Question

Algebraically manipulating the formula for finding the simple interest, / Pyt, find the time period in years that $16,000 was deposited in a savings account which eamed$4,000 in interest at a simple interest rate of 6%42 months60 months70 months50 monthsNone of these choices are correct.

Answer 1

Given the simple interest formula

`\begin{gathered} I=\text{ Prt} \\ \text{Where I=Interest; P=Principal; r=Rate and t=Time} \end{gathered}`

From the question, the money deposited is the Principal P=$16,000;

The earned interest I=$4000;

The simple interest rate, r= 6%;

and the time period in years, t= unknown

To find the time unknown time, we substitute for I, P, and r in the formula given. This will give

`\begin{gathered} I=\text{ Prt} \\ 4000=16000*(6)/(100)* t \\ 4000=(16000*6* t)/(100) \\ 4000=(96000t)/(100) \\ 4000=960t \\ t=(4000)/(960) \\ t=(50)/(12)\text{years} \end{gathered}`

Since the time given in the options are in months, we will convert time gotten to months

`\begin{gathered} \text{Convert }(50)/(12)\text{years to months} \\ \text{Note 1 year=12months} \\ \text{Therefore, }(50)/(12)\text{years}=(50)/(12)*12months \\ =50\text{months} \end{gathered}`

Therefore, the time period in months is 50months