Chicken kg 20 rupee Potato kg 1rupee Onion kg 1/4rupee iwant all of mix 20 kg in 20 rupees

This problem involves solving a system of linear equations to determine the quantities of chicken, potato, and onion that can be purchased within the given constraints. Let's denote:

- \( x \) as the kilograms of chicken,
- \( y \) as the kilograms of potatoes,
- \( z \) as the kilograms of onions.

Given:
- The cost of chicken is 20 rupees per kg.
- The cost of potatoes is 1 rupee per kg.
- The cost of onions is \( \frac{1}{4} \) rupee per kg.

The problem states that you want to buy a total of 20 kg of these items for exactly 20 rupees. This gives us the following equations:

1. \( x + y + z = 20 \) (Total weight of 20 kg)
2. \( 20x + y + \frac{1}{4}z = 20 \) (Total cost of 20 rupees)

We can now solve this system of equations.

### Step 1: Express \( y \) in terms of \( x \) and \( z \) from the first equation:
\[
y = 20 - x - z
\]

### Step 2: Substitute \( y = 20 - x - z \) into the second equation:
\[
20x + (20 - x - z) + \frac{1}{4}z = 20
\]

Simplify the equation:
\[
20x + 20 - x - z + \frac{1}{4}z = 20
\]
\[
19x - \frac{3}{4}z = 0
\]

### Step 3: Solve for \( x \) in terms of \( z \):
\[
x = \frac{3}{76}z
\]

### Step 4: Substitute \( x = \frac{3}{76}z \) back into the first equation:
\[
\frac{3}{76}z + y + z = 20
\]
\[
y = 20 - \frac{79}{76}z
\]

### Step 5: Ensure that \( y \geq 0 \) and \( z \geq 0 \):
This constraint ensures the physical meaning of the problem.

We find that the feasible solution is:
\[
z = 16 \text{ kg}, \quad x = \frac{3}{76} \times 16 \approx 0.63 \text{ kg}, \quad y = 20 - \frac{79}{76} \times 16 = 3.37 \text{ kg}
\]

This indicates you should purchase approximately:
- 0.63 kg of chicken
- 3.37 kg of potatoes
- 16 kg of onions

This would satisfy the conditions of 20 kg of total weight and 20 rupees total cost.

Would you like more details or have any questions?

### Relative Questions:
1. How do you solve a system of linear equations using substitution?
2. What are the methods to solve linear equations apart from substitution?
3. How can you check if a solution to a linear system is feasible in real-world scenarios?
4. What is the significance of constraints in optimization problems?
5. How do you interpret fractional solutions in practical contexts?

### Tip:
When solving real-world problems involving money and quantities, always ensure that your solutions are feasible by considering the context and rounding appropriately if necessary.

Learn how to solve a system of linear equations involving chicken, potato, and onion quantities under cost constraints. This problem requires balancing weight and cost limitations, employing algebraic methods for optimal purchase decisions.

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