Q6 A merchant has three varieties of rice having cost Rs.20/kg, Rs.24/kg and Rs.30/kg. He mixes them in some ratio such that on selling the mixture at Rs. 30/kg he earns a profit of 20%. If the mixture contains 2 kg of the third variety of rice and has integer number of kilograms of other two varieties also, then how many kilograms of the second variety can be there in the mixture ?

Let's break the problem down step by step:

### Information provided:
1. **Varieties of rice**:
   - First variety costs Rs. 20/kg.
   - Second variety costs Rs. 24/kg.
   - Third variety costs Rs. 30/kg.

2. The merchant sells the mixture at Rs. 30/kg and earns a **profit of 20%**.

3. The mixture contains **2 kg of the third variety** (which costs Rs. 30/kg).

4. We need to find out how many kilograms of the **second variety** can be in the mixture.

### Step 1: Determine the selling price of the mixture.

The mixture is sold at Rs. 30/kg, and the merchant earns a 20% profit. To find the cost price (C.P.) of the mixture, we use the formula for profit:

\[
\text{Selling Price (S.P.)} = \text{Cost Price (C.P.)} \times (1 + \text{Profit Percentage})
\]

Given:
- S.P. = Rs. 30/kg
- Profit Percentage = 20% = 0.2

\[
30 = \text{C.P.} \times (1 + 0.2)
\]

\[
30 = \text{C.P.} \times 1.2
\]

\[
\text{C.P.} = \frac{30}{1.2} = 25 \, \text{Rs./kg}
\]

Thus, the cost price of the mixture is Rs. 25/kg.

### Step 2: Set up the equation for the cost price of the mixture.

Let:
- \( x \) kg be the amount of the first variety (Rs. 20/kg),
- \( y \) kg be the amount of the second variety (Rs. 24/kg),
- \( 2 \) kg is the amount of the third variety (Rs. 30/kg) as given.

The total cost of the mixture is:

\[
\text{Total Cost} = (20x) + (24y) + (30 \times 2)
\]

Simplifying:

\[
\text{Total Cost} = 20x + 24y + 60
\]

The total weight of the mixture is \( x + y + 2 \) kg, and since the cost price of the mixture is Rs. 25/kg, we set up the equation for the total cost:

\[
\frac{20x + 24y + 60}{x + y + 2} = 25
\]

### Step 3: Solve for \( y \).

Multiply both sides of the equation by \( x + y + 2 \):

\[
20x + 24y + 60 = 25(x + y + 2)
\]

Expand the right-hand side:

\[
20x + 24y + 60 = 25x + 25y + 50
\]

Simplify the equation:

\[
20x + 24y + 60 - 25x - 25y - 50 = 0
\]

\[
-5x - y + 10 = 0
\]

Rearrange to get:

\[
y = -5x + 10
\]

### Step 4: Find integer values of \( x \) and \( y \).

For \( y \) to be an integer, \( x \) must be an integer. Let’s test values of \( x \):

- If \( x = 1 \), then \( y = -5(1) + 10 = 5 \).
- If \( x = 2 \), then \( y = -5(2) + 10 = 0 \).

Thus, the possible values are \( (x = 1, y = 5) \) or \( (x = 2, y = 0) \).

### Step 5: Conclusion.

The number of kilograms of the second variety that can be in the mixture is **5 kg** (when \( x = 1 \), \( y = 5 \)).

Would you like a further breakdown of any part of the solution or additional details?

### Related Questions:
1. What would the mixture cost per kg if the merchant didn't aim for any profit?
2. How would the answer change if the merchant wanted to earn a 10% profit instead of 20%?
3. What is the total weight of the mixture in the scenario above?
4. If the price of the second variety of rice increased to Rs. 26/kg, how would it affect the mixture's composition?
5. Can there be a scenario where all three varieties are mixed in equal proportions?

### Tip:
When dealing with mixture problems, always balance the total cost and total weight to maintain consistency in unit price calculations.

A merchant has three varieties of rice having cost Rs.20/kg, Rs.24/kg and Rs.30/kg. He mixes them in some ratio such that on selling the mixture at Rs. 30/kg he earns a profit of 20%. If the mixture contains 2 kg of the third variety of rice and has integer number of kilograms of other two varieties also, then how many kilograms of the second variety can be there in the mixture?

Solve a rice mixture problem where a merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg, and Rs.30/kg, selling the mixture at Rs.30/kg with a 20% profit. Learn how to calculate how many kilograms of the second variety are in the mixture. Ideal for Grades 9-11, this step-by-step guide covers algebraic methods and profit formulas.

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