SOLUTION: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?

Click here to see ALL problems on Mixture Word Problems

Question 347694: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
-----------------------
Equation:
acid + acid = acid
x + 0.10(54-x) = 0.20*54
---
Multiply thru by 100 and solve for "x":
100x + 10*54 - 10x = 20*54
90x = 10*54
x = 6 liters (amt. of pure acid needed in the mixture)
=============================
Cheers,
Stan H.
===========

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Let represent the amount of 10% solution that you have. Then represents the amount of pure acid needed.

Solve for then subtract the value of from 54.

John

My calculator said it, I believe it, that settles it