# Thread: Math Word Problem : Mixture. I don't get it. T-T

1. ## Math Word Problem : Mixture. I don't get it. T-T

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?

2. Originally Posted by strigy
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

$25\ quarts \cdot 0.6 = 15\ quarts$ of pure anti-freeze.

2. The desired mixture should contain

$25\ quarts \cdot 0.4 = 10\ quarts$ 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:

$x \cdot 0.6 = 5~\implies~x = \dfrac{25}3\ quarts$

3. I still don't get it. . Sorry. Thanks for the answer anyway.

4. 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!

5. Hello, strigy!

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?
Let $x$ = percent of weed killer (WK) in original mixture.

He starts with 26 lbs of mix which is $x\%$ weed killer.
. . It contains: . ${\color{blue}26x}$ lbs of WK.

One pound of mix is removed.
. . It contains: . ${\color{blue}x}$ lbs of WK.

One pound of pure (100%) WK is added.
. . It contains: . ${\color{blue}1}$ lb of WK.

Hence, the final mix contains: . $26x - x + 1 \:=\:{\color{red}25x + 1}$ lbs of WK.

But we know that the final mix will be 26 lbs which is 5% WK.
. . So it contains: . $(26)(0,05) \;=\;{\color{red}1.3}$ lbs of WK.

We just described the amount of WK in the final mix in two ways.

There is our equation! . . . $\boxed{25x + 1 \;=\;1.3}$

Solve for $x\!:\quad 25x \:=\:0.3 \quad\Rightarrow\quad x \:=\:\frac{0.3}{25} \:=\:0.012 \;=\;1.2\%$

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### a gardener has 26 kg of mixture

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