Lesson 9 problem set <br> 5.1

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

4 0

4 years ago

Need help please help

azamat

Answer:

B

Hope you get it right! :)

Step-by-step explanation:

7 0

3 years ago

ian invested an amount of money at 3% per annum compound interest. At the end of 2 years the value of the investment was £2652.2

Scorpion4ik [409]

Answer:

the amount of money Ian invested is P = £2,500

Step-by-step explanation:

The standard formula for compound interest is given as;

$A = P(1+r/n)^{nt} \\P = \frac{A}{(1+r/n)^{nt}} ...........1\\$

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case, Given that;

A = £2652.25

t = 2 years

n = 1 (semiannually)

r = 3% = 0.03

substituting the given values into equation 1;

$P = \frac{A}{(1+r/n)^{nt}} ...........1\\P = \frac{2652.25}{(1+0.03)^{2}} \\P = \frac{2652.25}{(1.03)^{2}} \\$

P = £2,500

the amount of money Ian invested is P = £2,500

8 0

3 years ago

Lesson 9 problem set 5.1