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Need to solve this problem: "How many liters of 12% acid solution must be mixed with 5 liters of a 20% acid solution to give a 14% solution?"

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1 Answer

For this question it's useful to define the unknown. Let's call x the number of liters of 12% acid solution to be used.

Since we're going to mix the solutions, our new solution will have (x+5) liters. 

How much acid is present in the new solution? Well, the original 5 liters contribute 20%, that is 

20*5/100 = 1 liter of acid

The added solution contributes 12%, that is

12*x/100 = 0.12*x liters

So we have a total of 1 + 0.12*x liters of acid in the (x+5) liters of solution (see above) and we want the acid to be 14% so:

1 + 0.12*x = 14*(x+5)/100

and now we just need to solve for x:

1 + 0.12*x = 0.14*(x+5)

1 + 0.12*x = 0.14*x + 0.7

1 - 0.7 = 0.14*x - 0.12*x

0.3 = 0.02*x

30 = 2*x

15 = x

And that's the correct answer!

 

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