Courses
Courses for Kids
Free study material
Offline Centres
More

# Two positive numbers $x$ and $y$ are inversely proportional. If $x$ increases by $20\%$, then percentage decrease in $y$ is:A. $20$B. $16\dfrac{2}{3}$C. $5$D. $1\dfrac{9}{11}$

Last updated date: 22nd Feb 2024
Total views: 338.1k
Views today: 5.38k
Verified
338.1k+ views
Hint: In this problem we have the two variables say $x$ and $y$, at the same time we have the relation between the two variables as inversely proportional. We will first represent this in mathematical form. Now we will replace the proportionality symbol with a constant and an equal to symbol. We will consider this as an equation and numbered as equation one. Now we have that if the variable $x$ increases by $20\%$, so we will calculate the changed value and denote it as ${{x}^{'}}$ and we will assume the variable that is proportional to ${{x}^{'}}$ as ${{y}^{'}}$. Now we will form a mathematical equation by using all the values we have up to now in the problem. After simplifying the obtained equation, we will get the required result.

Complete step by step solution:
Given that, $x$ and $y$ are inversely proportional.
We can represent the above statement mathematically as
$x\propto \dfrac{1}{y}$
Removing the proportionality symbol by introducing the proportionality constant say $k$ in the above equation, then we will get
\begin{align} & \Rightarrow x=k\times \dfrac{1}{y} \\ & \Rightarrow xy=k....\left( \text{i} \right) \\ \end{align}
Now we have that $x$ increases by $20\%$. So, the new changed value is
\begin{align} & {{x}^{'}}=x+20\%x \\ & \Rightarrow {{x}^{'}}=x+\dfrac{20}{100}x \\ & \Rightarrow {{x}^{'}}=x+0.2x \\ & \Rightarrow {{x}^{'}}=1.2x \\ \end{align}
Now the proportional value for ${{x}^{'}}$ from the equation $\left( \text{i} \right)$ is given by
$\Rightarrow {{x}^{'}}\times {{y}^{'}}=k$
From equation $\left( \text{i} \right)$ substituting the value $k=xy$ in the above equation, then we will get
\begin{align} & \Rightarrow 1.2x\times {{y}^{'}}=xy \\ & \Rightarrow {{y}^{'}}=\dfrac{y}{1.2} \\ \end{align}
Now the decrease in the variable $y$ will be
$\Rightarrow d=y-{{y}^{'}}$
Substituting ${{y}^{'}}=\dfrac{y}{1.2}$ in the above equation, then we will have
\begin{align} & \Rightarrow d=y-\dfrac{y}{1.2} \\ & \Rightarrow d=y\left( 1-\dfrac{1}{1.2} \right) \\ & \Rightarrow d=y\left( \dfrac{1.2-1}{1.2} \right) \\ & \Rightarrow d=\dfrac{y}{6} \\ \end{align}
Now the percentage decrease will be given by
\begin{align} & \Rightarrow d\%=\dfrac{d}{y}\times 100\% \\ & \Rightarrow d\%=\dfrac{\dfrac{y}{6}}{y}\times 100\% \\ & \Rightarrow d\%=\dfrac{100}{6}\% \\ & \Rightarrow d\%=16\dfrac{2}{3}\% \\ \end{align}
Hence Option – B is the correct answer.

Note: In this problem they have mentioned that the two variables are inversely proportional so we have represented it as $x\propto \dfrac{1}{y}$ and solved the problem. If they have mentioned that the two variables are directly proportional, then we must use the representation $x\propto y$ and solve the problem.