A farmer has enough food to feed 28 animals in his cattle for 9 days.

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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