SOLUTION: How much pure acid must be added to 25 gallons of 25% solution to obtain a 50% solution?
What I tried:
25(100x + 25) = 50(100x)
2500x + 625 = 5000x
625 = 2500x
x = 1/4 (.
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What I tried:
25(100x + 25) = 50(100x)
2500x + 625 = 5000x
625 = 2500x
x = 1/4 (.
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Question 689761: How much pure acid must be added to 25 gallons of 25% solution to obtain a 50% solution?
What I tried:
25(100x + 25) = 50(100x)
2500x + 625 = 5000x
625 = 2500x
x = 1/4 (.25)
But something doesn't seem quite right! Thank you in advance! Answer by solver91311(24713) (Show Source):
There are gallons of pure acid in gallons of pure acid. There are gallons of pure acid in 25 gallons of 25% solution. And there are gallons of pure acid in 25 plus gallons of 50% solution. Putting it all together:
Solve for
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
