The mixture was of zinc and cobalt and 430 tons of zinc were required to make 700 tons of the mixture. How many tons of cobalt were required to make 2800 tons of the mixture?

To find out how many tons of cobalt are required to make 2800 tons of the mixture, we need to first find the ratio of zinc to the total mixture and then apply this ratio to the new mixture amount.

### Step-by-Step Solution:

1. **Find the ratio of zinc in the mixture**:
   - Zinc in the mixture = 430 tons
   - Total mixture = 700 tons
   - The fraction of the mixture that is zinc:
   \[
   \frac{430}{700} = \frac{43}{70}
   \]

2. **Find the fraction of the mixture that is cobalt**:
   - Since the mixture consists of only zinc and cobalt, the fraction of cobalt in the mixture is:
   \[
   1 - \frac{43}{70} = \frac{70}{70} - \frac{43}{70} = \frac{27}{70}
   \]

3. **Apply this ratio to find the cobalt needed for 2800 tons of the mixture**:
   - Total mixture = 2800 tons
   - Tons of cobalt required:
   \[
   2800 \times \frac{27}{70} = 40 \times 27 = 1080 \text{ tons}
   \]

Therefore, **1080 tons of cobalt** are required to make 2800 tons of the mixture.

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you find the percentage composition of zinc and cobalt in the original mixture?
2. What if the mixture were made up of three metals? How would the calculation change?
3. How would you calculate the amount of zinc in 2800 tons of the mixture?
4. How can you verify that the mixture proportion is consistent?
5. How would you solve this if only the ratio of zinc to cobalt was given?

**Tip:** Always double-check the sum of fractions to ensure they add up to the whole when dealing with mixtures.

Learn how to solve a mixture problem involving zinc and cobalt. This problem shows how 430 tons of zinc are used in a 700-ton mixture and explains how to calculate the required amount of cobalt for a larger 2800-ton mixture. Suitable for Grades 6-8.

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