# A radiator contains 10 quarts of 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture 40% antitreeze?

Feb 15, 2017

$1 \frac{3}{7} \text{quarts}$ should be drained off and replaced with pure antifreeze.

$1 \frac{3}{7} \approx 1.4286$ to 4 decimal places

#### Explanation:

Current amount of antifreeze in quarts is $\frac{30}{100} \times 10 = 3$

Target is 40% -> 4 quarts
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Let the amount drained of and replaced with pure antifreeze be $x$

The amount left after draining off is $10 - x$

The amount of antifreeze in this is $\frac{30}{100} \left(10 - x\right)$

The replacement is pure antifreeze which is the amount $x$

so $\frac{30}{100} \left(10 - x\right) + x = 4$ quarts

$3 - \frac{3}{10} x + x = 4$

$3 + x \left(1 - \frac{3}{10}\right) = 4$

$x = \frac{4 - 3}{1 - \frac{3}{10}}$

$x = 1 \times \frac{10}{7} = 1 \frac{3}{7} \text{quarts}$

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$\textcolor{b l u e}{\text{Check}}$

$10 - 1 \frac{3}{7} = 8 \frac{4}{7}$

$= \left(\frac{30}{100} \times 8 \frac{4}{7}\right) + 1 \frac{3}{7}$

$= \left(\frac{3}{10} \times \frac{60}{7}\right) + \frac{10}{7}$

$= \frac{3 \times 6}{7} + \frac{10}{7}$

$= \frac{28}{7} = 4 \leftarrow \text{ 4 litres of pure antifreeze mixed into 10 quarts}$

$\textcolor{red}{\text{Correct}}$

Feb 16, 2017

Different approach

Drain off and replace $1 \frac{3}{7}$ quarts with pure antifreeze

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

This is a very effective method

You are blending two liquid with different concentration of antifreeze. One liquid has 30% concentration and the other 100% concentration.

If you have all of the original the antifreeze is 30%. If you have all the other the antifreeze content is 100%.

Connect the two points on a graph and you have a model for those conditions and all the other blend concentrations in between. From this we can and may determine the blend that gives us 40%. Sometimes it takes a little thought about how so set up the x-axis which represents the quantities of the blend.
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$\textcolor{b l u e}{\text{Answering the question}}$

The slope of part of the graph is the same as the slope of all of it.

$\left(\text{change in y")/("change in x")->("change blend concentration")/("change in added pure antifreeze}\right)$

$\frac{100 - 30}{10} \equiv \frac{40 - 30}{x}$

As we need $x$ turn everything upside down:

$\frac{10}{100 - 30} \equiv \frac{x}{40 - 30}$

$\frac{10}{70} = \frac{x}{10}$

$x = \frac{100}{70} \to \frac{100 \div 10}{70 \div 10} = \frac{10}{7} = 1 \frac{3}{7}$ in quarts