Question

A chemist wants to mix a 17% acid solution with a 38% acid solution to get 21 L of a 32% acid solution how many liters of the 17% solution and how many liters of the 38% solution should be mixed

Answer 1

6 liters of the 17% solution and 15 liters of the 38% solution.

Let x be the amount of the 17% solution and y be the amount of the 38% solution.

We know that together there are 21 liters of solution, so x+y=21.

0.17x represents the amount of acid from the first solution and 0.38y represents the amount of acid from the second solution.  0.32(21) represents the amount of acid in the new solution.

This gives us the system
x+y=21
0.17x+0.38y=0.32(21)

In the first equation, we will isolate x by subtracting y from both sides:
x+y-y=21-y
x=21-y

Now we will substitute this into the second equation:
0.17(21-y)+0.38y=0.32(21)

Using the distributive property,
3.57-0.17y+0.38y=6.72

Combining like terms,
3.57+0.21y=6.72

Subtract 3.57 from both sides:
3.57+0.21y-3.57=6.72-3.57
0.21y=3.15

Divide both sides by 0.21:
0.21y/0.21 = 3.15/0.21
y=15

Plug this into the first equation:
x+15=21
x+15-15=21-15
x=6

There are 6 liters of the 17% solution and 15 liters of the 38% solution.