Sheila bought a new computer for $2000 and has agreed to finance it at 12% interest with a $100 payments each month. When she makes her first payment next month, how much will she pay for interest alone?
3 answers
(2,000 * 0.12) / 12 = ?
I will assume the interest is compounded monthly, so the monthly rate is 1%
1% of 2000 = $20
so of the first payment of $100, $20 will go to interest and $80 will be actual repayment.
of course the next month, her interest will be less:
balance = 2000-80 = 1920
.01(1920) = $19.20
etc
An interesting question:
How many full payments of $100 will there be, and what will be the partial payment made one month after the last full payment of $100 ?
1% of 2000 = $20
so of the first payment of $100, $20 will go to interest and $80 will be actual repayment.
of course the next month, her interest will be less:
balance = 2000-80 = 1920
.01(1920) = $19.20
etc
An interesting question:
How many full payments of $100 will there be, and what will be the partial payment made one month after the last full payment of $100 ?
rate per month = .01 agreed = r
Luckily they only ask for the first month so it is 0.01 * 20 as Reiny and Ms. Sue said.
Later months get more complicated and it is like paying off a mortgage with the term unknown but the rate and payment known.
Payment = principal value [ r / {1-(1+r)^-n } ]
100 = 2000 [ .01 / {1 - (1.01)^-n } ]
5 = 1/ {1 - (1.01)^-n }
.2 = 1 - (1.01)^-n
.8 = 1.01^-n
log .8 = -n log 1.01
n = 22.4 months :)
so over the almost two years pay 2240 total :)
Luckily they only ask for the first month so it is 0.01 * 20 as Reiny and Ms. Sue said.
Later months get more complicated and it is like paying off a mortgage with the term unknown but the rate and payment known.
Payment = principal value [ r / {1-(1+r)^-n } ]
100 = 2000 [ .01 / {1 - (1.01)^-n } ]
5 = 1/ {1 - (1.01)^-n }
.2 = 1 - (1.01)^-n
.8 = 1.01^-n
log .8 = -n log 1.01
n = 22.4 months :)
so over the almost two years pay 2240 total :)
1 answer