Question 163193
Michael has 1 liter of a mixture containing 69% of boric acid. How much water must be added to make the mixture 50% boric acid?
:
Let x = amt of water required
then
(x+1) = the amt of the resulting mixture
:
Amt of boric acid equation:
;
.69(1) = .50(x+1)
;
.69 = .5x + .50
;
.69-.50 = .5x
:
.5x = .19
x = {{{.19/.5}}}
x = .38 liters of water required
In a fraction: {{{38/100}}} = {{{19/50}}}liter
:
:
Check solution
.69(1) = .50(1.38)
.69 = .69