Simple Interest Calculation
1) Maria borrowed $12,500 from the bank for 18 months at a simple interest rate of 7.5%. Find the total amount she repaid.
2) Edgar borrowed $4750 on January 5th at 6% simple interest for 240 days. He made a partial payment of $650 on March 7th and $1200 on May 22nd. (3 points each)
a. Find the balance of the loan on March 7th.
b. Find the balance on May 22nd
c. Find the date of maturity and the amount due on the date of maturity?
3) Andrew borrowed $18,00 from a bank. The bank gives him a 3.5% discounted loan for a period of 2 years.
a. How much interest does he pay for the use of the money?
b. How much money will he receive from the bank?
c. What is the actual interest rate paid?
4) Octavia invests $26,400 in a savings account that earns 2.15% interest compounded monthly. Find the balance after 7 years.
5) Which is the better choice: $10,000 deposited for a year at a rate of 2.5% compounded semi annually or $10,000 deposited for a year at a rate of 2.485% compounded weekly?
6) How much money should the Owens invest now at a rate of 2.25% compounded monthly in order to have $45,000 in 10 years?
7) Alex spent $12,000 on furniture for his apartment. He paid 15% down and financed the balance with a 36-month fixed installment loan with an APR of 7.5%.
a. Determine the total finance charge
b. Find the amount of the monthly payment for the loan.
8) Brianna is repaying her student loan. If she has 120 monthly payments of $166.53 for her $15,000 loan:
a. How much interest did the she pay for the loan?
b. Determine the APR of the loan to the nearest one-half of a percent.
9) Hayden borrowed $18,500 for a motorcycle loan. The bank offered him a 60 month loan with monthly payments of $375.11 each.
a. Find the APR of the loan.
b. Instead of making his 36th payment, Hayden decides to pay the remaining balance on the loan. How much interest will Hayden save?
c. What is the payoff amount on the loan?
10) Sylvia's credit card company determines her minimum monthly payment by adding all new interest to 4.35% of the outstanding principal. The credit card company charges an interest rate of 1.75% per month. On April 10th, Sylvia uses her credit card to pay for car repairs totalling $1854, she makes no other purchases during the month.
Calculation of Loan Repayment with Interest
a. Assuming Sylvia had no previous balance, determine her minimum payment due on May1st, her billing date.
b. Instead of making her minimum payment, Sylvia pays $350 on May 1st. If she makes no other charges, find the balance due on June 1st?
11) On the October 3 billing date, Victor had a balance due of $854.72 on his credit card. The transactions during the following month were: October 10 Charge: car insurance $145.22 October 15 Payment $200 October 25 Charge: gasoline $25.26 The interest rate on the card is 1.8% per month. Using the previous balance method, find the finance charge on November.
12) On the September 1 billing date, Martin had a balance due of $934.10 on his credit card. The transactions during the following month were: September 5 Payment $175.00 September 13 Charge: gas $49.14 September 22 Charge: dinner $80.66 September 29 Charge: electric bill $112.46 The interest rate on the card is 1.3% per month. Using the average daily balance method, find the finance charge on October 1 (September has 30 days).
13) The Yee's gross monthly income is $7245. They have 15 remaining payments of $345 on a car, 7 payments of $125 a television, and 52 payments of $245 on a student loan. What is the maximum monthly payment that they can get approved for on a home loan if the bank will approve a payment that is 28% of their adjustmed monthly income?
14) Juliana is buying a condominium that sells for $295,000. The bank is requiring a minimum down payment of 20%. The bank has offered Juliana 2 choices for the mortgage, a 15 year loan at 5% or a 30 year loan at 5.5%
a. Find the monthly payment for the 15 year loan
15. Gabriel is buying a house for $305,000. He puts 20% down then finances the balance for 30 years at 4.5% interest. The monthly homeowners association dues are $125 per month. Home insurance totals $2000 per year, and annual taxes are $3149. Assuming home owner association fees, taxes and insurance are added monthly to the total house payment, find the total monthly amount that Harrison will pay for the house.
16) You want to finance $26,450 for a car. Your bank offers you a 4 year loan at 2.5% interest.
a. Find your monthly car payment
b. Find the total amount of interest you will pay for the loan.
b. Find the monthly payment for the 30 year loan
Simple Interest Calculation
1.Total amount = Principal + Interest
Principal = $ 12,500
Rate of interest = 7.5% p.a.
