A bartender has a mixture of alcohol in which the ratio of alcohol and water is 7 : 3. He sells 40 liters of the mixture then he adds 12 liters of pure water. Now the ratio of alcohol and water is 9 : 5. What is the new quantity (in liters) of mixture?
Question
A bartender has a mixture of alcohol in which the ratio of alcohol and water is 7 : 3. He sells 40 liters of the mixture then he adds 12 liters of pure water. Now the ratio of alcohol and water is 9 : 5. What is the new quantity (in liters) of mixture?
Solution
Step 1: Let's first find out the quantity of alcohol and water in the initial mixture. We know that the ratio of alcohol to water is 7:3. So, the total quantity of the mixture is 7x + 3x = 10x.
Step 2: The bartender sells 40 liters of the mixture. So, the remaining quantity of the mixture is 10x - 40.
Step 3: In this remaining mixture, the quantity of alcohol remains the same because the bartender only adds water. So, the quantity of alcohol is 7x.
Step 4: The bartender then adds 12 liters of pure water to the mixture. So, the new quantity of water in the mixture is 3x - 40 + 12.
Step 5: Now, the ratio of alcohol to water in the new mixture is 9:5. So, we can set up the equation 7x/(3x - 40 + 12) = 9/5.
Step 6: Solving this equation, we get x = 20.
Step 7: Substituting x = 20 in 10x - 40, we get the new quantity of the mixture as 160 liters.
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