3
algebra

A chemist needs to mix a 20% acid solution with a 50% acid solution to create 30 liters of a 35% acid solution. How many liters of the 20% solution are needed?

A

10 liters

B

15 liters

C

18 liters

D

12 liters

Correct Answer: B

Choice B is the correct answer. Let xx = liters of 20% solution and yy = liters of 50% solution.

Step 1: Set up the system: {x+y=300.20x+0.50y=0.35(30)\begin{cases} x + y = 30 \\ 0.20x + 0.50y = 0.35(30) \end{cases}

Step 2: Simplify second equation: 0.20x+0.50y=10.50.20x + 0.50y = 10.5

Multiply by 10: 2x+5y=1052x + 5y = 105

Step 3: From first equation: y=30xy = 30 - x

Substitute: 2x+5(30x)=1052x + 5(30 - x) = 1052x+1505x=1052x + 150 - 5x = 1053x=45-3x = -45x=15x = 15

Solution: 15 liters of 20% solution (and 15 liters of 50% solution)

Verification: 15+15=3015 + 15 = 30 ✓ and 0.20(15)+0.50(15)=3+7.5=10.5=0.35(30)0.20(15) + 0.50(15) = 3 + 7.5 = 10.5 = 0.35(30)

💡 Strategic Tip: In mixture problems, (concentration × volume) gives the amount of pure substance.

Choice A is incorrect because 0.20(10)+0.50(20)=2+10=12 eq10.50.20(10) + 0.50(20) = 2 + 10 = 12 \ eq 10.5.

Choice C is incorrect because 0.20(18)+0.50(12)=3.6+6=9.6 eq10.50.20(18) + 0.50(12) = 3.6 + 6 = 9.6 \ eq 10.5.

Choice D is incorrect because 0.20(12)+0.50(18)=2.4+9=11.4 eq10.50.20(12) + 0.50(18) = 2.4 + 9 = 11.4 \ eq 10.5.