A puppy eats a sausage 5 seconds faster than a kitten eats an identical sausage. If they share one sausage, they can eat it in 6 seconds. How long does it take each animal to eat a sausage alone?

In work and time the time taken is inversely proportional to the work done. Work and time is a concept to solve the problems of work related question in a given time.The kitten eats a sausage in 15 seconds and puppy eats a sausage in 10 seconds.

What is concept of Work and time?

In work and time the time taken is inversely proportional to the work done. Work and time is a concept to solve the problems of work related question in a given time

Given-

Puppy and kitten eats one sausage combined in 6 sec.

A puppy eats a sausage 5 seconds faster than a kitten eats an identical sausage.

Let puppy eats a sausage in [tex]x[/tex] seconds and kitten eats a sausage in [tex]y[/tex] seconds.

As Puppy and kitten eats one sausage combined in 6 sec.Thus,

[tex]\dfrac{1}{x} +\dfrac{1}{y} =\dfrac{1}{6}[/tex]

Let above equation is equation one.

A puppy eats a sausage 5 seconds faster than a kitten eats an identical sausage. Thus

[tex]\dfrac{1}{x} -\dfrac{1}{y}=\dfrac{1}{5}[/tex]

Put this value of [tex]x[/tex] in the the equation one to get the value of y. Therefore,

[tex]\dfrac{1}{6} -\dfrac{1}{y}-\dfrac{1}{y}=\dfrac{1}{5}[/tex]

[tex]-\dfrac{1}{2y}=\dfrac{1}{5}-\dfrac{1}{6}[/tex]

[tex]-\dfrac{1}{2y}=-\dfrac{1}{30}[/tex]

[tex]y=15[/tex]

Thus the kitten eats a sausage in 15 seconds.

Put the value of y in equation one. thus,

[tex]\dfrac{1}{x} +\dfrac{1}{15} =\dfrac{1}{6}[/tex]

[tex]\dfrac{1}{x} =\dfrac{1}{6}-\dfrac{1}{15}[/tex]

[tex]\dfrac{1}{x} =\dfrac{9}{90}[/tex]

[tex]x=10[/tex]

Thus the puppy eats a sausage in 10 seconds.

Hence the kitten eats a sausage in 15 seconds and puppy eats a sausage in 10 seconds.

Learn more about the work and time here,

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