A farmer has two types of fertilizer. Fertilizer A contains 10% nitrogen and 6% phosphorous, and fertilizer B contains 5%, nitrogen and 8% phosphorous. How many kilograms of each fertilizer should be mixed to make a 200 kg mixture that contains 12 kg of nitrogen and 14 kg of phosphorous? Do using linear equation in one variable by using the fertilizer b as 200 - x. At last check the weight of phosphorus and nitrogen so that they are matching with the given weight with question

Let \( x \) be the amount of Fertilizer A (in kg) in the mixture. Then, the amount of Fertilizer B will be \( 200 - x \) kg, since the total weight of the mixture is 200 kg.

### Step 1: Set up equations based on nitrogen and phosphorous content

1. **Nitrogen Requirement:**
   - Fertilizer A has 10% nitrogen, so the nitrogen content from Fertilizer A will be \( 0.10 \times x \).
   - Fertilizer B has 5% nitrogen, so the nitrogen content from Fertilizer B will be \( 0.05 \times (200 - x) \).
   - The total nitrogen content needs to be 12 kg.
   
   Therefore, we can write the equation for nitrogen as:
   \[
   0.10x + 0.05(200 - x) = 12
   \]

2. **Phosphorous Requirement:**
   - Fertilizer A has 6% phosphorous, so the phosphorous content from Fertilizer A will be \( 0.06 \times x \).
   - Fertilizer B has 8% phosphorous, so the phosphorous content from Fertilizer B will be \( 0.08 \times (200 - x) \).
   - The total phosphorous content needs to be 14 kg.
   
   Therefore, we can write the equation for phosphorous as:
   \[
   0.06x + 0.08(200 - x) = 14
   \]

### Step 2: Solve the Nitrogen Equation

Expanding and simplifying the nitrogen equation:
\[
0.10x + 0.05 \times 200 - 0.05x = 12
\]
\[
0.10x + 10 - 0.05x = 12
\]
\[
0.05x = 2
\]
\[
x = \frac{2}{0.05} = 40
\]

So, \( x = 40 \) kg of Fertilizer A.

Then, the amount of Fertilizer B is:
\[
200 - x = 200 - 40 = 160 \, \text{kg}
\]

### Step 3: Verify the Nitrogen and Phosphorous Content

1. **Nitrogen Content:**
   - Fertilizer A contributes \( 0.10 \times 40 = 4 \) kg of nitrogen.
   - Fertilizer B contributes \( 0.05 \times 160 = 8 \) kg of nitrogen.
   - Total nitrogen = \( 4 + 8 = 12 \) kg (which matches the required amount).

2. **Phosphorous Content:**
   - Fertilizer A contributes \( 0.06 \times 40 = 2.4 \) kg of phosphorous.
   - Fertilizer B contributes \( 0.08 \times 160 = 12.8 \) kg of phosphorous.
   - Total phosphorous = \( 2.4 + 12.8 = 14 \) kg (which also matches the required amount).

### Final Answer:
- **Fertilizer A:** 40 kg
- **Fertilizer B:** 160 kg

Would you like further details on each calculation step?

---

1. How do you form equations based on percentage compositions?
2. Why was \(200 - x\) used for Fertilizer B?
3. Can this method be applied to other mixture problems?
4. What if the mixture weights were different but with the same requirements?
5. How do we verify if the solution satisfies all conditions?

**Tip:** Breaking down each percentage calculation step-by-step helps in preventing errors and understanding the setup of equations.

A farmer has two types of fertilizer. Fertilizer A contains 10% nitrogen and 6% phosphorous, and fertilizer B contains 5% nitrogen and 8% phosphorous. How many kilograms of each fertilizer should be mixed to make a 200 kg mixture that contains 12 kg of nitrogen and 14 kg of phosphorous? Do using linear equation in one variable by using the fertilizer b as 200 - x. At last check the weight of phosphorus and nitrogen so that they are matching with the given weight with question.

Explore how to determine the kilograms of Fertilizer A and Fertilizer B needed in a 200 kg mixture to meet nitrogen and phosphorous requirements. This solution uses a linear equation in one variable, focusing on percentage calculations and mixture algebra, suitable for Grades 8-10.

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