Two tins of equal dimensions have \(\frac{1}{5}\) and \(\frac{1}{6}\) portions filled with acid. If the remaining portions of the tins are filled with water and the resultant contents are mixed in a tumbler, then how many units of acid should be added to the tumbler so that the ratio of acid and water becomes 1 : 1 in the resulting solution?

Formula Used:

Ratio of acid to water = Amount of acid / Amount of water

If the ratio is 1 : 1, then Amount of acid = Amount of water.

Calculation:

Let the total capacity of each tin be the LCM of the denominators 5 and 6.

LCM(5, 6) = 30 units.

Amount of acid in Tin 1 = \(\frac{1}{5} \times 30\) = 6 units.

Amount of water in Tin 1 = 30 - 6 = 24 units.

Amount of acid in Tin 2 = \(\frac{1}{6} \times 30\) = 5 units.

Amount of water in Tin 2 = 30 - 5 = 25 units.

When the contents are mixed in a tumbler:

Total amount of acid = Amount of acid in Tin 1 + Amount of acid in Tin 2 = 6 + 5 = 11 units.

Total amount of water = Amount of water in Tin 1 + Amount of water in Tin 2 = 24 + 25 = 49 units.

Let 'x' units of acid be added to the tumbler so that the ratio of acid and water becomes 1 : 1.

New amount of acid = 11 + x units.

The amount of water remains the same = 49 units.

For the ratio of acid and water to be 1 : 1:

New amount of acid = Amount of water

11 + x = 49

x = 49 - 11

x = 38

Therefore, 38 units of acid should be added to the tumbler.

38 units of acid should be added to the tumbler.