Mathematical Solutions for Hire Purchase and Compound Interest
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This article provides mathematical solutions for hire purchase and compound interest problems. It includes examples of calculating deposit, balance, interest paid, monthly repayments, and total cost of a car. It also covers effective rate of interest and using tables for interest repayments.
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MATHEMATICAL SOLUTION [Author Name(s), First M. Last, Omit Titles and Degrees] [Institutional Affiliation(s)]
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PART A: EXAMPLE:A sapphire earing with marked price of $1800 is offered to the purchaser on the following terms: $200 deposit and balance to be paid over 24equal monthly instalments with interest charged at 11.5% flat rate. Find: a)Total interest paid b)The monthly repayments c)The total cost of car Solution: a)Cash price = $1,800 Deposit = $200 Balance to pay = $1,600 Interest(I)=(1600×11.5×2) 100 Interestpaid(I)=$368 b)The monthly repayments =Balance+interest no.ofperiods =$1968 24=$82 c)The total cost of car =Deposit+Loanbalance+Interest =200+1600+368=$21,368
Question 1 A car is purchased on hire purchase. The cash price is $21,000 and the terms are, a deposit of 10% of the cash price, then the balance to be paid off over 60 equal monthly instalments. Interest is charged at 12% p.a a)Calculate the deposit and the balance Deposit=10%of$21,00 =0.1×21,000= $2,100 Loanbalance=$¿) = $18,900 b)Calculate the interest paid on the balance Interestcharged=18,900×12×5 100 Interestctharged=$11,340 c)Calculate the monthly instalments Totalinstalments=Loanbalance+interestpaid = 18,900 + 11,340= $30,240 Monthlyinstalments=30,240 60= $504 d)Calculate the total cost of the car. TotalCost=¿ = $(2,100 +18,900 + 11,340) =$32,340 Question 2 The cash price of a suit is $2,000. If a customer pays a deposit of $300 and pays equal monthly instalments of $60 over 2.5 years, calculate: a)The amount of interest charged Solution
Loanbalance=(2,000−300)= $1,700 Monthly instalments =1,700+Interest 2.5×12 Since monthly instalments = $60, Therefore,$60=1,700+Interest 30 $1,800 = $1,700 – Interest Interest charged = $100 b)The total paid for the suit. Total amount = Deposit + Loan balance (Principal) + Interest = $(300 +1,700 +100) = $2,100 Question 3 An electric guitar is being bought on hire purchase for a $250 deposit and monthly instalments of $78.50 for 3 years. The cash price for this guitar is $2,500. Find the interest rate charged by the shop. Solution Deposit = $250 Monthly instalments = $78.50 Cash price = $2,500 Principal = Cash price – Deposit = $(2,500 – 250)= $2,250
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Monthly instalments =Principal+Interest Numberofperiod =2,250+Interest 12×3 But, monthly instalment =$78.50 $78.50 =78.50+Interest 36 $(36×78.50¿=2,250+Interest Interest =$(2,826 – 2,250) = $576 Since, Simple Interest =Principal×Rate×Time 100 $576 =2,250×Rate×3 100 Rate =576 67.5 Interest rate = 8.53% (2decimal places) PART B: Question 2 Find the value of A – G in the following table. Solution Cash price (CP) Deposit (D) Principal (CP-D) Rate (r )Interest (I)Installments (Monthly) TimeEffective interest rate $2,500$5002,500 –500 =$2,000 10% p.aA 2000×10×2 100 ¿$400 P+I Period 2,000+400 24 ¿$100 2 yearsB r÷n 100 10÷2 100×100 ¿5% $150$50$(150-50)6.75%100×6.75×6 100×12 ¿$3.375 C 100+3.375 6 6 months D 6.75÷6 100×100
¿$17.23¿1.125% $6,500$1,5006,500-1,500F 500×r×2 100=1000 100r=1000 r=10% 5000×I 24=250 ¿6000−5000 E $250 5000+I 24=250 2 yearsG 10÷2 100×100 ¿5% Summarized solution ABCDEFG =$400=5%=$17.23=1.125%=$250(given)=10%=5% PART C: Compound Interest Compound interest is when you deposit money in an account and rather getting the same amount of interest each year (simple interest), you receive interest on the new amount in your account at the end of every period. We use the formula: A=P(1+r n) nt Where, r = annual rate, n = number of times of interest is added per year, t = number of years, P=Principal, A=amount accrued in the bank EXAMPLE I borrow $1000 at 12% comp p.a I take the loan out for 3 years. Calculate the amount I will have in the bank if interest is: a)Compounded annually, n=1, interest is paid once a year, t=3years A=1000(1+0.12 3) 3×1 = $1,404.93
