Question

At the end of each year, Richard receives a

15% salary increase.
After two years, Richard's salary is £34,385.
What was his starting salary?
Give your answer in pounds (£).
Starting salary
Salary after one year
Salary after two years
4
£
£ 34,385
)+15%
+15%

Answer 1

let's approach this, this way

we have a principal amount P, with a annual rate of 15%, that after 2 years it became £34,385, what's P?

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \pounds 34385\\ P=\textit{original amount deposited}\\ r=rate\to 15\%\to (15)/(100)\dotfill &0.15\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}`

`34385 = P\left(1+(0.15)/(1)\right)^(1\cdot 2) \implies 34385=P(1.15)^2 \\\\\\ \cfrac{34385}{(1.15)^2}=P\implies 26000=P`

after two years? well, we already know, is above, after one year? well is just 26000 + 15% of itself, so

`\begin{array}c \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{15\% of 26000}}{\left( \cfrac{15}{100} \right)26000}\implies 3900~\hfill \underset{ \textit{after one year} }{\stackrel{ 26000~~ + ~~3900 }{29900}}`