SOLUTION: How many liter of a 90% acid solution much be mixed with a 25% acid solution to get 650 L of 80% acid solution?

Question 872723: How many liter of a 90% acid solution much be mixed with a 25% acid solution to get 650 L of 80% acid solution?
Found 2 solutions by Alan3354, dkppathak:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Do it like this one. Only the numbers are different.
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Amanda wants to make 8 gal. of a 20% saline solution by mixing together a 56% saline solution and a 8% saline solution. How much of each solution must she use?
================
e = amount of 8%
f = amount of 56%
---
e+ f = 8 (total solution)
8e + 56f = 20*8 (total saline)
---
e+ f = 8
e + 7f = 20
------------ Subtract
-6f = -12
f = 2
e = 6
Answer by dkppathak(439) (Show Source): You can put this solution on YOUR website!
How many liter of a 90% acid solution much be mixed with a 25% acid solution to get 650 L of 80% acid solution?
let X liters 90% acid solution added to Y liters of 25% acid solution
as per conditions X+Y =650 (1)
90% x+ 25%Y = 80%(X+Y)
90x+25Y = 80x+80Y
90X+25Y-80X-80Y =)
10X-55Y =0
10X=55Y
X=55/10 y= 5.5 Y by substituting the value of x in equation (1)
X+Y =650
5.5y +Y =650
6.5Y =650
Y=650/6.5 =6500/65 =100 liters
X+Y=650
X+100 =650
X=650-100 =550 liters
Answer
we should add 550 liters 90%acid solution to 100 liters of 25%acid solution

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