In: Finance
Jake owes $3,400 on his credit card. He is not charging any additional purchases because he wants to get this debt paid in full. The card has an APR of 13.2 percent. How much longer will it take him to pay off this balance if he makes monthly payments of $50 rather than $60?
Solution: | ||||
Answer is 36.74 months | ||||
longer will it take him to pay off this balance if he makes monthly payments of $50 rather than $60 | ||||
Working Notes: | ||||
credit card Loan amount $3,400 is the present value of all the payment done. | ||||
present value of annuity = Px[ 1-(1 /(1 + i)^n)]/ i | ||||
P=Monthly payment = $60 | ||||
i= interest rate per period = 13.20%/12 = 1.1% = 0.011 | ||||
n= no. Of period = n months = ?? | ||||
PV of annuity= credit card balance $3,400 | ||||
present value of annuity = Px[ 1-(1 /(1 + i)^n)]/ i | ||||
3400= 60 x (1-(1/(1+0.011)^n)/0.011 | ||||
(3400/60)x0.011 -1 = -(1/(1.011)^n) | ||||
-0.3766666667 = - (1/(1.011)^n) | ||||
0.3766666667 = (1/(1.011)^n) | ||||
(1.011)^n = (1/0.3766666667) | ||||
(1.011)^n =2.654867257 | ||||
taking log on both side | ||||
log(1.011)^n = Log(2.654867257) | ||||
n x Log (1.011) = Log(2.654867257) | using relation loga^b = b x Log a | |||
n = Log(2.654867257)/Log (1.011) | ||||
Notes: | Log values can be obtained from online calculators or excel | |||
For excel use this | =LOG(value) | |||
putting log values | ||||
n= 0.424042811/ 0.004751156 | Log(2.654867257) = 0.424042811 | |||
n= 89.25045774 months | Log (1.011) = 0.004751156 | |||
n= 89.25 months | ||||
P=Monthly payment = $50 | ||||
i= interest rate per period = 13.20%/12 = 1.1% = 0.011 | ||||
n= no. Of period = n months = ?? | ||||
PV of annuity= credit card balance $3,400 | ||||
present value of annuity = Px[ 1-(1 /(1 + i)^n)]/ i | ||||
3400= 50 x (1-(1/(1+0.011)^n)/0.011 | ||||
(3400/50)x0.011 -1 = -(1/(1.011)^n) | ||||
-0.252 = - (1/(1.011)^n) | ||||
0.252 = (1/(1.011)^n) | -0.2520000 | |||
(1.011)^n = (1/0.252) | -3.968253968 | |||
(1.011)^n =3.968253968 | ||||
taking log on both side | ||||
log(1.011)^n = Log(3.968253968) | ||||
n x Log (1.011) = Log(3.968253968) | using relation loga^b = b x Log a | |||
n = Log(3.968253968)/Log (1.011) | ||||
Notes: | Log values can be obtained from online calculators or excel | |||
For excel use this | =LOG(value) | |||
putting log values | ||||
n= 0.598599459/ 0.004751156 | Log(3.968253968) = 0.598599459 | |||
n= 125.9902767 months | Log (1.011) = 0.004751156 | |||
n= 125.99 months | ||||
longer will it take him to pay off this balance if he makes monthly payments of $50 rather than $60 | ||||
=Months at payment of $50 - Months at payment of $60 | ||||
=125.99 - 89.25 | ||||
=36.74 months | ||||
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