Question

A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze?

Answer 1

v30% mixture + v100% mixture = v60% mixture

0.3(10-x) + x = 0.6(10)
3 - 0.3x + x = 6
0.7x = 6 - 3
0.7x = 3
x = 3/0.7
x = 4.286 quarts 

Answer 2

The fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze is 1.43 quarts

What is Percentage?

A percentage is a number or ratio expressed as a fraction of 100.

What is system of equation?

In mathematics, a set of simultaneous equations, also known as a system of equations or an equation system, is a finite set of equations for which common solutions are sought.

Here,

Consider the amount of pure antifreeze that drained and replaced = x

Given that Radiator contains 10 quarts of  fluid

Then the amount of quarts after draining = 10 - x

Amount of antifreeze in that = $(10-x)\frac{30}{100}$

Then replaced with pure antifreeze = $(10-x)\frac{30}{100}+x$

Therefore,

$(10-x)\frac{30}{100}+x=4\\\frac{300}{100}-\frac{30}{100}x+x=4\\ 3- \frac{3}{10}x+x=4\\ 3-0.3x+x=4\\0.7x=1$

x=1.43 quarts

Hence, The fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze is 1.43 quarts

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