Question

1. A friend has an IRA with an APR of 6.25%. She stared the IRA at age 25 and deposits $50 per month.
          a) How much will her IRA contain when she retires at age 65?
          b) How much money did she contribute to the account?
          c) Use the answers from parts a) and b) to determine the amount of interest earned.

2. What if your friend from question #1 doubled her deposit and saved $100 per month? Do you think her money would double? Let's find out! Redo question #4, all the parts, with a monthly deposit of $100 per month. Were you surprised by the results?

Answer 1

1. (a) A= [{PMT*(1+(APR/n))^nY}-1]/(APR/n)

PMT= $50

APR=6.25/100

=0.0625

n=12

Y=40(65-25)

A=[$50*{((1+(0.0625/12))^12*40)-1}]/(0.0625/12)

=[$50*{((1+0.0052083)^480)-1}]/(0.0625/12)

=[$50*{(1.0052083^480)-1}]/(0.0052083)

=[$50*{12.1037109954-1}]/(0.0052083)

=[$50*{11.1037109954}]/(0.0052083)

=555.185549771/0.0052083

=$106,596

(b) Total deposits = $50 × 12 × 40 = $24,000

(c) Amount of interest earned=Amount- Total Deposits

=$106,596-$24,000

=$82,596

2.

A= [{PMT*(1+(APR/n))^nY}-1]/(APR/n)

PMT= $100

APR=6.25/100

=0.0625

n=12

Y=40(65-25)

A=[$100*{((1+(0.0625/12))^12*40)-1}]/(0.0625/12)

=[$100*{((1+0.0052083)^480)-1}]/(0.0625/12)

=[$100*{(1.0052083^480)-1}]/(0.0052083)

=[$100*{12.1037109954-1}]/(0.0052083)

=[$100*{11.1037109954}]/(0.0052083)

=1110.37109954/0.0052083

=$213,192

Yes, the money doubed from $106,596 to $213,192

(b) Total deposits = $100 × 12 × 40 = $48,000

(c) Amount of interest earned=Amount- Total Deposits

=$213,192-$48,000

=$165,192