ANSWER
Find out the how much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .
To proof
let us assume that the fluid should be drained and replaced with pure antifreeze be x .
As given
A radiator contains 10 quarts of fluid, 30% of which is antifreeze.
Now write 30% in the decimal form
[tex]= \frac{30}{100}[/tex]
Now write 60% in the decimal form
[tex]= \frac{60}{100}[/tex]
After drained and replaced the fluid with pure antifreeze than the quantity becomes = ( 10 - x )
The quantity of antifreeze is
[tex]= \frac{30\times (10 - x)}{100}[/tex]
The fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .
than the equation becomes
[tex]= x + \frac{30\times ( 10 - x )}{100} = \frac{60\times10}{100}[/tex]
solyving
100x +300 - 30x = 600
70x = 300
[tex]x = \frac{300}{70}[/tex]
x = 4.28 quarts (approx)
Hence 4.28 quarts (approx) fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .
Therefore the option (c.) is correct .
