You need a 50% alcohol solution. On hand, you have a 200 mL of a 20% alcohol mixture.

Question

How much pure alcohol will you need to add to obtain the desired solution?

Answer 1

20(200) + 100x = (200+x)50

100x -50x = 200(50) -20(200)

50x = 30(200)

x = 30(200)/50 = 120 mL of pure alcohol

mixed with 200 mL of 20%

yield a mixture of 320 mL of 50%

100-50=50

50-20=30

50/30 = 5/3

200/120= 20/12=5/3

5 times as much 20% as pure alcohol