Ques- A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the

mixture must be drawn off and replaced with water so that the mixture may be half water and half

syrup?

1/3

1/4

1/5

1/7

Let's solve this problem step by step.

### Step 1: Define the Initial Ratio and Total Parts
The vessel is initially filled with a mixture where the ratio of water to syrup is 3:5. This means there are 3 parts water and 5 parts syrup, totaling 8 parts.

Let the total volume of the mixture be \( V \) liters.

### Step 2: Amounts of Water and Syrup
- The amount of water in the vessel initially is \( \frac{3}{8}V \).
- The amount of syrup in the vessel initially is \( \frac{5}{8}V \).

### Step 3: Define the Fraction of Mixture Drawn Off
Let \( x \) be the fraction of the mixture drawn off and replaced with water.

When \( x \) fraction of the mixture is drawn off:
- The amount of water removed is \( x \times \frac{3}{8}V \).
- The amount of syrup removed is \( x \times \frac{5}{8}V \).

### Step 4: After Replacement with Water
After drawing off \( xV \) liters of the mixture, the remaining water and syrup quantities are:
- Water left: \( \frac{3}{8}V - x \times \frac{3}{8}V \)
- Syrup left: \( \frac{5}{8}V - x \times \frac{5}{8}V \)

When the drawn-off portion is replaced with water, the new amount of water becomes:
\[
\left(\frac{3}{8}V - x \times \frac{3}{8}V\right) + xV
\]
This simplifies to:
\[
\frac{3}{8}V(1 - x) + xV = \frac{3}{8}V + \frac{5}{8}Vx
\]

### Step 5: Set Up the Equation for Equal Parts
We want the final mixture to be half water and half syrup, meaning both water and syrup should each be \( \frac{V}{2} \).

Set up the equation for water:
\[
\frac{3}{8}V + \frac{5}{8}Vx = \frac{V}{2}
\]

### Step 6: Solve the Equation
Let's solve for \( x \):
\[
\frac{3}{8} + \frac{5}{8}x = \frac{1}{2}
\]
Multiply everything by 8 to clear the fraction:
\[
3 + 5x = 4
\]
Solve for \( x \):
\[
5x = 1
\]
\[
x = \frac{1}{5}
\]

### Final Answer
The fraction of the mixture that must be drawn off and replaced with water is **\(\frac{1}{5}\)**.

Would you like more details or have any questions?

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### Related Questions:
1. What if the initial ratio was 4:5 instead of 3:5?
2. How would the solution change if the goal was to make the mixture 2 parts water and 3 parts syrup?
3. What would happen if 1/7 of the mixture was drawn off instead of 1/5?
4. How can the solution be verified using another method?
5. What if the volume of the vessel was 10 liters, how much water would be in the final mixture?
6. What if only syrup could be added to adjust the ratio, how would that affect the solution?
7. Can this problem be solved using an algebraic substitution method?
8. How does changing the amount of the mixture drawn off affect the concentration of water and syrup?

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### Tip:
For problems involving replacement of a mixture, set up clear equations based on the ratios and carefully substitute values to avoid mistakes in the calculations.

Learn how to solve a mixture problem involving drawing off and replacing a fraction of the mixture with water. This problem covers ratio and proportion concepts and requires solving equations to achieve the desired mixture ratio. Suitable for students in Grades 9-12.

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