Chapter 8 linear appliCaTionS 265Let x represent the amount of 10% acid solution, so 100 – x represents the
amount of 30% acid solution.
10% 30% 25%
liters+ 100 – x^
liters=^100
liters0.10x + 0.30(100 – x) = 0.25(100)
0100 30 100 02 5 100
01030030 25
30..().()
..xx
xx+−=
+− =
− 002025
30 30
0205
5
020
25...xx
x
x=
−−
−=−
= −
−
=Add 25 liters of 10% solution to 100 − x = 100 − 25 = 75 liters of 30% solu-
tion to obtain 100 liters of 25% solution.
How much pure alcohol should be added to 6 liters of 30% alcohol solution
to obtain a 40% alcohol solution?
We represent pure alcohol as a 100% mixture. Let x represent the amount
of pure alcohol.
100% 30% 40%
liters+^6
liters= x + 6^
liters1.00x + 0.30(6) = 0.40(x + 6)
1000306040 6
100180 040240
06..() .( )
....
.xx
xx+=+
+= +
00180 240
060060
1x
x
x+=
=
=..
..Add one liter of pure alcohol to 6 liters of 30% alcohol solution to obtain a
40% alcohol solution.