Introduction
Title:Interest rate and credit card debt
Interest is payment from a borrower which more than
the money we borrow. Interest is the charge for the
privilege of borrowing money, typically expressed as
annual percentage rate. Interest can also refer to the
amount of ownership a stockholder has in a
company, usually expressed as a percentage like
1.5%,or a decimal 0.015.
There are two main types of interest that can be
applied to loans: simple and compound. Simple
interest is a set rate on the principle originally lent to
the borrower that the borrower has to pay for the
ability to use the money. Compound interest is
interest on both the principle and the compounding
interest paid on that loan. The latter of the two types
of interest is the most common.
Interest rates have a huge efect on loans. In short,
loans with high interest rates have higher monthly
payments or take longer to pay of than loans with
low interest rates.
In mathematics, a geometric series, also known as a
geometric sequence, is a sequence of numbers
where each term after the frst is found by
multiplying the previous one by a fxed, non-zero
number called the common ratio.
Each credit card has a maximum limit. A maximum
limit of RM20000 means total money spent plus
can ensure that the credit card users will not spent
too much money by using credit card. Each credit
card user will choose a payment plan to repay the
debt every month.
Methodology i = Annual interest rate12
Annual interest rate =18% i = monthly interest rate
We use 18% divided 12 and get a answer 1.5% .1.5% is our monthly interest rate.
Interest due = Balance,bn ×1.5%
Balance,bn = Balance,bn-1 - Paid to principal
bn =arn-1 a =8000
Write equation for b1,b2,b3,fnd the frequency and relationship
amog them by compare their diference and make a formula for bn
For example: equation b1 is b1 =b0 – [ R-(b0)(i)]
Find the equation forb2 and b3 with equation b1.
make three table with three amont larger than 520 such as
600.800.1000 to fnd monthly balance until balance bn is
less than payment.The last month,n will be the month paid of the payment
Interest due = Balance,bn ×1.5%
Paid to principal = Payment,R - Interest due Balance,bn = Balance,bn-1 - Paid to principal
We can sub n=0 to fnd equation about b0
When s=1,sub s=1 into equation bn = b0(1+i)(1+i+ 100)n-1
bn = b0(1+i)(1+i+ 1001 )n-1
Then ,sub bn =16000
b0 =8000 i= 0.015
16000=8000(1+0.015)(1+0.015 +1001 )n-1
Solve the equation to fnd n
Result
monthly interest rate
i = Annual interest rate12 i = 18%
12 i = 1.5%
i=monthly interest rate =1.5%
Every mont ,we will be charge 1.5% of interest rate from our balance.
This interest rate make us paid more mony then
When payment =RM520
3 107.91 520 412.09 6781.9 4 101.73 520 418.27 6363.6 5 95.45 520 424.55 5939.1 6 89.09 520 430.91 5508.2 7
9 69.4 520 450.6 4176.28
41.75 478.25 2305.27
14
34.58
520 485.4
2 1819.85
15 27.30 520 492.7 1327.15
16
19.91 520 500.09 827.06
17
12.41
520 507.5
9 319.47
When payment =RM520,month balance paid of =17
At 17 th month,payment from RM 520 increase to RM
839.47.This make the payment faster paid end and save the
interest rate charge by the bank. At 17th month , balance is
fully paid of.
2 112.8 600 487.2 7032.8
3 105.49 600 494.51 6538.3
4 98.07 600 501.93 6036.4
5 90.55 600 509.45 5526.9
6 82.9 600 517.1 5009.8
7 75.15 600 524.85 4485.15
8 67.28 600 532.72 3952.43
9 59.29 600 540.71 3411.72
10 51.18 600 548.82 2862.9
11 42.94 600 557.06 2305.84
13 26.11 600 573.89 1166.54
14 17.50 600 582.5 584.04
When payment =RM600,month balance paid of =14
At 14 th month,payment from RM 600 increase to RM
1184.04This make the payment faster paid end and save the
interest rate charge by the bank. At 14th month , balance is
fully paid of.
2 112.05 650 537.95 6932.05 3 103.98 650 546.02 6386.03
9
When payment =RM650,month balance paid of =13
At 13 th month,payment from RM 650 increase to RM
1094.48 .This make the payment faster paid end and save
the interest rate charge by the bank.At 13th month , balance
0 _ _ _ 8000
1 120 800 680 7320
2 109.8 800 690.20 6629.80
3 99.45 800 700.60 5929.30
4 88.94 800 711.10 5218.20
5 78.27 800 721.70 4496.50
6 67.45 800 732.60 3763.90
7 56.46 800 743.50 3020.50
8 45.31 800 754.70 2265.80
9 33.99 800 766.01 1499.80
10 22.5 800 777.50 722.30
When payment =RM800,month balance paid of =10
At 10th month,payment from RM 800 increase to RM1522.30
This make the payment faster paid end and save the interest
rate charge by the bank. At 10th month , balance is fully paid
of.