Time period = 18 months or 1.5 years
Simple interest = (P*R*T/100) = (12500*7.5*1.5/100) = $1,406.25
Total amount Mary repaid = 12500 + 1406.25 = $ 13,906.25
2.Principal = $4,750
Interest rate = 6% for 240 days
- Partial payment was made on March 7th to the extent of $ 650
Hence, balance loan amount = 4750 – 650 = $ 4,150
- Another partial payment to the tune of $ 1,200 was done on May 22
Hence,, balance loan amount = 4150 – 1200 = $ 2,950
- Maturity date is 240 days from January 5thwhich comes out as September 2nd.
Amount due on maturity = 2950 + (4750*6/100) = $3,235
3.
- Interest paid for the money = (3.5/100)*1800 = $63
- Money received from the bank = 1800 – 63 = $ 1,737
- Actual interest rate paid = (63/1737)*100 = 3.63%
4.Principal = $26,400
Rate of interest = 2.15% p.a. which is compounded monthly
Time period = 7 years or (12*7) = 84 months
The relevant formula is shown below.
A = P (1+ (R/n))nt
Where n highlights the number of compounding periods in an year and t is the time period in years
Amount balance after 7 years = 26400[1+ (2.15/1200)]84 = $30,683.63
5.Option 1: 2.5% compounded semi-annually
Principal = $ 10,000
One year = 2 period of 6 months
Amount due after one year = 10000*[1+ (2.5/200)]2 = $10,251.56
Option 2: 2.485% compounded weekly
Principal = $ 10,000
One year = 365/7 = 52 weeks
Amount due after one year = 10000*[1+ (2.485/5200)]52 = $10,251.55
Hence, from the above computation, it is apparent that the superior or better choice is option 1 or 2.5% interest compounded semi-annually.
6.Amount due = $45,000
Time period = 10 years or 120 months
Rate of interest = 2.25% p.a. compounded monthly
The relevant formula is shown below.
A = P (1+ (R/n))nt
Where n highlights the number of compounding periods in an year and t is the time period in years
Let the money to be invested now be X
Hence, 45000 = X[1+ (2.25/1200)]120
Solving the above, X = $35,940.8 or approximately $ 35,941
7.Using the APR table, the finance charge amounts to $11.98 per $ 100 borrowed
Total amount borrowed = (85/100)*12000 = $ 10,200
Hence, total finance charge for the above borrowing = 11.98*102 = $ 1,221.96
Total amount to be repaid = 10200 + 1221.96 = $11,421.96
Total equal monthly instalments = 36
Hence, amount of monthly payment for the loan = 11421.96/36 = $317.28
8.Total loan amount = $ 12,000
Total amount paid to discharge the above loan = 166.53*120 = $19,983.6
Hence, interest paid on the loan = 19983.6 – 12000 = $ 7.983.6
The APR computation is highlighted below.
APR = [(7983.6/12000)]/10 = 6.65%
Calculation of Loan Repayment with Interest
9.Total finance charge on the loan = (375.11*60) – 18500 = $ 4,006.6
Hence, APR = [(4006.6/18500)]/5 = 4.33%
The net present value of all payments should be equal to the loan amount i.e. $18,500. The APR would be used as the discounting factor. The requisite computation is shown below.
Total interest saved = (375.11*60) – (375.11*35) – 7043.79 = $ 2,333.96
The payoff amount for the loan = 375.11*35 + 7043.79 = $ 20,172.64
10.As per the policy, the minimum monthly payment is obtained by adding 4.35% of the outstanding principal and any new interest at the rate of 1.75% per month.
Outstanding principal as on May 1 = $ 1,854
New interest = (1.75/100)*1854*(20/30) = $ 21.63
Hence, minimum payment for the May 1 billing date = (4.35/100)*1854 + 21.63 = $ 102.28
Payment made on May 1 = $ 350
Hence, outstanding principal = 1854 – 350 + 21.63 = $ 1525.63
Interest levied for the month of May on the above unpaid balance = (1.75/100)*1525.63 = $26.70
Total balance due on June 1 = 1525.63 +26.70 = $1,552.33
11.As per the previous balance method, interest is levied on the amount outstanding at the end of the previous month irrespective of the transactions that have taken place in the current month.
Outstanding balance as on October 3 = $854.72
Applicable interest rate = 1.8% per month
Hence, finance charge on November 3 = (1.8/100)*854.72 = $ 15.38
12.The balances for the month of September based on the given information are computed below.
