Question

Grass in the lawn grows equally thick and at a uniform rate. It takes $24$ days for $70$ cows and $60$ days for $30$ cows to eat the whole grass. How many cows are needed to eat the grass in $96$ days

Answer 1

Hint: To find the number of cows required in the question we need to consider two variables one for grass present and one for the rate of growth of grass, write the given conditions as linear equations and solve them. After that put the obtained values in the third equation which is related to the question asked.

Complete step-by-step solution:
Let us consider the grass present in the lawn is $x$
The rate of growth of grass be $y$
Grass present in lawn added to rate of growth rate for given days is the total grass present
Now from the first condition we have
$x+24\times y=24\times 70$
From second condition
$x+60\times y=60\times 30$
the two equations are given by
$x+24y=1680.........(1)$
$x+60y=1800.......(2)$
Solve the two equations
The equation $x=1800-60y$ is to be substituted in $(1)$
We get the equation as
$ 1800-60y+24y=1680 $
$\Rightarrow 36y=120 $
$ \Rightarrow y=\dfrac{120}{36} $
$ \Rightarrow y=\dfrac{10}{3} $
After getting the value of $y$ Substitute the value in equation $(2)$
$ x+60\times \dfrac{10}{3}=1800 $
$ \Rightarrow x+20\times 10=1800 $
$ \Rightarrow x+200=1800 $
 $ x=1600 $
Now for the third condition let $z$ be the number of cows needed to eat the grass in $96$ days. The condition is given by
$x+96\times y=z\times 96$
Substitute the values of $x$ and $y$ to find the value of $z$
$ \Rightarrow 1600+96\times \dfrac{10}{3}=96\times z$
$ \Rightarrow 1600+32\times 10=96\times z $
$ \Rightarrow 1600+320=96\times z $
$ \Rightarrow 1920=96\times z $
$ \Rightarrow z=\dfrac{1920}{96} $
 $ \Rightarrow z=20 $
Hence the number of cows represented by $z=20$.

Note: The linear equations are solved to obtain the number of cows. As there are two unknowns, we need at least two equations to solve them. From the first two conditions we have the two linear equations. And we need to find the value of $z$ from the third equation.