Chapter 7, Problem 67SGR

To determine

To compare:The yearly interest rates of First Bank and Main Street Bank.

Answer to Problem 67SGR

Main Street Bank’s interest rate is higher than that of First Bank.

Explanation of Solution

Given information:First Bank’s savings accounts have an interest rate of 0.15% compounded monthly. Main Street Bank has a 1.5% yearly rate.

Formula used:

Compound interest formula for compounded monthly is given by,

  $I=P[{(1+\frac{\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}{100})}^{12t}-1]$

where, $I=$ Interest

  $P=$ Principal

  $R=$ Rate % per annum

  $t=$ time

Compound interest formula for compounded yearly is given by,

  $I=P[{(1+\frac{R}{100})}^t-1]$

where, $I=$ Interest

  $P=$ Principal

  $R=$ Rate % per annum

  $t=$ time

Calculation:

First Bank’s interest for 1 year is given by,

  $I=P[{(1+\frac{\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}{100})}^{12t}-1]$

  $\Rightarrow I=P[{(1+\frac{\raisebox{1ex}{$.15$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}{100})}^{12\cdot 1}-1]$

  $\Rightarrow I=P[{(1+\frac{.15}{1200})}^{12}-1]$

  $\begin{array}{l}$ \Rightarrow I=P[{(\frac{1200.15}{1200})}^{12}-1]$\Rightarrow I=P[1.00150103-1]\\ \Rightarrow I=0.00150103P\end{array}$

Main Street Bank’s interest for 1 year is given by,

  $I=P[{(1+\frac{R}{100})}^t-1]$

  $\Rightarrow I=P[{(1+\frac{1.5}{100})}^1-1]$

  $\begin{array}{l}$ \Rightarrow I=P[(\frac{101.5}{100})-1]$\Rightarrow I=P[1.015-1]\\ \Rightarrow I=0.015P\end{array}$

Clearly, Main Street Bank’s interest rate is higher than that of First Bank.