Question

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 108 liters containing 40% acid, using 22 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?

The chemist should use  liters of ___25% solution,  liters of___35% solution, and  liters of ___65% solution.

Answer 1

Let x amount of 25% acid,  y amount of 35% acid and z amount of 65% acid

From question: using 22 times as much of the 65% solution as the 35% solution.
z = 22y

Also, total volume of mixture = 108
x + y + z = 108
x + y + 22y = 108
x + 23y = 108 .....(i)

Also,
25% of x + 35% of y + 65% of z = 40% of 108
25x + 35y + 65z = 40*108
25x + 35y + 65*22y = 4320
25x + 1465y = 4320
From equation(i) x= 108 - 23y
25(108 - 23y) + 1465y = 4320
2700 - 575y + 1465y = 4320
889y = 1620
y = 1.82 letre

z = 22y = 22*1.82 = 40.04 letre
x = 108 - 23y = 108 - 23*1.82= 64.14

The chemist should use  liters of _64.14__25% solution,  liters of__1.82_35% solution, and  liters of __40.04_65% solution.