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A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

Updated On: Jan 15, 2026
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Solution and Explanation

Step 1: Initial Mixture

Given: A solution of 40 litres containing dye and water in a 2:3 ratio.
Calculations:

  • Dye = \(\frac{2}{5} \times 40 = 16\) litres
  • Water = \(40 - 16 = 24\) litres

Step 2: Water Addition

The solution's ratio is adjusted such that 1 part equals 8 litres.
New total volume = \(\frac{7}{5} \times 40 = 56\) litres.
Water added = \(56 - 40 = 16\) litres

Step 3: Removal of 1/4 of New Solution

Quantity removed = \(\frac{1}{4} \times 56 = 14\) litres.
The current ratio of dye to water is 2:5.
Thus, the removed quantities are:

  • Dye removed = \(\frac{2}{7} \times 14 = 4\) litres
  • Water removed = \(\frac{5}{7} \times 14 = 10\) litres

 

Step 4: Final Remaining Quantities

After removal:

  • Dye remaining = \(\frac{2}{7} \times 42 = 12\) litres
  • Water remaining = \(\frac{5}{7} \times 42 = 30\) litres

 

Step 5: Restoring the 2:3 Ratio

Let \(x\) litres of dye be added.
The desired ratio is: \(\frac{12 + x}{30} = \frac{2}{3}\)

Solving for \(\boldsymbol{x}\):

\(3(12 + x) = 60 \Rightarrow 36 + 3x = 60 \Rightarrow 3x = 24 \Rightarrow x = 8\)

Final Answer

8 litres of dye must be added to re-establish the 2:3 ratio.

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