Q20. If Rani mixes a liters of a juice with m% concentration, 2a liters of a juice with n% concentration, and 3a liters of a juice with (m+n)% concentration, the resultant mixture has a concentration of 50%.

Now, Rani wants to mix 40 liters of the first juice (with m% concentration) and 50 liters of the second juice (with n% concentration). Find the resultant concentration of juice in the new mixture.

Let's break down the problem step by step.

### Part 1: Finding a Relationship for the First Mixture

Rani initially mixes:
- \( a \) liters of juice with \( m\% \) concentration,
- \( 2a \) liters of juice with \( n\% \) concentration,
- \( 3a \) liters of juice with \( (m + n)\% \) concentration.

The total volume of the mixture is:
\[
a + 2a + 3a = 6a \text{ liters}.
\]

Now, let's calculate the amount of pure juice in each part:
- In the first juice: \( \frac{m}{100} \times a \text{ liters of pure juice} \),
- In the second juice: \( \frac{n}{100} \times 2a \text{ liters of pure juice} \),
- In the third juice: \( \frac{m + n}{100} \times 3a \text{ liters of pure juice} \).

The total amount of pure juice in the mixture is:
\[
\frac{m}{100} \times a + \frac{n}{100} \times 2a + \frac{m + n}{100} \times 3a.
\]

We know that the resultant mixture has a concentration of 50%, so the total amount of pure juice in the 6a liters must be:
\[
\frac{50}{100} \times 6a = 3a \text{ liters of pure juice}.
\]

Equating the two expressions for the total amount of pure juice:
\[
\frac{m}{100} \times a + \frac{n}{100} \times 2a + \frac{m + n}{100} \times 3a = 3a.
\]

Dividing both sides by \( a \), we get:
\[
\frac{m}{100} + \frac{2n}{100} + \frac{3(m + n)}{100} = 3.
\]

Simplifying the equation:
\[
\frac{m}{100} + \frac{2n}{100} + \frac{3m}{100} + \frac{3n}{100} = 3,
\]
\[
\frac{4m}{100} + \frac{5n}{100} = 3,
\]
\[
4m + 5n = 300.
\]
This is the relationship between \( m \) and \( n \).

### Part 2: Mixing 40 Liters of the First Juice and 50 Liters of the Second Juice

Now, Rani wants to mix:
- 40 liters of the first juice (with \( m\% \) concentration),
- 50 liters of the second juice (with \( n\% \) concentration).

The total volume of this new mixture is:
\[
40 + 50 = 90 \text{ liters}.
\]

The amount of pure juice in the first juice is:
\[
\frac{m}{100} \times 40 = \frac{40m}{100} = \frac{2m}{5} \text{ liters}.
\]

The amount of pure juice in the second juice is:
\[
\frac{n}{100} \times 50 = \frac{50n}{100} = \frac{n}{2} \text{ liters}.
\]

The total amount of pure juice in the new mixture is:
\[
\frac{2m}{5} + \frac{n}{2} \text{ liters}.
\]

The concentration of the new mixture is:
\[
\frac{\left( \frac{2m}{5} + \frac{n}{2} \right)}{90} \times 100.
\]

Now we can use the relationship \( 4m + 5n = 300 \) to simplify the result. Solving for \( n \) in terms of \( m \):
\[
n = \frac{300 - 4m}{5}.
\]

Substitute this into the formula for the concentration:
\[
\text{Concentration} = \frac{\frac{2m}{5} + \frac{\frac{300 - 4m}{5}}{2}}{90} \times 100.
\]

Simplifying:
\[
\text{Concentration} = \frac{\frac{2m}{5} + \frac{300 - 4m}{10}}{90} \times 100,
\]
\[
\text{Concentration} = \frac{\frac{4m + 300 - 4m}{10}}{90} \times 100,
\]
\[
\text{Concentration} = \frac{300}{900} \times 100 = \frac{1}{3} \times 100 = 33.33\%.
\]

### Final Answer:
The resultant concentration of juice in the new mixture is **33.33%**.

Would you like further details or have any questions?

### Related Questions:
1. What happens if Rani uses a different amount of the first juice?
2. How would the concentration change if she adds an additional 10 liters of water?
3. Can you derive the formula for any arbitrary concentrations \( m \) and \( n \)?
4. How does the ratio of the volumes of the juices affect the final concentration?
5. What if the concentration of the third juice was unknown—how would you calculate it?

### Tip:
When solving mixture problems, always check the total volume and concentration of pure substances to ensure correct proportions.

Q20. If Rani mixes a liters of a juice with m% concentration, 2a liters of a juice with n% concentration, and 3a liters of a juice with (m+n)% concentration, the resultant mixture has a concentration of 50%. Now, Rani wants to mix 40 liters of the first juice (with m% concentration) and 50 liters of the second juice (with n% concentration). Find the resultant concentration of juice in the new mixture.

Learn how to solve a mixture problem where Rani mixes 40 liters of juice with m% concentration and 50 liters with n% concentration. The problem demonstrates the use of algebraic relationships and concentration formulas, ideal for students in Grades 10-12.

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