Tutorial 6 Flashcards
1) A person repays a loan of £5,500 by monthly instalments in arrears of £200 for 3 years.
Calculate the flat rate and subsequent APR for this loan arrangement.
Flat rate = (12x3x200 – 5500)/3x5500 = 10.3%
APR roughly 2xflat rate, rounded to nearest 0.1%
5500 = 2400a3¬(12)
Try i =20% – 2400x2.1065x0.2/0.183714 = 5503.77
Try i = 20.1%, i(12) = 18.4559%= 2400(1-1.201^-3)/0.184559 = £5497.31
0.2 + (5500-5503.77)/(5497.31-5503.77)x(0.201-0.2) = 0.201 = 20.1% pa
2) A loan is repaid by equal instalments of £1,200 for 20 years. Calculate the following,
assuming the instalments are payable annually in arrears and i = 6%.
i) The original loan amount
ii) The interest element of the 5th payment
iii) The amount of capital of the 10th payment
iv) The amount of capital paid in the last 5 years
v) The capital and interest element of the last payment
vi) Total interest paid over the period of the loan
i) 1200a20¬ = 1200(1-1.06^-20)/0.06 = £13,763.91
ii) 0.06xL,4 = (1200a16¬ )*6%
L4 = 1200(1-1.06^-16)/0.06 = £12,127.07
Interest = 12127.07x0.06 = £727.62
iii) X – 0.06xL,9 = 1200 - (1200a11¬)*6%
1200 – 0.06x[1200(1-1.06^-11)/0.06 = £632.15
iv) L,15 - L,20 = 1200a5¬ = 1200(1-1.06^-5)/0.06 = £5,054.84
v) Interest = 0.06L,19 Capital = X- 0.06L,19 or simply L,19 or 1200v
L,19 =1200a1¬ = 1200(1-1.06^-1)/0.06 = 1132.08 = capital
Interest = 0.06x1132.08 = 67.92
vi) Total repayments = 20x1200 = £24,000
Original Loan = £13,763.91
Total interest = 24000-13763.91 = £10,236.09
3) A loan of £10,000 is repayable by 8 equal annual instalments. Calculate the following,
assuming payments are in arrears and i is 10% pa for the first 5 years and 8% thereafter.
i) The annual repayment
ii) The loan outstanding after the 3rd instalment
iii) The interest element of the 6th instalment
iv) The total interest paid over the term of the loan
i) 10000 = X(a5¬ @10% + v^5a3¬ @8%)
10000=X[(1-1.1^-5)/0.1+1.1^-5(1-1.08^-3)/0.08] — X = £1,854.96
ii) X(a2¬ @10% + v^2a3¬ @8%)
=1854.96[(1-1.1^-2)/1.1 +1.1^-2(1-1.08^-3)/0.08 = £7,170.11
iii) i@8% x 1854.96a3¬ @8%
0. 08x1854.96(1-1.08^-3)/0.08 = £382.43
iv) 8X – 10000
=8x1854.96-10000 = £4,839.68
Question to work through in tutorial
An individual has been offered the option of two loans to finance the purchase of a car for £12,000.
1) Monthly repayments (in advance) of £X for 3 years at i = 6%
2) Monthly repayments (in arrears) of £Y for 5 years at i = 7.5%
Which loan should he choose (based on total overall payments)?
Assuming the individual chooses the loan with the lowest total overall payments, after 1 year, the
individual finds that he is struggling with the repayments and negotiates with the bank to reduce his payments to £100 per month, what is the term of the new loan (assume i unchanged)?
Part 1.1 – 12000 = 12X𝑎̈3¬(12)@6% d(12) = 0.058128
X=12000/[12(1-1.06^-3)/0.058128)] = £362.44
Part 1.2 – 12000 = 12Ya5¬(12)@7.5% i(12) = 0.072539
Y=12000/[12(1-1.075^-5)/0.072539] = £239.05
Choose loan – min(12x3x362.44,12x5x239.05) = min(13047.84,14343)
Choose loan 1
Part 2
Calculate outstanding loan after 1 year
so 12x362.44𝑎̈2¬(12)@6% = L = 12x362.44(1-1.06^-2)/0.058128 = £8,230.74
So 8230.74 = 12x100x𝑎̈n¬(12)@6%
8230.74 =1200(1-1.06^-n)/0.058128 rearrange
1.06^-n = 1-0.398697
n = 8.73 years
So say 8 yrs and 9 months
Overall 1 year + 8 years 9 months = 9 yrs and 9 months
