A tank holds 100 gallons of a thoroughly mixed solution which is 50% alcohol. Twenty-five gallons are drained from the tank and replaced with a solution that is 20% alchohol. The solution in the tank is thoroughly mixed. This procedure is repeated 2 more times. What percentage of the final solution is alcohol? Express your answer to the nearest whole number. Please help me and explain <3 ◕‿‿◕
+526 In 100 gallons, alcohol = 50%
25 gallons is replaced with 20% alcohol
So, resulting sol \(=(75×{50\over 100})+(25×{20\over 100})\)
\(=37.5+5\)
\(=42.5\)% alcohol
The process is done again,
Resultant sol \(=(75×{42.5\over 100})+(25×{20\over 100})\)
\(=31.875+5\)
\(=36.875\)% alcohol
After the process is repeated again,
Resultant sol \(=(75×{36.875\over 100})+(25×{20\over 100})\)
\(=27.65625+5\)
\(=32.65625\) \(=33\)% alcohol
∴The final solution has 33% alcohol
+526 In 100 gallons, alcohol = 50%
25 gallons is replaced with 20% alcohol
So, resulting sol \(=(75×{50\over 100})+(25×{20\over 100})\)
\(=37.5+5\)
\(=42.5\)% alcohol
The process is done again,
Resultant sol \(=(75×{42.5\over 100})+(25×{20\over 100})\)
\(=31.875+5\)
\(=36.875\)% alcohol
After the process is repeated again,
Resultant sol \(=(75×{36.875\over 100})+(25×{20\over 100})\)
\(=27.65625+5\)
\(=32.65625\) \(=33\)% alcohol
∴The final solution has 33% alcohol
