SOLUTION: A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?
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Question 129672This question is from textbook Algebra structure and meathod
: A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?
This question is from textbook Algebra structure and meathod
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?
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20% solution DATA:
Amount = x liters; active ingredient = 0.20x L
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60% solution DATA:
Amount = "10-x" liters ; active = 0.60(10-x)= 6-0.6x liters
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Mixture DATA;
Amt. = 10 L : active = 0.30*10 = 3 liters
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EQUATION:
active + active = active
0.2x + 6 - 0.6x = 3
-0.4x = -3
x = 7.5 liters (amount of 20% solution needed for the mix)
10-x = 2.5 liters (amount of 60% solution needed for the mix)
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Cheers,
Stan H.
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