Question:
A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle?
Solution:
Let x be the required number of days. Then, we have:
| No. of days | 9 | x |
| No. of animals | 28 | 36 |
Clearly, more number of animals will take less number of days to finish the food.
So, it is a case of inverse proportion.
Now, $9 \times 28=x \times 36$
$\Rightarrow x=\frac{9 \times 28}{36}$
$\Rightarrow x=7$
Therefore, the food will last for 7 days.
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