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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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Concise solution Comment 4 of 4 |
Imagine pouring all the jars together.  You'd have 10 liters of 50% solution which is 5 liters of acid total.

Jar 1 contains 4*.45 = 1.8 liters of acid
Jar 2 contains 5*.48 = 2.4 liters of acid
Jar 3 must contain the remaining .8 liters of acid so must be an n=80% solution.

This then becomes a standard mixture problem (twice actually.) Consider Jar 1
4*.45 + (x/y)*.8 = (4+(x/y))*.5
x/y = 2/3

n+x+y = 80+2+3=85

  Posted by Jer on 2016-01-26 07:31:10
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