You can put this solution on YOUR website! How much antifreeze must be added to 20 L of a 60% solution to
make a 75% solution?
:
Let a = amt of pure antifreeze required to accomplish this
:
A typical mixture equation
.60(20) + a = .75(a+20)
12 + a = .75a + 15
a - .75a = 15 - 12
.25a = 3
a = 3/.25
a = 12 L of antifreeze required
:
:
See if that checks out
.6(20) + 12 = .75(12+20)
12 + 12 = .75(32)
24 = 24, confirms our solution of a = 12
