SOLUTION: How many liters (L) of a 10% silver iodide solution must be mixed with 7 L of a 4% silver iodide solution to get a 6% solution?

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Question 44886: How many liters (L) of a 10% silver iodide solution must be mixed with 7 L of a 4% silver iodide solution to get a 6% solution?
Found 2 solutions by stanbon, AnlytcPhil:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
How many liters (L) of a 10% silver iodide solution must be mixed with 7 L of a 4% silver iodide solution to get a 6% solution?
Start: 7L of 4% solution has 0.04(7)=0.28 liters of active ingredient
Add: x liters of 10% solution which has 0.10x liters of active ingredient
End up with: x+7 liters of 6% solution which has 0.06(x+7) liters of active
ingredient.
EQUATION:
active + active = active
0.28 + 0.10x=0.06x+0.42
Multiply thru by 100 to get:
28 +10x = 6x + 42
4x = 14
x=7/2= 3.5 liters (amount of 10% solution you need to add)
Cheers,
Stan H.
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
How many liters (L) of a 10% AgI solution 
must be mixed with 7 L of a 4% AgI solution 
to get a 6% AgI solution? 

Let X be the answer.  Then you can read your problem 
this way:

When X liters (L) of a 10% AgI solution are mixed 
together with 7 L of a 4% AgI solution, we get X+7 
liters of a 6% AgI solution?  What is X?

The equation comes from this reasoning:

When the 10% of X L that is pure AgI in the first 
stuff is added to the 4% of 7 L that is pure AgI 
in the second, we get the 6% of the resulting X+7 L 
which is pure AgI.    

So 

10% of X plus 4% of 7 equals 6% of X+7

or

0.10X + 0.04(7) = 0.06(X+7)

Solve that and you'll get X = 3.5 L 

Edwin
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