September 1 to September 4 = $ 934.10
September 5 to September 12 = 934.10 -175 = $ 759.1
September 13 to September 21 = 759.1 + 49.14 = $ 808.24
September 22 to September 28 = 808.24 + 80.66 = $ 888.90
September 29 to September 30 = 888.90 + 112.46 = $ 1001.36
Hence, average daily balance for month of September = [(934.1*4) +(759.1*8) + (808.24*9) + (888.90*7) + (1001.36*2)]/30 = $ 843.61
Thus, finance charge on October 1 = (1.3/100)*843.61 = $ 10.97
13.Consider the total gross income earned by Yee in 52 months = 7245*52 = $ 376,740
Total repayments for the car loan = 345*15 = $ 5,175
Total repayments for the television = 125*7 = $ 875
Total repayments for the student loan = 245*52 = $ 12740
Hence, adjusted income over 52 months = 376740 – 5175 – 875 – 12740 = $ 357,950
Therefore, monthly adjusted income = (357950/52) = $ 6,883.65
Maximum monthly payment for the home loan = 0.28*6883.65 = $ 1,927.4
14.Price of condominium = $ 295,000
Down payment required = 0.2*295,000 = $ 59,000
Hence, principal amount for the mortgage = 295000 – 59000 = $ 236,000
Principal amount = $ 236,000
Time period = 15 years or 180 months
Rate of interest = 5% per annum or (5/12 = 0.4167%) per month
The relevant formula for equal monthly instalment determination is given below.
EMI = [P x R x (1+R)N]/[(1+R)N -1]
Thus, EMI = 236000*0.004167*(1.004167)180/(1.004167180-1) = $ 1866. 27
Principal amount = $ 236,000
Time period = 30 years or 360 months
Rate of interest = 5.5% per annum or (5.5/12 = 0.4583%) per month
The relevant formula for equal monthly instalment determination is given below.
EMI = [P x R x (1+R)N]/[(1+R)N -1]
Thus, EMI = 236000*0.004583*(1.004583)360/(1.004583360-1) = $ 1,339.98
15.Price of house = $ 305,000
Mortgage principal = (0.8*305000) = $ 244,000
Time period for loan repayment = 30 years or 360 months
Rate of interest = 4.5% p.a. or (4.5/12 = 0.375%) per month
The relevant formula for equal monthly instalment determination is given below.
EMI = [P x R x (1+R)N]/[(1+R)N -1]
Thus, EMI = 244000*0.00375*(1.00375)360/(1.00375360-1) = $ 1,236.31
However, besides the EMI for the mortgage, there are certain additional payments highlighted below.
Monthly association due = $ 125
Monthly home insurance expense = (2000/12) = $ 166.67
Monthly tax payment = (3149/12) = $ 262.42
Hence, total monthly amount that Harrison will pay for the house = 1,236.31 + 125 + 166.67 + 262.42 = $ 1,790.4
16.Total principal for the car loan = $ 26,450
Loan duration = 4 years or 48 months
Rate of interest = 2.5% p.a. or (2.5/12 = 0.2083%) per month
The relevant formula for equal monthly instalment determination is given below.
EMI = [P x R x (1+R)N]/[(1+R)N -1]
Thus, EMI = 26450*0.002083*1.00208348/(1.00208348 – 1) = $ 579.63
Total amount paid to discharge the loan over 4 years = 579.63 *48 = $27,822.06
However, the principal to be repaid was $ 26,450. Hence, any incremental payment would be on account of interest payment.
Thus, total interest paid = $27,822.06 - $ 26,450 = $ 1,372.06
To export a reference to this article please select a referencing stye below:
My Assignment Help. (2020). Examples Of Loan And Interest Calculations. Retrieved from https://myassignmenthelp.com/free-samples/mgf1107-explorations-to-mathematics.
"Examples Of Loan And Interest Calculations." My Assignment Help, 2020, https://myassignmenthelp.com/free-samples/mgf1107-explorations-to-mathematics.
My Assignment Help (2020) Examples Of Loan And Interest Calculations [Online]. Available from: https://myassignmenthelp.com/free-samples/mgf1107-explorations-to-mathematics
[Accessed 22 December 2024].
My Assignment Help. 'Examples Of Loan And Interest Calculations' (My Assignment Help, 2020) <https://myassignmenthelp.com/free-samples/mgf1107-explorations-to-mathematics> accessed 22 December 2024.
My Assignment Help. Examples Of Loan And Interest Calculations [Internet]. My Assignment Help. 2020 [cited 22 December 2024]. Available from: https://myassignmenthelp.com/free-samples/mgf1107-explorations-to-mathematics.