perplexus.info :: Weights and Scales : Three Jar Trial

Three Jar Trial (Posted on 2016-01-25)

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.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Solution

| Comment 1 of 4

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.

Posted by Brian Smith on 2016-01-25 12:00:55

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