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A man adds 1.6 liters of a 60% antifreeze solution to the water already in his car radiator.
When the radiator is filled the solution is 8% antifreeze.
How many liters does the radiator hold? (Hint: The water is a 0% antifreeze solution)
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1.6 liters of the 60% antifreeses contains 0.6*1.6 = 0.96 liters of the pure antigreese.
Let x be the added volume of the water.
The added water does not contain antifreese, at all.
So we write an equation, saying that the pur antifreese amount is THE SAME before and after mixing
0.96 = 0.08*(1.6+x)
It is easy to solve
= 1.6 + x
12 = 1.6 + x
x = 12 - 1.6 = 10.4.
So, 10.4 liters is the added volume of water; the total volume of the liquid in the tank after mixing is 1.6 + 10.4 = 12 liters.
Solved.
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It can be solved by the other way (MENTAL reasoning without using equations).
Originally, there were 0.6*1.6 = 0.96 liters of the pure antifreese in the tank.
After adding water, the amount of the pure antifreese did not change, but the concentration became 0.08.
Hence, the total volume after mixing is = 12 liters.
It is the same solution as the Algebra solution above, but presented in wording form.
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To see many other similar (and different) problems of the same type, look into the lesson
- Special type mixture problems on DILUTION adding water
in this site.
