A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Can somebody please explain how it's done?
A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Can somebody please explain how it's done?
1. The actual mixture contains
$\displaystyle 25\ quarts \cdot 0.6 = 15\ quarts$ of pure anti-freeze.
2. The desired mixture should contain
$\displaystyle 25\ quarts \cdot 0.4 = 10\ quarts$ of pure anti-freeze.
3. That means you have to remove that amount of the actual mixture (call it x) which contains exactly 5 quarts of pure anti-freeze:
$\displaystyle x \cdot 0.6 = 5~\implies~x = \dfrac{25}3\ quarts$
Oh, wait scrap what I said. I think I get it now!!!!!!!! omg THANK YOU.
Err..okay I have another one:
A gardener has 26lb of a mixture of fertilizer and weed killer. If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5 % weed killer. What percent of the original mixture was weed killer?
There doesn't need to be an answer, just steps on how to get there. Thanks!
Hello, strigy!
Let $\displaystyle x$ = percent of weed killer (WK) in original mixture.A gardener has 26lb of a mixture of fertilizer and weed killer.
If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5% weed killer.
What percent of the original mixture was weed killer?
He starts with 26 lbs of mix which is $\displaystyle x\%$ weed killer.
. . It contains: .$\displaystyle {\color{blue}26x}$ lbs of WK.
One pound of mix is removed.
. . It contains: .$\displaystyle {\color{blue}x}$ lbs of WK.
One pound of pure (100%) WK is added.
. . It contains: .$\displaystyle {\color{blue}1}$ lb of WK.
Hence, the final mix contains: .$\displaystyle 26x - x + 1 \:=\:{\color{red}25x + 1}$ lbs of WK.
But we know that the final mix will be 26 lbs which is 5% WK.
. . So it contains: .$\displaystyle (26)(0,05) \;=\;{\color{red}1.3}$ lbs of WK.
We just described the amount of WK in the final mix in two ways.
There is our equation! . . . $\displaystyle \boxed{25x + 1 \;=\;1.3}$
Solve for $\displaystyle x\!:\quad 25x \:=\:0.3 \quad\Rightarrow\quad x \:=\:\frac{0.3}{25} \:=\:0.012 \;=\;1.2\%$