Question

A farmer had fodder for 40 animals for 60 days. He bought some more animals and the forder lasted for 50 days. How many animals did he buy?​

Answer 1

Given:-

• No. of Animals = 60
• Fodder was lasted for 40 animals = 60 days

To Find:-

• No. of Animals he had brought & the food lasted for 50 days

Answer:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

It is given that the fodder for 40 animals was lasted for 60 days .

We have to calculate the no. of animals he had brought in which the fodder lasted for 50 days . So,

• It is case of direct variation

Let the total animals in 2nd case be x

$:\implies$ 40/50 = x/60

$:\implies$ 50 × x = 60 × 40 (direct variation)

$:\implies$ x = 2400/50

$:\implies$ x = 48

when farmer had 40 animals the fodder was lasted for 60 days .

If farmer had 48 animals the fodder lasted for 50 days.

So , the Number of animals brought by the farmer = 48-40

$:\implies$ Number of animals brought by the farmer = 8

• So, the farmer brought 8 animals .

Answer 2

Given:-

• A farmer had fodder for 40 animals for 60 days.

• He bought some more animals and the forder lasted for 50 days.

To Find:-

• No. of Animals he buys

Solution:-

Let the total no. of animals be x.

Then, simply the equation formed is:-

${\sf{\implies\:\dfrac{40}{50}=\dfrac{x}{60}}}$

On Cross Multiplication,

${\sf{\implies\:40\times60=50\times x}}$

${\sf{\implies\:x=\dfrac{40\times60}{50}}}$

${\sf{\implies\:x=\dfrac{2400}{50}}}$

${\sf{\implies\:x=\dfrac{240}{5}}}$

${\bf{\implies\:x=48}}$

Total no. of animals = 48

And,

No. of animals bought = 48 - 40 = 8

Hence, 8 animals are bought by the farmers.

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