Question

A farmer has enough food to feed 20 animals in his cattle for 6 days.
How long what the food last if there were 10 animals in the cattle? this is a question of indirect proportion pls hlap meee​

Answer 1

Answer:

let the number of days be x

given that he has enough food to feed 20 animals in his cattle for 6 days.later on,10 more animals be added.

therefore the number of days required to feed 30 animals.

20*6=(20+10)*x

120=30*x

x=120/30

=4

therefore a number of days = 4

hope this helps

Answer 2

Answer:

4 days

Step-By-Step Explanation:

If the number of animals increases, then it will take fewer days to last.

Then the two quantities are in inverse proportions.

Let the required number of days be p.

$\because x_1y_1 = x_2y_2$

Where,

$x_1 = 20, \: y_1 = 6, \: x_2 = 30$ and $y_2 = p$ (let)

=> 20 × 6 = 30 × p

=> 120 = 30p

=> p = 120/30

=> p = 4

Hence the required number of days = 4.