Question:
A chemist has three different acid solutions. The first acid solution contains
25% acid, the second contains 40%, and the third contains 60%. He wants to use all three solutions to obtain a mixture of 60 liters containing 45% acid, using 3 times as much of the 60% solution as the 40%solution. How many liters of each solution should be used?
Solution:
Let the volume (in litres) of acids used be:
x1=amount of 25% acid
x2=amount of 40% acid
x3=3*x2=amount of 60% acid (as per instructions)
We also know that x1+x2+x3=60 L, or substituting x3=3*x2, then
x1+x2+3*x2=60L
=> x1=60-4*x2 L
To make a 45% acid, we have
0.25x1+0.40x2+0.60x3=0.45*60=27 L (of pure acid)
Substituting x1 and x3
0.25(60-4*x2)+0.40x2+0.60(3*x2) = 27
Distribute and collect terms
15-x2 + 0.40x2 + 1.80x2 = 27
(-1+0.40+1.80)x2=27-15
1.1x2=12
x2=10 L
x1=60-4x2=20L
x3=3*x2=30L
Check:
total volume=10+20+30 = 60L [checks]
Concentration
(0.25*x1+0.40*x2+0.60*x3)L/60L
=(0.25*20+0.40*10+0.60*30)L/60L
=(5+4+18)/60
=27/60
=40% [ checks]
