Question

A farmer has enough food to feed 20 animals in his cattle for 6 days. How

long would the food last if there were 10 more animals in his cattle?​

Answer 1

Answer:

Let the food will now last for x days. Then,

Number of animals 20 30

Number of days 6 x

Clearly, more is the number of animals, less will be the number of days for food to last.

So, it is a case inverse proportion.

∴ 20 × 6 = 30x ⇒ x =

30

20×6

= 4

Hence, the food will now last for 4 days.

Step-by-step explanation:

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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,

$x_1 = 20, \: y_1 = 6, \: x_2 = 3$ and $y_2 = p$ (let)

=> 20 × 6 = 30 × p

=> 120 = 30p

=> p = 120/30

=> p = 4

Hence the required number of days = 4.