Question 349169
Let {{{a}}}liters = amount of pure acid to be added
Let {{{b}}}liters = amount of 5% solution needed
In words:
(total liters of acid in final solution)/(total liters of final solution)
= 80%
----------------------
given:
(1) {{{a + b = 95}}}
{{{(a + .05b)/95 = .8}}}
{{{a + .05b = 76}}}
(2) {{{100a + 5b = 7600}}}
Multiply both sides of (1) by {{{5}}} and
subtract from (2)
(2) {{{100a + 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:
{{{(a + .05b)/95 = .8}}}
{{{(75 + .05*20)/95 = .8}}}
{{{(75 + 1)/95 = .8}}}
{{{76/95 = .8}}}
{{{76 = 76}}}
OK