One solution is 10% acid and another is 22% acid. How many cubic centimeters of each should be mixed to obtain 100 cubic centimeters of a solution that is 12% acid?
One solution is 10% acid and another is 22% acid. How many cubic centimeters of each should be mixed to obtain 100 cubic centimeters of a solution that is 12% acid?
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To learn, in general, how to set up and solve this sort of exercise, try here. Then:
. . .cc's solution:
. . . . .10%: x
. . . . .22%: 100 - x
. . . . .mix: 100
. . .percent acid:
. . . . .10%: 0.1
. . . . .22%: 0.22
. . . . .mix: 0.12
. . .cc's acid:
. . . . .10%: 0.1(x)
. . . . .22%: 0.22(100 - x)
. . . . .mix: 0.12(100)
Add the inputs, set equal to the output, and solve for the variable.
The answer is two numbers the quantities of each of the two solutions that go into the mixture, if you gave some idea of the work or thinking that went into your suggested answer it would be easier for us to spot what difficulties you are having with this problem.
Let the volume in cc of solution 1 be $\displaystyle x$, and of solution 2 be $\displaystyle y$. Then as we have 100cc of mixture:
$\displaystyle x+y=100$.
$\displaystyle x$ cc of solution 1 contains $\displaystyle 0.1x$ cc of acid, $\displaystyle y$ cc of solution 2 contains $\displaystyle 0.22y$ cc of acid, so the $\displaystyle 100$ cc of mixture contains $\displaystyle 0.11x+0.22y$, and this is 12% of $\displaystyle 100$ cc so:
$\displaystyle 0.11x+0.22y=12$
So you need to solve these equations for $\displaystyle x$ and $\displaystyle y$ to get the required answer.
CB
Lets say we mixed "a" cc of 10% soln and "b" cc of 22%soln
Amount of acid in First = 10a/100 = a/10
Amount of acid in second = 22b/100
Now its given that
Total amount of mixture
a+b = 100
Amount of acid in mixture
a/10 +22b/100 = 12
Solve the two equations.
solve the two equations!
Hello, needhelp1!
Another approach . . .
One solution is 10% acid and another is 22% acid.
How many cc's of each should be mixed to obtain 100 cc's of a solution that is 12% acid?
Let $\displaystyle x$ = amount of solution $\displaystyle A$ (10% acid).
Let $\displaystyle 100-x$ = amount of solution $\displaystyle B$ (22% acid).
$\displaystyle x$ cc's of $\displaystyle A$ is 10% acid.
. . It contains: $\displaystyle 0.10x$ cc's of acid.
$\displaystyle 100-x$ cc's of $\displaystyle B$ is 22% acid.
. . It contains: $\displaystyle 0.22(100-x)$ cc's of acid.
The mixture contains: $\displaystyle {\color{blue}0.10x + 0.22(100-x)}$ cc's of acid.
The mixture is 100 cc's which is 12% acid.
. . It contains: $\displaystyle 0.12(100) = {\color{blue}12}$ cc's of acid.
We just described the final amount of acid in two ways.
There is our equation . . . $\displaystyle \boxed{\;0.10x + 0.22(100-x) \:=\:12\;}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The information can be organized in a chart . . .
.$\displaystyle \begin{array}{|c|| c|c|c|}\hline
& \text{Sol'n} & \text{Percent} & \text{Acid} \\ \hline \hline
A & x & 10\% & 0.10x \\ \hline
B & 100-x & 22\% & 0.22(100-x) \\ \hline \hline
\text{Mix} & 100 & 12\% & 0.12(100) \\ \hline \end{array}$
And the equation is waiting for us in the last column . . .