word problems

Question

What quantity of 70 % acid solution must be mixed with a 20 % solution to produce 300 mL of a 50 % solution?

Answer 1

Using allegation:

 

HAVE .......... WANT ........... NEED

70%....................................30...

.......................50%....................

20%....................................20...

 

So, you need 30 parts of the 70% and 20 parts of the 20%.  This adds to 50 parts which is equal to the 300 mls. Thus, each "part" is equal to 300ml/50 parts = 6 mls.  Thus, 6 ml x 30 = 180 mls of 70% and 6 ml x 20 = 120 mls of 20% to give you 300 mls of 50%.

 

Done another way:

(x ml)(70%) + (y mls)(20%) = (300 mls)(50%)

x + y = 300 ml and x = 300 -y (substitute in first equation)

(300-y)(70) + 20Y = 15,000

21,000 - 70y + 20y = 15,000

50y = 6000

y = 120 ml of 20%

x = 300 ml - 120 ml = 180 ml of 70%