Annoying Loan Reyament Questions (1 Viewer)

[BigBear_25]

Can someone help me with these loan repayment questions

1) A loan of $6000 over 5 years at 15% p.a interest, charged monthly, is paid back in 5 yearly instalments. How much is each instalment, and how much money is paid back altogether?

2) Bill thinks he can afford a mortgage payment of $800 a month. How much can he borrow, to the nearest $100, over 25 years at 11.5% p.a. interest.

Thank you in advanced.

P.S: these questions are from Maths in Focus 2

 

[rama_v]

Is the answer to the first question $1835.68 per year?

 

[insert-username]

1) A loan of $6000 over 5 years at 15% p.a interest, charged monthly, is paid back in 5 yearly instalments. How much is each instalment, and how much money is paid back altogether?

Let R = the value of the repayment
Interest Rate = 15/12% = 1.25% per month

Year 1 = 6000*1.0125¹² - R

Year 2 = 6000*1.0125²⁴ - R*1.0125¹² - R

Year 3 = 6000*1.0125³⁶ - R*1.0125²⁴ - R*1.0125¹² - R

Year 5 = 6000*1.0125⁶⁰ - R*1.0125⁴⁸ - R*1.0125³⁶ - R*1.0125²⁴ - R*1.0125¹² - R

0 = 6000*1.0125⁶⁰ - R(1 + 1.0125¹² + 1.0125²⁴ + 1.0125³⁶ + 1.0125⁴⁸)

R = (6000*1.0125⁶⁰)/(1 + 1.0125¹² + 1.0125²⁴ + 1.0125³⁶ + 1.0125⁴⁸)

R = (6000*1.0125⁶⁰)/[(1.0125⁶⁰ - 1)/(0.1607545)

= $1835.68 per year.

Therefore $9178.40 is repaid altogether


I_F

 

[BigBear_25]

Thanks for that.

Does any one no how to do the second question.

 

[SoulSearcher]

2) Bill thinks he can afford a mortgage payment of $800 a month. How much can he borrow, to the nearest $100, over 25 years at 11.5% p.a. interest.

Here, you just have to find the original loan the person took out. Let the original amount be P, rate is 1 + (0.115/12) and time period be 300 months
At the beginning of the loan period,
A₀ = P
After first repayment,
A₁ = P * (1+23/2400) - 800
A₂ = A₁*(1+23/2400) - 800
= P*(1+23/2400)² - 800((1+23/2400) + 1)
A₃ = P*(1+23/2400)³ - 800((1+23/2400)² + (1+23/2400) + 1)
...
A₃₀₀ = P*(1+23/2400)³⁰⁰ - 800((1+23/2400)²⁹⁹ (1+23/2400)²⁹⁸+ ... + (1+23/2400) + 1)
but A₃₀₀ = 0, as the loan is repayed, and ((1+23/2400)²⁹⁹ (1+23/2400)²⁹⁸+ ... + (1+23/2400) + 1) is a geometric series with 300 terms, first term 1 and ratio (1+23/2400), so
0 = P*(1+23/2400)³⁰⁰ - 800((1+23/2400)³⁰⁰ - 1)/(23/2400) - 1
P*(1+23/2400)³⁰⁰ = 800((1+23/2400)³⁰⁰ - 1)/((23/2400)-1)
P = { 800((1+23/2400)³⁰⁰ - 1)/((23/2400)-1) } / { (1+23/2400)³⁰⁰ }
= $78,703.83
= $78,700 to the nearest hundred