An automobile radiator contains a 10-qt mixture of water and antifreeze that is 40% antifreeze. How much should you drain from the radiator and replace with pure antifreeze so the liquid in the radiator will be 80% antifreeze?

after defining variables equation, with x = amount drained from radiator and added as pure antifreeze:

0.40(10-x) + x = 10(0.8)
4 - 0.40x + x = 10(0.8)
4 - 10(0.8) = -0.6x
x = (4-8)/(-0.6) = 6.667qts

Check:
0.40(10 - 6.667) + 6.667 = 10(0.80) = 8
%antifreeze = 100%(8/10) = 80%

lets go stoners math quiz.