Question

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?

Answer 1

 Step 1: Initial Mixture

Given: A solution of 40 litres with dye and water in the ratio 2:3.
Therefore:

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

 

 Step 2: Water Added

The solution is changed to a new ratio where 1 part = 8 litres.
New total volume = \(7 \times 8 = 56\) litres.
Water added = \(56 - 40 = 16\) litres

 Step 3: 1/4 of New Solution Removed

Removed quantity = \(\frac{1}{4} \times 56 = 14\) litres.
The current ratio of dye and water = \(2:5\)
So, removed:

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

 

 Step 4: Final Quantities Left

After removal:

• Dye = \(16 - 4 = 12\) litres
• Water = \(40 - 4 - 10 = 30\) litres

 

 Step 5: Making Ratio Again 2:3

Let \(x\) litres of dye be added.
We want: \(\frac{12 + x}{30} = \frac{2}{3}\)

Cross-multiplying:
\(3(12 + x) = 60 \Rightarrow 36 + 3x = 60 \Rightarrow 3x = 24 \Rightarrow x = 8\)

 Final Answer

8 litres of dye must be added to restore the ratio to 2:3.