SOLUTION: A 6 Liter solution is 60% HCL acid. How much of this must be taken out and be replaced with water to reduce the solution into 45% HCL acid?

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Question 935308: A 6 Liter solution is 60% HCL acid. How much of this must be taken out and be replaced with water to reduce the solution into 45% HCL acid?
Found 3 solutions by mananth, TimothyLamb, lwsshak3:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
You remove x liters from the solution
replace it with water
remaining amount - 6-x of 60%
percent ---------------- quantity
water 0.00% ---------------- x liter
Acid 60.00% ------ 6 - x liter
Mixture 45.00% ---------------- 6
Total 6 liter
0.00% x + 60.00% ( 6 - x ) = 45.00% * 6
0 x + 60 ( 6 - x ) = 270
0 x + 360 - 60 x = 270
0 x - 60 x = 270 -360
-60 x = -90
/ -60
x = 1.5 liter water
4.5 liter 60.00% Acid

m.ananth@hotmail.ca


Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
x = liters of 60% HCL acid/water = 40% HCL water/acid
y = liters of 100% water
z = liters of 45% HCL acid/water = 55% HCL water/acid
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z = 6
0.40x + 1.00y = 0.55z
x + y = z
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put the system of linear equations into standard form
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z = 6
0.40x + 1.00y - 0.55z = 0
x + y - z = 0
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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x = liters of 60% HCL acid/water = 40% HCL water/acid = 4.5
y = liters of 100% water = 1.5
z = liters of 45% HCL acid/water = 55% HCL water/acid = 6
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answer:
replace 1.5 liters of 60% HCL acid with pure water
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Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A 6 Liter solution is 60% HCL acid. How much of this must be taken out and be replaced with water to reduce the solution into 45% HCL acid?
***
let x=amount of 60% acid solution to be removed and replaced with water
6-x=amount of 60% acid solution remaining
..
60%(6-x)=45%*6
3.6-.6x=2.7
.6x=.9
x=1.5 liters