Question

In an office there were initially 'n' employees. The HR manager first hired p% employees, then after a month q% employees left the office, then there were finally 'n' employees remained in the office, the value of $p-q$ is
(a) pq
(b) $\dfrac{pq}{100}$
(c) $\dfrac{p}{q}$
(d) $\dfrac{100p}{q}$

Answer 1

Hint: To find the value of $p-q$ , we have to form an equation with given conditions. We are given that the total number of employees initially is n. Hence, we can write the number of employees after hiring p% employees as $n+p\%\text{ of }n$. We have to then simplify this. Then, the number of q% employees will be q% of the number of employees hired. We have to form an equation by subtracting q% employees that left from the employees after hiring p%. We will then equate this to the employees left at the end. We have to simplify and rearrange the terms so that the value of $p-q$ can be found.

Complete step by step answer:
We are given that the total number of employees initially is n. We are also given that p% of employees were hired. Hence, we can write the number of employees after hiring p% employees as
$n+p\%\text{ of }n$
We can write the above form as
 $\Rightarrow n+p\%\times n$
We know that a percentage is converted to a number by dividing it by 100.
$\begin{align}
  & \Rightarrow n+\left( \dfrac{p}{100}\times n \right) \\
 & \Rightarrow n+\dfrac{np}{100} \\
\end{align}$
We are also given that after a month q% employees left the office. This number of employees is calculated as
$\begin{align}
  & q\%\text{ of }\left( n+\dfrac{np}{100} \right) \\
 & \Rightarrow q\%\times \left( n+\dfrac{np}{100} \right) \\
 & \Rightarrow \dfrac{q}{100}\left( n+\dfrac{np}{100} \right) \\
\end{align}$
We are also given that the employees left at the end is n. Hence, we can form an equation by subtracting q% employees that left from the employees after hiring p%. We will equate this to the employees left at the end.
$\Rightarrow n+\dfrac{np}{100}-\dfrac{q}{100}\left( n+\dfrac{np}{100} \right)=n$
We have to solve the equation in such a way that we will get the value of $p-q$ .
Let us take the first term from LHS to the RHS.
$\Rightarrow \dfrac{np}{100}-\dfrac{q}{100}\left( n+\dfrac{np}{100} \right)=0$
Let us simplify the second term of LHS by applying distributive law.
$\begin{align}
  & \Rightarrow \dfrac{np}{100}-\dfrac{q}{100}\times n-\dfrac{q}{100}\times \dfrac{np}{100}=0 \\
 & \Rightarrow \dfrac{np}{100}-\dfrac{nq}{100}-\dfrac{qnp}{10000}=0 \\
\end{align}$
Let us take common n outside from the numerator of the LHS and 100 from its denominator.
$\Rightarrow \dfrac{n}{100}\left( p-q-\dfrac{pq}{100} \right)=0$
Let us take $\dfrac{n}{100}$ to the RHS.
$\Rightarrow p-q-\dfrac{pq}{100}=0$
Now, we have to take the third term of LHS to the RHS.
$\Rightarrow p-q=\dfrac{pq}{100}$

So, the correct answer is “Option b”.

Note: Students must know how to form equations according to the given conditions. They must read the question carefully as there can be a chance of making mistakes by adding the employees after hiring p% and the q% employees that are left. Students must know how to convert a percentage into a number and also how an equation is solved.