A scientist wants to dilute a 60% acid solution by adding some 20% solution. If she starts with 80 ml of the 60% solution, how many milliliters of the 20% solution does she need to add to get a resulting 45% acid solution?
Solution
Let x be the amount of the 60% solution
Let y be the amount of the 20% solution
Let x + y be the amount of the mixture
60% of x solution + 20% of y solution = 45% of amount of mixture
We use 60% of the x solution or 60% of 80 ml.
60% of 80 = 0.60 times 80 = 48
We use 20% of the y solution or 20% of y
20% of y = 0.20 times y = 0.20y
The mixture is x + y and we need 45% of the mixture or
45% of (x + y)
45% of (x + y) = 0.45(x + y)
Putting it all together we get:
48 + 0.20y = 0.45(48 + y)
48 + 0.20y = 21.6 + 0.45y
48 - 21.6 = 0.45y - 0.20y
26.4 = 0.25y
y = 26.4 / 0.25 = 105.6
Indeed,
0.60 times 80 + 0.20 times 105.6 = 48 + 21.12 = 69.12
0.45 times (48 + 105.6) = 0.45 times 153.6 = 69.12
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Sep 18, 19 01:16 PM
Factoring using the box method. Common pitfalls to avoid when using this method.