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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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Comments: ( Back to comment list | You must be logged in to post comments.)
Determing n | Comment 2 of 4 |
There are 10 liters of solution, so they collectively contain 5 liters of acid.

The acid in Jar 3 = 5 - 4*.45 - 5*.48 = 5 - 1.8 - 2.4 = 0.8 liters

  Posted by Steve Herman on 2016-01-25 16:07:43
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