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Amit was given an increment of \[20\% \] on his salary. If his new salary is Rs. \[30,600\], what was his salary before the increment?

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Last updated date: 14th Jul 2024
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Answer
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Hint: Here in this question, we have to find the salary of Amit before increment. To find this, take an initial salary as x and multiply this with the fraction of percentage of increment we get the increment in the salary and further equating the sum of initial salary and increment salary with the new salary of Amit on simplification we get the required solution.

Complete step-by-step solution:
Consider the data in given equation
Amit gets a salary increment of \[20\% \] on his salary.
Amit’s new salary is Rs. \[30,600\].
We have to find the salary of Amit before the increment.
Let us take the salary before increment = \[x\]
Now find the amount of increment in the salary i.e., \[20\% \] of \[x\]
\[ \Rightarrow \,\,\,20\% \times x\]
Write the percentage term i.e., \[20\% \] into the fraction.
\[ \Rightarrow \,\,\,\dfrac{{20}}{{100}} \times x\]
Divide both numerator and denominator by 20, then
\[ \Rightarrow \,\,\,\dfrac{1}{5} \times x\]
\[ \Rightarrow \,\,\,\dfrac{x}{5}\]
Given that, Amit’s new salary is Rs. \[30,600\] which equals the sum of his initial salary and amount of increment in the salary i.e.,
\[ \Rightarrow \,\,x + \dfrac{x}{5} = 30,600\]
Take 5 as LCM in LHS, then we have
\[ \Rightarrow \,\,\dfrac{{5x + x}}{5} = 30,600\]
\[ \Rightarrow \,\,\dfrac{{6x}}{5} = 30,600\]
Multiply both side by 5, then
\[ \Rightarrow \,\,6x = 30,600 \times 5\]
\[ \Rightarrow \,\,6x = 153000\]
Divide both side by 6, then
\[ \Rightarrow \,\,x = \dfrac{{153000}}{6}\]
On simplification, we get
\[\therefore \,\,x = 25,500\]
Therefore, the salary of Amit before increment is 25,500.

Note: Here the question is related to the percentage. By using the specific methods and rules we can convert the number. As we know that the percentage term is written as \[x\% = \dfrac{x}{{100}}\], here, marked x as the total value. Using the simple arithmetic operation i.e., multiplication and division to get the required solution.