Question

A farmer has enough food to deed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle ?

Answer 1

Let total animals can be feed with food in x days

Animals : Days :: Animals : Days

28 : 9 :: 28 + 8 = 36 : x

(more animals, less days)
As it represents inverse ratio, So $\frac{x}{28}=\frac{9}{36}$
$\Rightarrow x=\frac{9}{36}\times 28$

$\therefore$ No. of days food will lost $x=7$ days