A tank holds x gallons of a saltwater solution that is 20%

Initial volume of salt in the saltwater solution = \(\frac{20}{100}x\) = \(0.2x\)

Initial volume of water in the saltwater solution = \(\frac{80}{100}x\) = \(0.8x\)

If \(1/4\) of the water is evaporated, then \(1 - 1/4 = 3/4\) is remaining.

Remaining water = \(\frac{3}{4}(\frac{80}{100}x)\) = \(\frac{3}{5}x\) = \(0.6x\)

After adding 10 gallons of water and 20 gallons of salt to the solution ->

New volume of salt = \(0.2x + 20\)
New volume of water = \(0.6x + 10\)

Total volume of the new salt water solution = \((0.2x + 20) + (0.6x + 10) = 0.8x + 30\)

We know that the new solution has a salt concentration of \(33 \frac{1}{3}\)% = \(33.3333......\)%
\(33.3333......\)% = \(\frac{33.33333.....}{100} = 0.33333...... = \frac{1}{3}\)

So, \(1/3\) volume of the new saltwater solution is equal to the volume of salt we calculated after adding 20 gallons of salt to the initial solution.

\(0.2x + 20 = \frac{1}{3}(0.8x + 30)\)

\(3(0.2x + 20) = 0.8x + 30\)

\(0.6x + 60 = 0.8x + 30\)

\(0.6x - 0.8x = 30 - 60\)

\(-0.2x = -30\)

\(x = \frac{30}{0.2}\)

\(x = \frac{300}{2}\)

\(x = 150\)

Correct option : D