Chapter 7, Problem 35PT

To determine

The better rate for savings account

Answer to Problem 35PT

2% compounded yearly offers a better rate for savings account.

Explanation of Solution

Given information:One offers 2% compounded yearly interest rate and another offers 0.2% compounded monthly interest 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:

Interest for 1 year compounded monthly 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}{$0.2$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}{100})}^{12\cdot 1}-1]$

  $\Rightarrow I=P[{(1+\frac{0.2}{1200})}^{12}-1]$

  $\begin{array}{l}$ \Rightarrow I=P[{(\frac{1200.2}{1200})}^{12}-1]$\Rightarrow I=P[1.00200187-1]\\ \Rightarrow I=0.00200187P\end{array}$

Interest for 1 year compounded yearly is given by,

  $I=P[{(1+\frac{R}{100})}^t-1]$

  $\Rightarrow I=P[{(1+\frac{2}{100})}^1-1]$

  $\begin{array}{l}\Rightarrow I=P[(\frac{102}{100})-1]\\ \Rightarrow I=P[1.02-1]\\ \Rightarrow I=0.02P\end{array}$

Clearly, 2% compounded yearly offers a better rate for savings account.