A 9-quart cooling system is checked and found to be filled with a solution that is $10 \%$ antifreeze. The desired strength of the solution is $70 \%$ antifreeze. How many quarts of solution need to be drained and replaced with pure antifreeze to reach the desires strength?
Answer
So, \(\boxed{6}\) quarts of the solution need to be drained and replaced with pure antifreeze to reach the desired strength of 70% antifreeze.
Step 1 ：Let's denote the amount of solution to be drained and replaced with pure antifreeze as \(x\) quarts.
Step 2 ：The amount of antifreeze in the original solution is 10% of 9 quarts, which is \(0.9\) quarts.
Step 3 ：When we remove \(x\) quarts of the 10% solution, we are removing \(0.1x\) quarts of antifreeze.
Step 4 ：So, the amount of antifreeze left in the solution after draining is \(0.9 - 0.1x\) quarts.
Step 5 ：Then we add \(x\) quarts of pure antifreeze to the solution.
Step 6 ：So, the total amount of antifreeze in the solution is now \((0.9 - 0.1x + x)\) quarts.
Step 7 ：We want this to be 70% of the total 9 quarts, so we set up the following equation: \(0.9 - 0.1x + x = 0.7 * 9\)
Step 8 ：Solving this equation for \(x\) gives us the amount of solution that needs to be drained and replaced with pure antifreeze.
Step 9 ：Combine like terms: \(0.9 + 0.9x = 6.3\)
Step 10 ：Subtract 0.9 from both sides: \(0.9x = 5.4\)
Step 11 ：Divide both sides by 0.9: \(x = 6\) quarts
Step 12 ：So, \(\boxed{6}\) quarts of the solution need to be drained and replaced with pure antifreeze to reach the desired strength of 70% antifreeze.
