	A farmer has enough grains to feed 50 cattle for 12 days. He buys 10 cattle more. For how many days will the grain last now?

The problem can be solved using the concept of inverse variation. The total amount of food available is constant, so the number of cattle and the number of days the food lasts are inversely proportional.

### Step-by-step solution:

1. The farmer originally has enough grain to feed **50 cattle** for **12 days**. Let the total amount of grain be denoted as \( G \). The total grain consumed by the 50 cattle in 12 days is:
   \[
   G = 50 \times 12 = 600 \text{ cattle-days of grain}.
   \]

2. After buying 10 more cattle, the farmer now has \( 50 + 10 = 60 \) cattle.

3. Let \( x \) represent the number of days the grain will last for the 60 cattle. The grain consumed by the 60 cattle over \( x \) days will still equal the total grain, \( G = 600 \), so:
   \[
   60 \times x = 600.
   \]

4. Solving for \( x \):
   \[
   x = \frac{600}{60} = 10 \text{ days}.
   \]

### Final answer:
The grain will now last for **10 days**.

---

Do you want more details or have any other questions?

Here are 5 related questions for further practice:
1. What if the farmer had bought 20 more cattle instead of 10? How many days would the grain last?
2. If the grain lasted 8 days with 60 cattle, how many cattle did the farmer originally have?
3. How many days would the grain last if the farmer decided to sell 10 cattle instead of buying more?
4. If the farmer had only enough grain for 30 cattle for 15 days, how much grain would he need for 45 cattle for the same number of days?
5. How many days would the grain last if the farmer had enough for 50 cattle for 6 days but needed to feed 75 cattle?

**Tip:** Inverse variation problems involve a constant total quantity (like grain), so when one variable increases (like the number of cattle), the other (days the grain lasts) decreases proportionally.

A farmer has enough grains to feed 50 cattle for 12 days. He buys 10 cattle more. For how many days will the grain last now?

Explore how to solve an inverse variation problem where a farmer needs to calculate how long his grain will last after increasing the number of cattle from 50 to 60. Learn the concept of inverse variation, step-by-step solution, and key formulas used. Ideal for students in Grades 6-8.

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