Chicken kg 20 rupee Potato kg 1rupee Onion kg 1/4rupee I want 20 kg in 20 rupees

To buy 20 kg of items with a budget of 20 rupees, we need to consider the prices per kilogram of each item:

- Chicken: 20 rupees per kg
- Potato: 1 rupee per kg
- Onion: 0.25 rupees per kg

Let's break down how we can distribute the quantities within the 20-rupee budget while achieving a total of 20 kg:

### Possible Combination:

1. **Potato**: \( x \) kg
   - Cost: \( x \times 1 \) rupee per kg

2. **Onion**: \( y \) kg
   - Cost: \( y \times 0.25 \) rupees per kg

3. **Chicken**: \( z \) kg
   - Cost: \( z \times 20 \) rupees per kg

Given constraints:
- \( x + y + z = 20 \) kg (total weight)
- \( x \times 1 + y \times 0.25 + z \times 20 = 20 \) rupees (total cost)

### Solving the Equations:

Let's try \( z = 1 \) kg of chicken, so:

- Cost of chicken = \( 1 \times 20 = 20 \) rupees
- Remaining budget = \( 20 - 20 = 0 \) rupees
- Remaining weight = \( 20 - 1 = 19 \) kg

Given the budget is exhausted by the chicken, we cannot buy any potatoes or onions in this scenario.

Let's reduce the quantity of chicken and explore another possibility.

### Another Option:
Suppose you don't buy chicken at all:

Let’s try only buying potatoes and onions:
- \( x \times 1 + y \times 0.25 = 20 \) rupees
- \( x + y = 20 \) kg

We can calculate \( y \) in terms of \( x \):

- \( x + 0.25y = 20 \)
- \( y = 20 - x \)

Substitute in the cost equation:

- \( x + 0.25(20 - x) = 20 \)
- \( x + 5 - 0.25x = 20 \)
- \( 0.75x = 15 \)
- \( x = 20 \)

This suggests:
- Potatoes: 20 kg
- Onions: 0 kg
- Total cost: \( 20 \times 1 = 20 \) rupees

So, you can buy **20 kg of potatoes** for exactly **20 rupees**.

### Conclusion:

- Buy **20 kg of potatoes** to use exactly 20 rupees.

Would

Learn how to solve budget constraints using linear equations. This problem involves determining the quantity of items to buy within a given budget. Explore practical applications of algebraic concepts in budgeting. Suitable for high school students.

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