SOLUTION: How many liters of water should be added to a 25% antifreeze to obtain 30 liters
of a 20% solution
could someone please explain how to work this problem out
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of a 20% solution
could someone please explain how to work this problem out
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Question 1144522: How many liters of water should be added to a 25% antifreeze to obtain 30 liters
of a 20% solution
could someone please explain how to work this problem out Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52800) (Show Source):
Let x be the volume of the original 25% mixture to add water to it.
Then the amount of the pure antifreeze in it is 0.25*x liters.
The amount of the pure antifreeze in the final 30 liters of the 20% solution (after adding water) is 0.2*30 = 6 liters
and it is with no change:
0.25x = 6 liters.
Hence, x = = 24 liters.
Thus 24 liters of the original 25% solution should be use, and 6 = 30-24 liters of the water should be added. ANSWER
Tutor @ikleyn has provided a response with a detailed algebraic solution, along with links to resources where you can find many other examples of this method for solving this kind of problem.
Here is a non-algebraic solution method which, if you understand it, will get you to the answer to problems like this much faster, and with far less effort, than the formal algebraic method.
Here is the solution using this method for your example.
(1) You are starting with 25% antifreeze solution and adding 0% antifreeze solution (water) to get a 20% antifreeze solution.
(2) Look where 20 lies in relation to the 25 and the 0. Looking at the three numbers on a number line can help: 25, 20, and 0. You should see that 20 is 1/5 of the way from 25 to 0.
(3) That means 1/5 of the mixture should be the water you are adding.
Since the total mixture is to be 30 liters, the amount of water to be used is 30*(1/5) = 6 liters.
