Question

If $x$ varies inversely as $y$, and $x=11$ when $y=15$ ,how do you find $x$ when $y=3$?

Answer 1

Hint: This question tells us the relation between the two variables given to us with the help of that relation we are supposed to find the volume of a variable when another is given. So to solve the question we need to take a constant which would help us in finding the problem easily.

Complete step-by-step answer:
$x=\dfrac{k}{y}$ , $k$ is a constant ,and $x$ and $y$ are the variable.
 Now, we shall put the values of $x$ and $y$as $11$ and $15$ respectively, to find the value of $k$.
$k=x\times y$
Values to be put on the above equation :
$k=11\times 15$
$\Rightarrow 165$
So the value of $k$ for the above equation is $165$.
Now, we need to find the value of $x$ for $y=3$, We know the value of k , so on using the same formula as $x=\dfrac{k}{y}$ we will get the value for $x$ .
$x=\dfrac{k}{y}$
On substituting the values of all the variable
$\Rightarrow x=\dfrac{165}{3}$
On calculating we get,
$\Rightarrow 55$
$\therefore $ The value of $x$ for $y$ which is equal to $3$ is $55$.

Note: Important point to solve this kind of question is to read the question properly and what is the relation given between the two variables. This type of question has another approach which is shorter than the first one. Question here tells us that two variables are inversely proportional to each other , so this shows that the product of the two variables is a constant $(x\times y=\text{constant)}$. So if we equate the first product with the second product we would get the value of x for a certain value of variable $y$ without calculating the value of constant (k) .
${{x}_{1}}=11,{{y}_{1}}=15{{y}_{2}}$
Consider ${{x}_{1}}=11,{{y}_{1}}=15$and ${{y}_{2}}=3$then the value of ${{x}_{2}}$ will be
${{x}_{2}}=\dfrac{{{x}_{1}}\times {{y}_{1}}}{{{y}_{2}}}$
$\Rightarrow \dfrac{11\times 15}{3}$
$\Rightarrow \dfrac{165}{3}$
$\Rightarrow 55$
$\therefore $ The value $x$ comes out to be $55$.