When payment =RM1000
Month,n Interest due
Payment,R Paid to principal
(RM) (RM) (RM) (RM)
0 _ _ _ 8000
1 120 1000 880 7120
2 106.8 1000 893.2 6226.8
3 93.4 1000 906.6 5320.2
4 79.8 1000 920.2 4400
5 66 1000 934 3466
6 51.99 1000 948.01 2517.99
7 37.77 1000 51.99 2466
8 36.99 1000 963.01 1502.99
9 22.54 1000 977.46 525.53
When payment =RM1000,month balance paid of =9
At 9 th month,payment from RM 1000 increase to
RM1525.53.This make the payment faster paid end and save
the interest rate charge by the bank. At 9th month , balance
is fully paid of.
b1 =b0 – [ R-(b0)(i)] b1 = b0 – ( R – ib0 )
Substitute b1 into b2 , b2 = ( 1 + i ) [ ( 1 + i )b0 – R] – R = ( 1 + i )2b0 - ( 1 + i )R – R
b3 = b2 – ( R – ib2 ) = ( 1 + i )b2 – R
Substitute b2 into b3 , b3 = ( 1 + i ) [ ( 1 + i )2b0 - ( 1 + i )R –
R ] – R
= ( 1 + i )3b0 - ( 1 + i )2R - ( 1 +
i )R – R
After expressed b1 , b2 and b3 in term of R, i, n, and b0, I observe relationship between them to form a formula for bn. bn = b0( 1 + i )n - ( 1 + i )n-1R - ( 1 + i )n-2R - … - ( 1 + i )R – R
= b0( 1 + i )n– R [( 1 + i )n-1 + ( 1 + i )n-2 + … + ( 1 + i ) – 1]
= b0( 1 + i )n– R [1 + ( 1 + i ) + ( 1 + i )2 + ( 1 + i )3 + … +
( 1 + i )n-1]
Formula of sum: a=1 , r=1+i
Sn =a(r
n
−1)
r−1
=1[(1+i)−1]
(1+i)−1
=(1+ii)n−1
Formula for bn
bn = b0( 1 + i )n– R[(1+i)i¿¿n−1]¿ bn =balance in nth month
R =Payment
n = month
bn = b0( 1 + i )n– R[(1+i)i¿¿n−1]¿
When payment = RM 520
bn =8000 ( 1 + 0.015 )n– 520[(1+0.0150.015)¿¿n−1]¿
bn =0
8000 ( 1 + 0.015 )n– 520[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–520[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–520[(1+0.015)¿¿n−1]¿ =0
120(1+0.015)n–520¿ ¿ =- 520
(1+0.015)n [120-520] = -520 (1+0.015)n[(-400)] =-520 (1+0.015)n = 1.3
n log10 1.015 = log10 1.3
n= log 1.015log 1.3
n =17.62
The month when balance is paid of while payment RM520 is about 18 month
When payment = RM 600
bn =8000 ( 1 + 0.015 )n– 600[(1+0.0150.015)¿¿n−1]¿
bn =0
8000 ( 1 + 0.015 )n– 600[(1+0.015)¿¿n−1]
120(1+0.015)n–600[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–600[(1+0.015)¿¿n−1]¿ =0
120(1+0.015)n–600¿ ¿ =-600
(1+0.015)n [120-600] = -600 (1+0.015)n[(-480)] =-600 (1+0.015)n = 1.25
n log10 1.015 = log10 1.25
n= log 1.015log 1.25
n =14.99
The month when balance is paid of while payment RM520 is about 15 month
When payment = RM 650
bn =8000 ( 1 + 0.015 )n– 650[(1+0.0150.015)¿¿n−1]¿
bn =0
8000 ( 1 + 0.015 )n– 650[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–650[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–65 0[(1+0.015)¿¿n−1]¿ =0
120(1+0.015)n–650¿ ¿ =- 650
(1+0.015)n [120-650] = -650 (1+0.015)n[(-530)] =-650 (1+0.015)n = 1.23
n= log 1.015
n =13.90
The month when balance is paid of while payment RM520 is about 14 month
When payment = RM 800
bn =8000 ( 1 + 0.015 )n– 800[(1+0.0150.015)¿¿n−1]¿
bn =0
8000 ( 1 + 0.015 )n– 800[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–520[(1+0.015)¿¿n−1]
0.015 ¿ =0
120(1+0.015)n–800[(1+0.015)¿ ¿n−1]¿ =0
120(1+0.015)n–800¿ ¿ =- 800
(1+0.015)n [120-800] = -800 (1+0.015)n[(-680)] =-800 (1+0.015)n = 1.18
n log10 1.015 = log10 1.18
n= log 1.015log 1.18
n =11.11
The month when balance is paid of while payment RM520 is about 11 month
When payment = RM 1000
bn =8000 ( 1 + 0.015 )n– 1000[(1+0.0150.015)¿¿n−1]¿
8000 ( 1 + 0.015 )n– 1000[(1+0.015)¿¿n−1]
(n-1) log10 (207200)= log10 400203
(n-1) =19.933 n= 20.933
When 1% is charge,Current payment is doubled after 28 month.
When 2% is charge,Current payment is doubled after 20 month.
