SOLUTION: Pure acid is to be added to a 5% acid solution to obtain 95L if 80% solution. What amounts of each should be used?

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Question 349169: Pure acid is to be added to a 5% acid solution to obtain 95L if 80% solution. What amounts of each should be used?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let aliters = amount of pure acid to be added
Let bliters = amount of 5% solution needed
In words:
(total liters of acid in final solution)/(total liters of final solution)
= 80%
----------------------
given:
(1) a+%2B+b+=+95
%28a+%2B+.05b%29%2F95+=+.8
a+%2B+.05b+=+76
(2) 100a+%2B+5b+=+7600
Multiply both sides of (1) by 5 and
subtract from (2)
(2) 100a+%2B+5b+=+7600
(1) -5a+-+5b+=+-475
95a+=+7125
a+=+75
b+=+95+-+75
b+=+20
75 liters of pure acid and 20 liters of 5% acid solution are needed
check answer:
%28a+%2B+.05b%29%2F95+=+.8
%2875+%2B+.05%2A20%29%2F95+=+.8
%2875+%2B+1%29%2F95+=+.8
76%2F95+=+.8
76+=+76
OK