Question

There are three bottles of mixture of syrup and water in the ratios \(2:3\), \(3:4\) and \(7:5\). \(10\) litres of first and \(21\) litres of second bottles are taken. How much quantity of mixtures from third bottle is to be taken so that the final syrup and water ratio of mixture from three bottles will be \(1:1\) if the final mixture from all the three bottles is added and mixed in a big container?

• A. \(20\) litres
• B. \(25\) litres
• C. \(30\) litres
• D. \(35\) litres

Hint:

Instead of calculating water separately, you can just calculate the difference between syrup and water in each bottle.
Bottle 1: \(S-W = -2\).
Bottle 2: \(S-W = -3\).
Bottle 3: \(S-W = \frac{2}{12}x\).
For 1:1 ratio, sum of differences must be zero: \(-2 - 3 + \frac{x}{6} = 0 \implies x = 30\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To achieve a \(1:1\) ratio in the final mixture, the total quantity of syrup must be equal to the total quantity of water from all three bottles.
Step 2: Key Formula or Approach:
Calculate absolute volumes of Syrup (\(S\)) and Water (\(W\)) for each bottle and solve the equation \(S_{total} = W_{total}\).
Step 3: Detailed Explanation:
1. Bottle 1: Volume = \(10\) L, Ratio \(S:W = 2:3\).
Syrup \(S_1 = \frac{2}{2+3} \times 10 = 4\) L.
Water \(W_1 = \frac{3}{2+3} \times 10 = 6\) L.
2. Bottle 2: Volume = \(21\) L, Ratio \(S:W = 3:4\).
Syrup \(S_2 = \frac{3}{3+4} \times 21 = 9\) L.
Water \(W_2 = \frac{4}{3+4} \times 21 = 12\) L.
3. Bottle 3: Let the volume taken be \(x\) L, Ratio \(S:W = 7:5\).
Syrup \(S_3 = \frac{7}{7+5} \times x = \frac{7x}{12}\).
Water \(W_3 = \frac{5}{7+5} \times x = \frac{5x}{12}\).
4. Equate Total Syrup and Total Water:
\[ 4 + 9 + \frac{7x}{12} = 6 + 12 + \frac{5x}{12} \]
\[ 13 + \frac{7x}{12} = 18 + \frac{5x}{12} \]
\[ \frac{7x}{12} - \frac{5x}{12} = 18 - 13 \]
\[ \frac{2x}{12} = 5 \implies \frac{x}{6} = 5 \implies x = 30 \text{ litres.} \]
Step 4: Final Answer:
We need to take \(30\) litres from the third bottle.