<PRE><B>Question 465528</B>
How much 20% anti-freeze should be mixed with 60% anti-freeze to get 10 liters of 28% anti-freeze?
:
Let x = amt of 60% stuff
then
(10-x) = amt of 20%
:
Write a mixture equation using decimals
.60x + .20(10-x) = .28(10)
.60x + 2 - .20x = 2.8
.6x - .2x = 2.8 - 2
.4x = .8
x = {{{.8/.4}}}
x = 2 liters of 60% stuff
then
10 - 2 = 8 liters of 20%
:
:
Check this: .6(2) + .2(8) = .28(10)
1.2 + 1.6 = 2.8</PRE>