Question 142043
The radiator in Michelle's car contains 16 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 50% antifreeze?
:
Let x = amt drained and the amt added
:
Antifreeze amt added = 1.0x
Antifreeze mixture drained = .30x
:
Write an "amt of antifreeze" equation:

Original amt - drained amt + added amt = resulting amt
:
.30(16) - .30x + 1.0x = .50(16)
;
4.8 + .7x = 8
;
.7x = 8 - 4.8
x = {{{3.2/.7}}}
x = 4.57 L to be drained and replaced with pure antifreeze
:
:
Check solution on a calc:
4.8 - (4.57*.3) + 4.57 = 7.999 ~ 8, which is .50(16)