i need help with this

Question

You need a 40% alcohol solution. On hand, you have a 150 mL of a 20% alcohol mixture. You also have 50% alcohol mixture. How much of the 50% mixture will you need to add to obtain the desired solution?

Answer 1

Let x = amount of 50% solution to add.

Focus on how much alcohol is in each "beaker".

The beaker of 50% solution has 0.5x mL of alcohol.

The beaker of 20% solution has 0.2 (150)mL of alcohol.= 30 mL of alcohol.

The final beaker contains x + 150 mL of which 40% is alcohol, or 0.4(x + 150).

.5x + 30 = .4(x + 150)

Multiply every term by 10 to clear the equation of decimals.

5x + 300 =4(x + 150)

5x + 300 = 4x + 600

x = 300

Check.

300 mL of 50% solution gives 150 mL

20% of the 150mL solution we have already seen is 30 mL.

So when those beakers are combined 450 mL of which we hope 40% is alcohol.(or 180 mL)

40% of 450 = .4(450) = 180 mL (check)