A mixture contains 40% alcohol and 60% water. If 5 liters of the mixture is taken out and replaced with 5 liters of pure alcohol, what will be the new percentage of alcohol in the mixture?
Practice Questions
1 question
Q1
A mixture contains 40% alcohol and 60% water. If 5 liters of the mixture is taken out and replaced with 5 liters of pure alcohol, what will be the new percentage of alcohol in the mixture?
50%
55%
60%
65%
After removing 5 liters of the mixture, the remaining alcohol is 0.4 * (total volume - 5) + 5 liters of pure alcohol.
Questions & Step-by-step Solutions
1 item
Q
Q: A mixture contains 40% alcohol and 60% water. If 5 liters of the mixture is taken out and replaced with 5 liters of pure alcohol, what will be the new percentage of alcohol in the mixture?
Solution: After removing 5 liters of the mixture, the remaining alcohol is 0.4 * (total volume - 5) + 5 liters of pure alcohol.
Steps: 10
Step 1: Start with a mixture that is 40% alcohol and 60% water.
Step 2: Determine the total volume of the mixture. Let's assume the total volume is 100 liters for easy calculation.
Step 3: Calculate the amount of alcohol in the original mixture. 40% of 100 liters is 40 liters of alcohol.
Step 4: Calculate the amount of water in the original mixture. 60% of 100 liters is 60 liters of water.
Step 5: Remove 5 liters of the mixture. Since the mixture is 40% alcohol, the amount of alcohol removed is 40% of 5 liters, which is 2 liters.
Step 6: After removing 5 liters, the remaining alcohol is 40 liters - 2 liters = 38 liters.
Step 7: Now, add 5 liters of pure alcohol to the remaining mixture. So, the new amount of alcohol is 38 liters + 5 liters = 43 liters.
Step 8: The total volume of the mixture remains 100 liters (after replacing the 5 liters).
Step 9: Calculate the new percentage of alcohol in the mixture. The new percentage is (43 liters of alcohol / 100 liters of mixture) * 100 = 43%.
Step 10: Therefore, the new percentage of alcohol in the mixture is 43%.
