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.

Each of the three jars has a Volume, Acid Content, and Concentration to keep track of.

Before distributing Jar 3:

| Volume | Acid Content | Concentration

-------+--------+--------------+---------------

Jar 1 | 4 | 1.8 | 0.45

Jar 2 | 5 | 2.4 | 0.48

Jar 3 | 1 | N | N

After distributing Jar 3: (Letting Z=x/y)

| Volume | Acid Content | Concentration

-------+--------+--------------+---------------

Jar 1 | 4+Z | 1.8+Z*N | 0.5

Jar 2 | 6-Z | 2.4+(1-Z)*N | 0.5

Then two equations can be formed:

(1.8+Z*N)/(4+Z) = 1/2

(2.4+(1-Z)*N)/(6-Z) = 1/2

Each equation can be solved for N:

N = (Z+0.4)/(2Z)

N = (1.2-Z)/(2-2Z)

Equating these two expressions yields Z=2/3, and therefore N=80%. The value of N+X+Y is 85.