An ore contain `8%` zinc. How many kg of the ore will be required to get 2.8 kg of zinc ?

Answer

To solve the problem of how many kg of ore will be required to get 2.8 kg of zinc from an ore that contains 8% zinc, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the percentage of zinc in the ore**:
The ore contains 8% zinc. This means that in every 100 kg of ore, there are 8 kg of zinc.

2. **Calculate the amount of zinc in 1 kg of ore**:
Since the ore contains 8% zinc, we can find out how much zinc is present in 1 kg of ore.
\[ \text{Amount of zinc in 1 kg of ore} = 8\% \text{ of } 1 \text{ kg} = \frac{8}{100} \times 1 = 0.08 \text{ kg} \]

3. **Determine how much ore is needed for 2.8 kg of zinc**:
We know that 1 kg of ore contains 0.08 kg of zinc. To find out how many kg of ore is needed to obtain 2.8 kg of zinc, we can set up the equation:
\[ \text{Amount of ore required} = \frac{\text{Amount of zinc required}}{\text{Amount of zinc in 1 kg of ore}} = \frac{2.8 \text{ kg}}{0.08 \text{ kg}} \]

4. **Calculate the amount of ore required**:
Now, we perform the division:
\[ \frac{2.8}{0.08} = 35 \text{ kg} \]

Thus, **35 kg of ore will be required to get 2.8 kg of zinc**.

### Final Answer:
35 kg of ore is required to obtain 2.8 kg of zinc.

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