Janet saves $3,000 a year at an interest rate of 4.2 percent. What will her savings be worth at the end of 35 years?

A. $229,317.82               

B. $230,702.57               

C. $230,040.06

D. $234,868.92               
E.   $236,063.66

Answer:

The formula for calculating the Future value of savings at the end of n years is

= P *[ [ ( 1 + r ) ⁿ- 1 ] / r ]

Where P = Periodic Deposit i.e., Fixed amount of Annual deposit

r = rate of interest   ; n = no. of years

A per the information given in the question we have

P = $ 3,000    ; r = 4.2 % = 0.042   ;   n = 35

Applying the above values in the formula we have:

= 3000 * [ [ ( 1 + 0.042 ) ³⁵   - 1 ] / 0.042 ]                    

= 3000 * [ [ ( 1.042 ) ³⁵   - 1 ] / 0.042 ]

= 3000 * [ [ 4.2205610 – 1 ] / 0.042 ]

= 3000 * [ 3.2205610 / 0.042 ]

= 3000 * 76.6800196

= 230040.0588

= $ 230040.06 ( when rounded off to two decimal places )

Thus the Solution is Option C. $ 230,040.06

Thus if Janet saves $3,000 a year at an interest rate of 4.2 percent her savings be worth $ 230,040.06 at the end of 35 years.

Note: The value of ( 1.042 ) ³⁵   is calculated using the Excel formula =POWER(Number,Power)

=POWER(1.042,35)= 4.2205610