Mathematics: Mr Willkinson invests £3000 at a compound interest of 2.2% per anum he wants his investment to earn more than £800 interestwork out in the least time in years it will take

24xy^2 = 6xy(4y)42xy = 6xy(7)answerGCF of 24xy^2 and 42xy is 6xy...

Mr Willkinson invests £3000 at a compound interest

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kaylee0424

09.02.2022

if his investment were to earn an interest £800, then his accumulated amount will be 300 + 800 or namely £3800.

let's check at what point he'll be getting an accumulated amount of £3800, and then we know that compounding it further will simply result in more bucks, thus we know that'd be the least it'd take, keeping it further will simply give more.

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\£3800\\ P=\textit{original amount deposited}\dotfill &\£3000\\ r=rate\to 2.2\%\to \frac{2.2}{100}\dotfill &0.022\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\end{cases}$

$3800=3000\left(1+\frac{0.022}{1}\right)^{1\cdot t}\implies \cfrac{3800}{3000}=(1.022)^t\implies \cfrac{19}{15}=1.022^t \\\\\\ \log\left( \cfrac{19}{15} \right)=\log(1.022^t)\implies \log\left( \cfrac{19}{15} \right)=t\log(1.022) \\\\\\ \cfrac{\log\left( \frac{19}{15} \right)}{\log(1.022)}=t\implies \stackrel{\textit{10 years, 313 days, 21hrs and 36 mins }}{10.86\approx t}$

so depends, I guess if he just waits one more minute that may do it =).

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Mr Willkinson invests £3000 at a compound interest of 2.2% per anum he wants his investment to earn more than £800 interest
work out in the least time in years it will take

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