How much water must be added to 60 liters of 60 percent alcohol mixture to reduce it to a mixture of 50 percent alcohol concentration? What if 30 percent alcohol mixture is added instead of water?
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To solve this problem, we need to calculate the amount of water or a 30% alcohol mixture that needs to be added to the 60 liters of the 60% alcohol mixture.
Let's start by finding out how much alcohol is present in the 60 liters of the 60% alcohol mixture.
Amount of alcohol in the mixture = 60 liters * 0.60 = 36 liters
Now we need to calculate the final amount of alcohol we want in the mixture. Since we want a 50% alcohol mixture, we can calculate that as follows:
Final amount of alcohol in the mixture = (60 liters + x liters) * 0.50
where 'x' represents the amount of water or a 30% alcohol mixture to be added.
Now we can set up the equation:
36 liters = (60 liters + x liters) * 0.50
Simplifying the equation:
36 liters = 30 liters + 0.50x liters
Subtracting 30 liters from both sides:
6 liters = 0.50x liters
Dividing both sides by 0.50:
x = 6 liters / 0.50
x = 12 liters
Therefore, if water is added, 12 liters of water should be added to the 60 liters of the 60% alcohol mixture to obtain a mixture with a 50% alcohol concentration.
If you want to add a 30% alcohol mixture instead of water, we need to consider that a 30% alcohol mixture means that for every 100 liters of the mixture, there is 30 liters of alcohol and 70 liters of other liquid (which is not alcohol).
Therefore, in order to calculate the amount of the 30% alcohol mixture needed, we can set up the equation similarly:
36 liters = (60 liters + x liters) * 0.30
Simplifying the equation:
36 liters = 0.30x + 18 liters
Subtracting 18 liters from both sides:
18 liters = 0.30x
Dividing both sides by 0.30:
x = 18 liters / 0.30
x = 60 liters
Therefore, if a 30% alcohol mixture is added instead of water, 60 liters of the 30% alcohol mixture should be added to the 60 liters of the 60% alcohol mixture to obtain a mixture with a 50% alcohol concentration.