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? [This Question was asked as TITA]

• A. 8 litres
• B. 5 litres
• C. 6 litres
• D. 7 litres

Answer 1

The Correct Option is A

Original quantity of dye and water in the original solution i.e., \(16\) \(litres\) \(\bigg(i.e. = 40×\frac{2}{5}\bigg)\) and \(24\) \(litres\) \((i.e. = 40-16)\)

Quantity of water added = \(16\) \(litres\) (As \(1\) part = \(8\) \(litres\)). 

Quantity of dye and water removed = \(\frac{1}{4}×\frac{2}{7}(56)\) i.e., \(4\) litres and \(\frac{1}{4}×\frac{5}{7}×(56)\) i.e., \(10\)l litres. 

Final quantity of dye and water is \(12\) litres and \(30\) litres.

\(∴\) Quantity of dye to be added to make the ratio of dye and water again \(2: 3\) i.e., \(8\) litres.

Answer 2

Total solution \(= 40\) litres
Ratio of dye and water \(= 2:3\)
Dye \(= 16 \) litres and water \(= 24\) litres
\(⇒ \)New ratio \(= 2:5\)
\(⇒ \)dye remains same \(= 16\) litres
\(⇒\) Water \(= 24 + 16 = 40\) litres
When \(\frac 14\)mixture is removed
Then, Dye \(= 12\) litres and water \(= 30\) litres
Now ratio \(=2:3\)
\(⇒\) \(3\) unit \(= 30\)
\(⇒ 1\) unit \(= 10\)
\(⇒2\) units = \(20\) litres
Now, extra Dye \(= 20-12=8\) litres
Hence, \(8\) litres extra dye is added.

So, the correct option is (A): \(8\) litres