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Three Jar Trial (Posted on 2016-01-25) Difficulty: 3 of 5
Consider three jars, labelled Jar 1, Jar 2 and Jar 3.

Jar 1 contains four liters of a solution that is 45% acid.
Jar 2 contains five liters of a solution that is 48% acid.
Jar 3 contains one liter of a solution that is n% acid.

From jar 3, x/y liters of the solution is added to jar 1, and the remainder of the solution in jar 3 is added to jar 2, where x and y are relatively prime positive integers.

At the end, each jar 1 and jar 2 contain solutions that are 50% acid.

Determine n+x+y.

No Solution Yet Submitted by K Sengupta    
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re: Determing n | Comment 3 of 4 |
(In reply to Determing n by Steve Herman)

Very nice!  I missed that in my more formal solution.


With n=0.8, x/y can be found quickly then:
Jar 1 afterwards has 1.8 + (x/y)*0.8 liters of acid in 4+(x/y) liters of solution.  Then 1/2 = (1.8 + (x/y)*0.8)/(4+(x/y)).  Cross multiplying and solving the simple linear equation leads directly to x/y = 2/3.

This makes my solution look needlessly complicated!

  Posted by Brian Smith on 2016-01-25 23:23:51
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