b)Compounded daily, n=365, interest is paid 365days a year,t=3years A=1000(1+0.12 3×365) 365×3 = $ 1,433.24 c)Compounded monthly, n=12, interest is paid 12times a year A=1000(1+0.12 12) 12×3 = $1,430.77 PART D: Compound interest from a table Often a bank will provide a table rather than you having to use a formula in Part C above. EXAMPLE Using the financial table, calculate the monthly installments required to pay off a 25-year loan of $110,000 at 4% p.a monthly reducible interest. A = 5.28 ×110 × 12 months × 25 years = 174,240 Question 1
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Calculate the annual instalments needed to pay off a $110,000 home loan at 7% p.a reducible interest if the installments are to be paid in equal installments over 15 years. Give your answer to nearest correct cents. Solution Number of lots =(110,000 1000)=110 Installments = 110 ×109.79 ×1 ×15 = $181,153.50 Question 2 Mae received a 30-year $89,000 loan at 2% p.a monthly reducible interest if the term of the loan was reduced to 25 years, calculate:
Solution a)The amount of each monthly instalment over a 30-year loan term Number of lots¿89,000 1000=89lots Monthly installment = 89 × 3.7 × 30 × 12 = $ 118,548 b)The monthly needed for a 25-year term Monthly installment = 89 × 3.7 × 25 ×12 = $98,790 c)The amount saved, from this 5-year reduction in the term of the loan Amount saved = $(118,548 – 98,790) = $19758. Question 2 Using the financial table, calculate the monthly installments required to pay off a 25-year loan of $1000 at 4% p.a monthly reducible interest.
Solution Number of lots =1000 1000=1 Monthly installment = 1 × 5.28 ×12 ×25 = $1,584 Question 3 Calculate the annual installment needed to pay off a $110,000 home loan at 7% p.a reducible interest if installments are to be paid in equal amounts over 15 years. Give your answer to the nearest cent
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Solution Number of lots =110,000 1000= 110 lots Annual installments = 110 × 109.79 ×1 ×15 = $181,153.50 PART E: Effective rate of interest ( Compound interest) If I want to compare a bank account that compounds at different rates, then we can use the following formula Effectiveinterestrate=(1+r/n 100)n −1 EXAMPLE Neil invested $3,500 at 4.3% p.a compounded daily. Find the effective interest rate Effectiveinterestrate=(1+4.3/365 100) 365 −1
= 0.0439 ×100% = 4.39% Question 1 A loan attracts interest at a rate of 1.8% compounding annually. What is the effective rate correct to one decimal place? Solution Effectiveinterestrate=(1+1.8/1 100) 1×1 −1 Effective interest rate = (1.018 – 1) ×100% = 1.8% Question 2 Two different lending institutions are offering different rates on their loans. Betta Bank is offering 4.4% compounding monthly. Lucky lending is offering 4.2% compounding quarterly. a)What is the effective rate of Beta Bank correct to two decimal places? Solution Effectiveinterestrate=(1+4.4/12 100)12 −1 Effectiveinterestrate=(1+0.3667 100)12 −1 Effective interest rate =(1.003667)12−1¿×100% = (1.0449 – 1)×100%= 4.49% (2 decimal places) b)What is effective rate of Lucky lending correct to two decimal places?
Solution Effectiveinterestrate=(1+4.2/0.75 100)0.75 −1 Effectiveinterestrate=(1+5.6 100)0.75 −1 Effectiveinterestrate=(1.056)0.75−1 = (1.0417 – 1) ×100% = 4.17% (2 decimal places) c)Which has a better offer? Lucky lending has a better offer of 4.17% compared to Betta Bank at the rate of 4.49% PART F: Using tables for interest repayments Banks will you a table to help you calculate Home loans. Banks offer different interest rates depending on the interest rate you pay and the length of your loan. For example, in the table below you can see that a 20-year loan at an annual 6% interest, will incur a monthly repayment of $7.16 for each $1000 that is borrowed. That means that if we borrowed $1,500, we would be borrowing 15 lots of $1000.So to work out the total amount of the loan, we would calculate $7.16×12 months×20 years×15 lots. So, we’d end up paying $10776 interest over the 20 years period.
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Question 1 Julia takes out a loan of $500, 00 on a 25year old agreement of 7% p.a. Calculate: a)The total amount of the loan Solution Number of lots =500,000 1000=500lots Total amount = (7.07 × 12 × 25 ×500) = $1,060,500 b)The total interest paid on the loan Interest paid = $(1,060,500 – 500,000) = 560,500 References Disney, R., & Gathergood, J. (2013). Financial literacy and consumer credit portfolios.Journal of Banking & Finance,37(7), 2246-2254.
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