Question

The ratio of As salary to Bs was 4:5. As salary is increased by \[10\% \] and Bs by \[20\% \], what is the ratio of their salaries now?
A.\[{\text{11:15}}\]
B.\[{\text{11:55}}\]
C.\[{\text{11:66}}\]
D.\[{\text{11:33}}\]

Answer 1

Hint: Here, increment of salary\[ = x \times \dfrac{n}{{100}}\], where the primary amount of salary = x and the percentage increase of the salary = n.
So, increased salary \[ = x + \left( {x \times \dfrac{n}{{100}}} \right)\].

Complete step-by-step answer:
Let, the salary of ‘B’ is x.
Here, Salary of As: Bs = 4:5
So, \[\dfrac{{Salary{\text{ }}of{\text{ }}A}}{x} = \dfrac{4}{5}\]
\[Salary{\text{ }}of{\text{ }}A = \dfrac{{4x}}{5}\]
Now, if salary of A is increased by \[10\% \],
The new salary of A \[ = \dfrac{{4x}}{5} + \left( {\dfrac{{4x}}{5} \times 10\% } \right)\]
  \[ = \dfrac{{4x}}{5}\left( {1 + \dfrac{10}{{100}}} \right)\]
\[
   = \dfrac{{4x}}{5} \times \dfrac{{11}}{{10}} \\
   = \dfrac{{44x}}{{50}} \\
   = \dfrac{{22x}}{{25}} \\
  \]
Now, the salary of B \[ = x + \left( {x \times 20\% } \right)\]
  \[ = x + \left( {x \times \dfrac{{20}}{{100}}} \right)\]
\[ = x\left( {1 + \dfrac{1}{5}} \right) = \dfrac{{6x}}{5}\]
Hence, the ratio of their new salaries are
\[\dfrac{A}{B} = \dfrac{{\dfrac{{22x}}{{25}}}}{{\dfrac{{6x}}{5}}} = \dfrac{{22x}}{{25}} \times \dfrac{5}{{6x}} = \dfrac{{11}}{5} \times \dfrac{1}{3} = \dfrac{{11}}{{15}}\]
Hence, the ratio is \[11:15\].

Note: We have to notice whether there is increase or decrease in salary.
For increase, the increment will be added with the primary salary and for decrease, the decrement will be subtracted from the primary value.