Maths Quest Maths B Year 11 for Queensland Chapter 8 Applications of exponential and log functions in financial maths
WorkSHEET 8.21
WorkSHEET 8.2 Applications of exponential and log functions in financial maths
Name ______
Maths Quest Maths B Year 11 for Queensland Chapter 8 Applications of exponential and log functions in financial maths
WorkSHEET 8.21
1 / Agro Fertilisers has borrowed $12 400 from Roo Rail Loans for extensions that will enable them to increase the quantity and quality of their produce. The loan is to be paid back in 12 monthly instalments of $1046.82 at an interest rate of 1.002% per month.(a)What is the amount still owed after the first and second years?
(b)What is the total interest charged over the full term of the loan? / (a)Growth factor (R) = 1 +
= 1 +
= 1.01002
Balance (A1)= A0 RQ
= 12 400 1.01002 1046.82
= $11 477.43
Balance (A2)= A1RQ
= 11 477.431.01002
1046.82
= $10 545.61
(b)Total interest (I) = 1046.82 12 12 400
= $161.84 / 1
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2 / Astootin Vesta borrowed $14 300 which is to be repaid by monthly instalments of $405 at an interest rate of 0.05% per month with interest debited monthly.
(a)What amount is still owing after the first two payments? (Show each month’s calculations separately.)
(b)By how much has the principal reduced after the first two repayments?
(c)How much interest has been paid after these two payments? / (a)Growth factor (R) = 1 +
= 1 +
= 1.0005
Balance (A1)= A0 RQ
= 14 300 1.0005 405
= 13 902.15
Balance (A2)= A1RQ
= 13 902.151.0005 405
= $ 13 504.10
(b)Reduction= P A2
= 14 300 13 504.10
= $795.90
(c)Interest paid (I2)= 2 405 795.90
= 810 795.90
= $14.10 / 1
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3 / Use the annuities formula
An= PRn– to find A, given:
(a)P = 300, r = 1, n = 36, Q = 9
(b)P = 1450, R = 1.001, n = 20, Q = 75 / (a)An= PRn–
R = 1 + = 1 + = 1.01
P = 300, R = 1.01, n = 36, Q = 9
A36= 3001.0136
A36= 3001.4308
A36 = 429.2306 387.6919
A36 = $ 41.54
(b)An= PRn
P = 1450, R = 1.001, n = 20, Q = 75
A20 = 14501.00120
A20 = 1450 1.0202
A20 = $35.05 / 1
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4 / Hyp Rice Lenders offer loans of $30 000 to farmers at a rate of 6.5% p.a. (debited monthly). The loan is repaid in monthly instalments of $504.30 over 6 years.
(a)What is the growth rate (R)?
(b)How much of the loan repayments is interest?
(c)Use An= PRn to calculate the amount owing after 36 payments have been made. / (a)Interest rate per month (r)=
0.5417%
Growth rate (R)= 1 +
= 1.005417%
(b) Total repayments= 504.30 72
= $36 309.60
Interest paid= 36 309.60 30 000
= $6309.60
(c)
An= PRn
P = 30 000, R = 1.005417, n = 36, Q = 504.30
A36 = 30 0001.00541736
A36 = 36 440.58373785
A36 = 36 440.58373785 19 986.37855516
A36 = $16 454.21 / 1
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5 / Luke has borrowed $1200 from Mae Kim Honey. The loan is to be repaid after 4 years with an annual interest rate of 5.4% (interest debited fortnightly). Repayments are by fortnightly instalments of $12.84.
(a)Calculate the amount still owed by Luke after 3 years.
(b)Calculate the amount of interest paid during the 79th instalment. / (a)Interest rate per fortnight (r) =
0.208%
Growth rate (R)= 1 +
= 1.00208%
An= PRn
P = 1200, R = 1.00208, n = 78, Q = 12.84
A78= 12001. 0020878
A78 = 1411.133
A78 = 1411.133 1086.118
A78 = $325.02
(b)n = 79
A79= 12001. 0020879
A79= $312.85
Reduction = 325.02 312.85
= $12.17
Interest= 12.84 12.17
= $0.67 / 1
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6 / Mrs Walker wishes to purchase a car for $12400 from Crash Deal Car Sales. Her bank offers her a loan to be taken out over 10 years with monthly instalments. If there is a 6% p.a. interest rate calculate the instalments (Q) needed to complete the loan in the specified period. Use the following formula:
Q = . / Interest rate per month (r) =
= 0.5%
Growth rate (R)= 1 +
= 1.005%
P = 12 400, R = 1.005, n = 120
Q =
Q =
Q = $137.67 / 1
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7 / Use the annuities formula to find the number of monthly repayments (n) which would need to be made to repay the loan fully under the conditions stated below:
r = 9% p.a., P = $70 000, Q = $629.81
(a)use n = 230
(b)use n = 240
(c)which of (a) or (b) is the best estimate of the number of instalments needed to finalise the loan. / (a)Interest rate per month (r) =
= 0.75%
Growth rate (R)= 1 +
= 1.0075%
Let n = 230
An= PRn
r = 9% p.a., P = $70 000, Q = $629.81
A230 = 70 0001.0075230
A230 = 70 0005.5765
A230 = 390 355.81 384 311.03
A230 = $6044.78
(b)Let n = 240
A240 = 70 0001.0075240
A240 = 70 000 6.009151524
A240 = 420 640.61 420 641.83
A240= –$1.22
A240 is as close to zero as we need to go due to round off errors, use of $629.81 instead of $629.8082, during calculations.
240 monthly repayments equates to 20 years of repayments to finalise the loan.
(c)Option (b). / 1
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8 / A loan of $5400 at 6% p.a. can be fully repaid in 20 years. If monthly instalments are debited at $38.69 per month
(a)how much would the time of repayment be reduced, and
(b)how much money could be saved if instalments remain the same but are made fortnightly? / (a)Interest rate per fortnight (r)=
0.23%
Growth rate (R)= 1 +
= 1.0023%
Start with a guess. We know that 240 repayments will definitely discharge the loan, so a value of less than this is required.
Using n = 160 produces a Q value of $40.38, which is too high.
Using n = 180 produces a value of $36.67.
Using this approach, n = 168 produces a
Q $38.79
r = 6% p.a., P = $5400
Q =
Q =
Q $38.79
In terms of years, the loan would be full repaid in 6 years and 12 fortnightly payments.
This represents a reduction in time of approximately 13.5 years and 72 fewer repayments, a saving of:
(b)72 38.69 = $ 2785.68 / 1
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9 / For the loan in question 7, what would be the resultant savings in time and money if the repayments remained monthly but could be doubled? / r = 6% p.a., P = $5400
Q =
Q =
Q $77.41
The loan would be settled in approximately 86months or = 7 years and two months, saving 12years and 10 months.
Savings = 24038.69 8677.38
= 9285.60 6654.68
= $2630.92 / 2
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10 / Explain why there is a saving in time using the method in question 8 (monthly doubled repayments) compared to the method proposed in question 7 (fortnightly repayments). How much time is saved? / There is one extra payment of $77.38 per year using the fortnightly repayment scheme. This means that over a 6-year loan, the time is reduced by at least 6 monthly repayments of $77.38. / 2