As a cleaning agent, a solution that is 24% vinegar is often used. How much pure (100%) vinegar and 5% vinegar must be mixed to obtain 50 oz of a 24% solution?

X: amt of pure vinegar, measured in oz

y: amt of 5% sol'n, measured in oz

x+y = 50 oz = total amt of 24% sol'n obtained

Left side Right side

Amount of pure vinegar = Amt of pure vinegar

1.00x + 0.05y = 0.24 (x+y) = 0.24 (50) = 12 (oz)

Eliminate either x or y using the statement x+y=50 oz: x = 50-y

Then

1.00 (50-y) + 0.05y = 12 (oz)

50-y + 0.05y = 12

50 - 0.95y = 12 (oz) (amount of pure vinegar in the 24% solution)

28 = 0.95y

y = 29.47 oz (of the 5% solution)

Then x = 50-y = 50-29.47 = 20.53 oz (of the undiluted vinegar)