Question 1204463
.
Pure acid is to be added to a 10​% acid solution to obtain  45L of a 40​% acid solution.
What amounts of each should be​ used?
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        There is an elegant method to solve this problem  (and many other similar problems)

        without using equations.  It uses your common sense and your logic,  ONLY.



<pre>
45 liters of the 40% acid contain  0.4*45 = 18 liters of the pure acid and  45 - 12 = 27 liters of water.


When we add pure acid to the 10% acid mixture, we do not add water.  

Hence, 27 liters of water was in 10% acid solution initially.


Again: the initial 10% acid mixture contained 27 liters of water, and water was 9 
of 10 parts of the volume of this 10% mixture.


Hence, the pure acid in this 10% acid mixture was 3 liters , while the water was 27 liters.


To get 18 liters of the pure acid in the final 40% mixture, 18-3 = 15 liters of the pure acid should be added
to the initial 10% mixture.


<U>ANSWER</U>.  15 liters of the pure acid and 3+27 = 30 liters of the 10% mixture shoud be used.
</pre>

Solved.


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