Question

Calculate the balance in Mr. Warren's account at the end of the year, if he deposits $\$100$each at the ends of months $1$ and $6$, and $\$200$ each at the ends of months $7$ and $9$.

His bank pays $8\%$ per year, compounded quarterly, and simple interest on the inter period deposits.

Answer 1

Calculating the balance at the end of the year:

Step-1: Given data:

Given,

Rate of the interest $=8\%$

Ends of months $1$ and $6$ he deposit $=\$100$

Ends of months $7$ and $9$ he deposit $=\$200$

Compounded quarterly and simple interest on the inter period deposit.

Interest calculated in $3,6,9,12$ months

It means through out the period (three months) the balance is same then use compound interest formula for that particular amount or use simple interest formula for the inter period deposits.

Step-2: Calculating the balance amount after first quarterly:

End of the 1^st month he deposit $\$100$.

Interest calculated in $3$month is simple interest for $\$100$. Because the particular amount are in account is two months only.so we cant able to use compound interest formula.

Then,

Simple interest $I=pnr$ $p$ is principal amount; $n$ is number of years, $r$ rate of interest in decimal.

$p=100;n=2\text{months}=\frac{2}{12}\text{years}r=0.08$

$$\begin{gathered} =\frac{100\times 2\times 8}{100\times 12} \\ =1.3333 \end{gathered}$$

Balance after first quarterly period is $100+1.3333=\$101.3333$

Step-3: Calculating the balance amount after second quarterly:

Balance amount after first quarterly $=\$101.3333$. [Starting of $4^{t{h}^{}}$ month]

This amount remains same for after three months So we use compound interest formula

Compound interest for quarterly $A=p{\left(1+\frac{r}{4}\right)}^{4\times n}$;$A$ is Amount, $n$ is number of years $r$ rate of interest in decimal.

Here,

$p=101.3333;n=3\text{months}=\frac{3}{12}\text{years}r=0.08$

$$\begin{gathered} \Rightarrow A=101.3333{\left(1+\frac{0.08}{4}\right)}^{4\times \frac{3}{12}} \\ =101.3333\times {\left(1+0.02\right)}^1 \\ =101.3333\times \left(1.02\right) \\ =103.36 \end{gathered}$$

Balance after first second period is $=\$103.36$

Ends of months $6$ he deposit $=\$100$

Step-4: Calculating the balance amount after third quarterly:

Starting balance for $7$ month is $=103.36+100=\$203.36$

Ends of months $7$ he deposit $=\$200$

The balance amount after third quarterly $=$Amount compound interest for $\$203.36$$+$ Amount simple interest for $\$200$

Amount compound interest for $\$203.7$

$p=203.7;n=3\text{months}=\frac{3}{12}\text{years}r=0.08$

$$\begin{gathered} \Rightarrow A=203.36{\left(1+\frac{0.08}{4}\right)}^{4\times \frac{3}{12}} \\ =203.36\times {\left(1+0.02\right)}^1 \\ =203.36\times \left(1.02\right) \\ =207.4272 \end{gathered}$$

Amount simple interest for $\$200$

$p=200;n=2\text{months}=\frac{2}{12}\text{years}r=0.08$

$$\begin{gathered} I=\frac{200\times 2\times 8}{100\times 12} \\ =2.6667 \end{gathered}$$

Amount$=I+p$

$$\begin{gathered} =200+2.6667 \\ =202.6667 \end{gathered}$$

Balance after third quarterly period is $=207.4272+202.6667=\$410.0939$

Ends of months $9$ he deposit $=\$200$

Step-5: Calculating the balance at the end of year:

Month	starting balance	Deposit	Interest	Ending balance
$1$	$\$0$	$\$100$	$\$0$	$\$100$
$3$	$\$100$	$\$0$	$\$1.3333$	$\$101.3333$
$6$	$\$101.3333$	$\$100$	$\$2.0267$	$\$203.36$
$7$	$\$203.36$	$\$200$	$\$0$	$\$403.36$
$9$	$\$403.36$	$\$200$	$\$6.7339$	$\$610.0939$
$12$	$\$610.0939$	$\$0$	$\$12.2019$	$\$622.2958$

Starting balance for $9$ month is $=410.0939+200=\$610.0939$

Here,

$p=610.4407;n=3\text{months}=\frac{3}{12}\text{years}r=0.08$

$$\begin{gathered} \Rightarrow A=610.0939{\left(1+\frac{0.08}{4}\right)}^{4\times \frac{3}{12}} \\ =610.0939\times {\left(1+0.02\right)}^1 \\ =610.0939\times \left(1.02\right) \\ =622.2958 \end{gathered}$$

Rounding off to two decimal place $=\$622.30$

Hence, the balance at the end of the year is $\$622.30$.