Question 621350: 1) In a chemistry class, 8 liters of a 4%-saline solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! The key to these mixture problems is to find a way to write an equation of the form:
(amount of one solution)*(its percent) + (amount of another solution)*(its percent) + ... = (total of all amounts)*(desired percent)
(Note: All percents should be written as decimals or fractions. Never do calculations with percents as percents!)
With this mixture problem we have two solutions we want to mix. We know the percents of these two solutions and the percent for the desired mixture. We also know how much of the first solution to use. What we don't know is the amount of the second solution or the amount the the final mixture. If we call the amount of the 10% solution "x", then the total amount would be the sum of the other two amounts: 8+x. We are now ready to write our equation:
8*0.04 + x*0.10 = (8+x)*-0.06
Now that we have an equation, we can use algebra to solve it. As usual, start with simplifying each side:
0.32 + 0.10x = 0.48 + 0.06x
Next we gather the x terms on one side. Subtracting 0.06x from each side:
0.32 + 0.04x = 0.48
Now we solve for x. Subtracting 0.32 from each side:
0.04x = 0.16
Divide both side by 0.04:
x = 4
So we mix 4 liters of the 10% solution with the 8 liters of 4% solution to get 8+4 or 12 liters of 6% solution.
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