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?

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]