Mixture word problem

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