Understanding the Mixture Problem

This problem involves a mixture of water and syrup where the ratio changes after a part of the mixture is drawn off and replaced with water. We need to find out what fraction of the mixture was drawn off and replaced.

Initial Mixture Composition

The vessel initially contains a mixture of water and syrup in the ratio 5:11.

• Parts of water = 5
• Parts of syrup = 11
• Total parts = $5 + 11 = 16$

The initial fraction of water in the mixture is $\frac{5}{16}$.

The initial fraction of syrup in the mixture is $\frac{11}{16}$.

The Process: Drawing Off and Replacing

Let's assume a fraction, say $x$, of the mixture is drawn off from the vessel. When a fraction $x$ of the mixture is removed, the amounts of water and syrup removed are also in the same ratio as the original mixture.

Amount of water removed = $x \times (\text{Initial amount of water})$

Amount of syrup removed = $x \times (\text{Initial amount of syrup})$

After drawing off the fraction $x$, the remaining quantity of the mixture is $(1-x)$ times the initial quantity. The amounts of water and syrup remaining are:

• Amount of water remaining = $(1-x) \times (\text{Initial amount of water})$
• Amount of syrup remaining = $(1-x) \times (\text{Initial amount of syrup})$

Then, the same amount (the drawn-off part, which is fraction $x$ of the initial total volume) is replaced with water. This means an amount equal to $x$ times the initial total volume of water is added, and no syrup is added.

Final Mixture Composition

The final mixture is required to have syrup and water in the ratio 3:2. This means the ratio of water to syrup in the final mixture is 2:3.

• Parts of water in the final mixture = 2
• Parts of syrup in the final mixture = 3
• Total parts in the final ratio = $2 + 3 = 5$

The final fraction of water in the mixture is $\frac{2}{5}$.

The final fraction of syrup in the mixture is $\frac{3}{5}$.

Note that the total volume of the mixture in the vessel remains the same throughout the process (amount drawn off is equal to the amount replaced).

Setting Up the Equation for Ratio Change

We can solve this problem by focusing on one component. Let's focus on the amount of syrup. Syrup is removed when the mixture is drawn off, but no syrup is added when water is replaced.

Let the initial total quantity of the mixture be $Q$.

Initial amount of syrup = $\frac{11}{16} Q$

Amount of syrup remaining after drawing off fraction $x$ of the mixture = $(1-x) \times \left(\frac{11}{16} Q\right)$.

Since only water is added, the amount of syrup in the final mixture is the same as the amount of syrup remaining after drawing off the mixture.

The final mixture has syrup and water in the ratio 3:2. The total quantity of the final mixture is still $Q$.

Final amount of syrup = $\frac{3}{5} Q$

Now, we equate the final amount of syrup to the amount of syrup remaining after the draw-off:

$(1-x) \times \left(\frac{11}{16} Q\right) = \frac{3}{5} Q$

Solving for the Drawn Off Part

We need to solve the equation $(1-x) \times \left(\frac{11}{16} Q\right) = \frac{3}{5} Q$ for the value of $x$. Since $Q$ is the total quantity and is not zero, we can cancel $Q$ from both sides of the equation:

$(1-x) \times \frac{11}{16} = \frac{3}{5}$

Now, isolate $(1-x)$ by multiplying both sides by $\frac{16}{11}$:

$1-x = \frac{3}{5} \times \frac{16}{11}$

$1-x = \frac{3 \times 16}{5 \times 11}$

$1-x = \frac{48}{55}$

Now, solve for $x$:

$x = 1 - \frac{48}{55}$

$x = \frac{55}{55} - \frac{48}{55}$

$x = \frac{55 - 48}{55}$

$x = \frac{7}{55}$

So, the part of the mixture that must be drawn off and replaced with water is $\frac{7}{55}$.

Comparing with Options

Let's check the calculated value against the given options:

• Option 1: $\frac{27}{35}$
• Option 2: $\frac{7}{55}$
• Option 3: $\frac{14}{45}$
• Option 4: $\frac{36}{65}$

Our calculated value, $\frac{7}{55}$, matches Option 2.

Revision Table: Key Concepts for Mixture Problems

Additional Information on Mixture Problems

Mixture problems often involve ratios, percentages, or concentrations. A common type is drawing off a part and replacing it with another substance, which changes the overall composition. Here are some tips for solving such problems:

• Identify Initial and Final States: Clearly note the ratios or concentrations of components at the beginning and at the end.
• Understand the Change Process: Determine what is being added or removed and how it affects each component.
• Focus on One Component: It is often easier to track the quantity or fraction of one component, especially if one component is added or not added during the process. In this problem, tracking syrup was effective because only water was added.
• Set Up an Equation: Relate the initial composition, the change process, and the final composition using an algebraic equation.
• Assume a Total Volume (if helpful): While not always necessary (as seen in our solution where Q cancelled out), assuming a convenient total volume (like the LCM of denominators) can sometimes make calculations simpler, especially if specific quantities were involved.

Understanding how ratios and fractions change upon removal and addition is key to mastering these types of quantitative aptitude problems.