to increase it to a 20% antifreeze solution?
Can anyone help me please?
Let x = the number of quarts of a 12% antifreeze solution removed. And
it is also the number of quarts of pure antifreeze replaced.
(12%)(5) - (12%)(x) + (100%)(x) = (20%)(5)
If you multiply each side by 100, you get:
(12)(5) - (12)(x) + (100)(x) = 20(5)
Can you finish it from there?
Hello, festey1!
If you're careful, you can "talk" your way to the equation . . . well, almost.
We will remove \(\displaystyle x\) quarts of the solution, then add \(\displaystyle x\) quarters of pure A/F.
We start with 5 quarts of solution which contains 12% A/F.
It contains: .\(\displaystyle 0.12 \times 5 \,=\,0.6\) quarts of A/F.
We remove \(\displaystyle x\) quarts of the solution which is 12% A/F.
. . So we remove: .\(\displaystyle 0.12x\) quarts of A/F.
We replace it with \(\displaystyle x\) quarts of pure A/F.
. . This contains: .\(\displaystyle x\) quarts of A/F.
Hence, the final mixture contains: .\(\displaystyle 0.6 - 0.12x + x\) quarts of A/F. .[1]
But we know that the final mixture will be 5 quarts which is 20% A/F.
. . It will contain: .\(\displaystyle 5 \times 0.20 \:=\:1\) quart of A/F. .[2]
We just described the amount of A/F in the final mixture in two ways.
There is our equation . . . . \(\displaystyle 0.6 - 0.12x + x \:=\:1\)
. . Now solve for \(\displaystyle x.\)
If you're careful, you can "talk" your way to the equation . . . well, almost.
How much must be removed from a 5-qt container containing 12% antifreeze
and replaced by pure antifreeze to increase it to a 20% antifreeze solution?
We will remove \(\displaystyle x\) quarts of the solution, then add \(\displaystyle x\) quarters of pure A/F.
We start with 5 quarts of solution which contains 12% A/F.
It contains: .\(\displaystyle 0.12 \times 5 \,=\,0.6\) quarts of A/F.
We remove \(\displaystyle x\) quarts of the solution which is 12% A/F.
. . So we remove: .\(\displaystyle 0.12x\) quarts of A/F.
We replace it with \(\displaystyle x\) quarts of pure A/F.
. . This contains: .\(\displaystyle x\) quarts of A/F.
Hence, the final mixture contains: .\(\displaystyle 0.6 - 0.12x + x\) quarts of A/F. .[1]
But we know that the final mixture will be 5 quarts which is 20% A/F.
. . It will contain: .\(\displaystyle 5 \times 0.20 \:=\:1\) quart of A/F. .[2]
We just described the amount of A/F in the final mixture in two ways.
There is our equation . . . . \(\displaystyle 0.6 - 0.12x + x \:=\:1\)
. . Now solve for \(\displaystyle x.\)