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Percentage Problem
- Thread starter derek70
- Start date
Hello, derek70!
Actually, this is not a "very simple" problem.
It's a dreaded Mixture Problem . . . brrr!
Let \(\displaystyle A\) = original amount of solution (in, say, quarts).
Since 100% of it is the Ingredient,
. . the original solution contains \(\displaystyle A\) quarts of Ingredient.
We add \(\displaystyle x\) quarts of a 10% solution.
. . It contains \(\displaystyle 0.10x\) quarts of Ingredient.
Hence, the final mixture contains: .\(\displaystyle A + 0.10x\) quarts of Ingredient. .[1]
But we know that the final mixture will be \(\displaystyle A + x\) quarts which is 85% Ingredient.
Hence, the final mixture contains: .\(\displaystyle 0.85(A+x)\) quarts of the Ingredient. .[2]
We just described the amount of Ingredient in the final mixture in two ways.
There is our equation! . \(\displaystyle \hdots\quad A + 0.1x \:=\:0.85(A + x)\)
Solve for \(\displaystyle x:\;\;A + 0.1x \:=\:0.85A + 0.85x \quad\Rightarrow\quad 0.75x \:=\:0.15A\)
. . . . . . . . . . \(\displaystyle x \:=\:\dfrac{0.15A}{0.75} \quad\Rightarrow\quad x \:=\:0.2A \)
We should add \(\displaystyle 0.2A\) quarts of the 10% solution.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
We have \(\displaystyle A\) quarts of solution which contains \(\displaystyle A\) quarts of Ingredient.
We add \(\displaystyle 0.2A\) quarts which is 10% Ingredient.
. . It contains: .\(\displaystyle (0.1)(0.2A) \,=\,0.02A\) quarts of Ingredient.
The final mixture contains \(\displaystyle A + 0.02A \:=\:1.02A\) quarts of Ingredient.
The final mixture consists of \(\displaystyle A + 0.2A \:=\:1.2A\) quarts.
The final percentage is: .\(\displaystyle \dfrac{1.02A}{1.2A} \:=\:0.85 \:=\:85\%\;\;\checkmark\)
Actually, this is not a "very simple" problem.
It's a dreaded Mixture Problem . . . brrr!
If we start with a 100% solution of a liquid
and combine a second solution that is a 10% solution,
how much of the 10% solution is added to result in an 85% solution?
Let \(\displaystyle A\) = original amount of solution (in, say, quarts).
Since 100% of it is the Ingredient,
. . the original solution contains \(\displaystyle A\) quarts of Ingredient.
We add \(\displaystyle x\) quarts of a 10% solution.
. . It contains \(\displaystyle 0.10x\) quarts of Ingredient.
Hence, the final mixture contains: .\(\displaystyle A + 0.10x\) quarts of Ingredient. .[1]
But we know that the final mixture will be \(\displaystyle A + x\) quarts which is 85% Ingredient.
Hence, the final mixture contains: .\(\displaystyle 0.85(A+x)\) quarts of the Ingredient. .[2]
We just described the amount of Ingredient in the final mixture in two ways.
There is our equation! . \(\displaystyle \hdots\quad A + 0.1x \:=\:0.85(A + x)\)
Solve for \(\displaystyle x:\;\;A + 0.1x \:=\:0.85A + 0.85x \quad\Rightarrow\quad 0.75x \:=\:0.15A\)
. . . . . . . . . . \(\displaystyle x \:=\:\dfrac{0.15A}{0.75} \quad\Rightarrow\quad x \:=\:0.2A \)
We should add \(\displaystyle 0.2A\) quarts of the 10% solution.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
We have \(\displaystyle A\) quarts of solution which contains \(\displaystyle A\) quarts of Ingredient.
We add \(\displaystyle 0.2A\) quarts which is 10% Ingredient.
. . It contains: .\(\displaystyle (0.1)(0.2A) \,=\,0.02A\) quarts of Ingredient.
The final mixture contains \(\displaystyle A + 0.02A \:=\:1.02A\) quarts of Ingredient.
The final mixture consists of \(\displaystyle A + 0.2A \:=\:1.2A\) quarts.
The final percentage is: .\(\displaystyle \dfrac{1.02A}{1.2A} \:=\:0.85 \:=\:85\%\;\;\checkmark\)