Maths@CSHS

Bank accounts, tax, Hire purchase and Revision on Financial Mathematics will be on a new blog:

mathsfm083.edublos.org 

Open the following for theory and questions on Inflation and Depreciation

inflation-and-depreciation8

inflation3

depreciation4

 

Open the following to get theory on Annuities formula and Perpetuities: 
http://docs.google.com/Presentation?id=dg436rpv_48fvjrvwcd

                                                                      QUESTIONS

 REDUCING BALANCE LOANS/ANNUITIES FORMULA/PERPETUITIES

 1 A reducing balance loan of $90 000 is borrowed at the monthly adjusted interest rate of 11% per annum. Monthly instalments of $857 are to be paid over a period of 30 years. Find:

a the amount still owing after 10 years

b the amount still owing after 20 years

c the total amount paid back on the loan after the full period 30 years

d the amount of interest paid on the loan

 

2 A building society offers $120 000 home loans at the compound interest rate of 10.25% per annum adjusted monthly.

a If repayments are $1100 per month, calculate the amount still owing on the loan after 12 years.

b If the loan is to be fully repaid after 12 years, calculate the amount of the monthly repayment.

 

3 A couple negotiates a 25-year mortgage of $150 000 at a fixed rate of 7.5% per annum compounded monthly for the first 7 years, then at the market rate for the remainder of the loan. They agree to monthly repayments of $1100 for the first 7 years. Calculate:

a the amount still owing after the first 7 years

b the new monthly repayments required to pay off the loan if after 7 years the market rate has risen to 8.5% per annum.

 

4 Dan arranges to make repayments of $450 per month to repay a loan of $20 000 with interest being charged at 9.5% per annum compounded monthly. Find:

a the number of monthly repayments required to pay out the loan (to the nearest month)

b the amount of interest charged

 

 

5 Lee purchases an annuity for $140 000, with interest of 6.25% per annum compounded monthly. If he receives payments of $975 per month, how long will the annuity last? Give your answer to the nearest month.

 

6 Raj purchases an annuity for $85 500, with interest of 7.25% per annum compounded quarterly.

a If he receives quarterly payments for 10 years, how much will he receive each quarter?

b If he receives a regular quarterly payment of $5000, how long will the annuity last? Give your answer to the nearest quarter.

 

7 Bree has $25 000 in an account which pays interest at a rate of 6.15% per annum compounding monthly.

a If she makes monthly deposits of $120 to the account, how much will she have in the account at the end of 5 years?

b If she makes monthly withdrawals of $120 from the account, how much will she have in the account at the end of 5 years?

 

8 Jarrod saves $500 per month in an account which pays interest at a rate of 6% per annum compounding monthly.

a If he makes monthly deposits of $500 to the account, how much will he have in the account at the end of 10 years?

b Suppose that after 10 years of making deposits, Jarrod starts withdrawing $500 each month from the account. How much will he have in the account at the end of another 10 years?

 

9 Mr and Mrs Kostas decide to bor row $25 000 to help them finance the construction of their swimming pool. They consider two loan repayment options:

Loan option A: Monthly repayments of 7.5% per annum compounded monthly

Loan option B: Quarterly repayments at 7.5% per annum compounded quarterly

They wish to pay off the loan over 5 years. Calculate to the nearest dollar for each loan:

a the total repayment

b the total interest paid and hence decide which, if either, is the better loan.

 

10 If Mr and Mrs Kostas of Question 11 choose Loan option A, how much interest do they pay if the interest rate is increased by 0.5%?

 

11 An amount of $35 000 is borrowed in 2005 for 20 years at 10.5% per annum compounded monthly.

a What are the repayments for the loan?

b How much interest is paid on the loan over the 20-year period?

c How much is still owing at the end of 4 years?

After four years, the interest rate rises to 13.75% per annum.

d What are the new repayments that will see the amount repaid in a total of 20 years?

e How much extra must now be repaid on the loan over the term of 20 years?

 

12 A couple put a $20 000 down-payment on a new home and arranged to pay off the rest in monthly instalments of $625 for 30 years at a monthly compounded interest rate of 8.5% per annum.

a What was the selling price of the house to the nearest cent?

b How much interest will they pay over the term of the loan?

c How much do they owe after 6 years?

After 6 years the interest rates climb by 0.9%. The couple must now extend the period of their loan in order to pay it back in full.

d How much do they still owe after the original 30-year period?

e Will they ever repay the loan at their original monthly repayment of $625?

f Calculate the new monthly repayment amount required if the couple still wishes to pay off the loan in 30 years.

 

13  Shelly buys some track shoes for $149.99. She puts the shoes on her credit card because she cannot afford to pay for them outright.

a If Shelly pays interest on the shoes for three months, how much are they now costing her? (Assume the interest is charged at 18% per annum, compounded monthly, and that no payments are made over the 3-month period.)

b If Shelly had paid $50 off the principal amount at the end of the first and second months, after the interest had been added, how much would the final payment have been at the end of the third month? How much would Shelly save by doing this, rather than by adopting the first method?

 

14 A flat rate loan over 6 years at 12.75% per annum amounted to a repayment of $12 500.

a How much was originally bor rowed?

b Calculate the quarterly repayments.

c Compare the savings of a reducing balance loan by working out the quarterly

repayments with interest set at 12.75% per annum compounded quarterly.

d How much is saved over the full 6-year period by adopting a reducing balance loan?

 

15 A personal loan of $7500 is taken out at 11.5% per annum over 4 years.

a Calculate the total amount to be repaid:

i if the loan was a flat rate loan

ii if the loan was a reducing balance loan with monthly repayments

b What flat rate payment of interest would the monthly repayment in a part ii be equivalent to?

 

16 In order to invest in the stockmarket, Jamie takes out an interest only loan of $50 000. If the interest on the loan is 8.15% per annum, compounding monthly, what will be his monthly repayments?

 

17 Jackson takes out an interest only loan of $30 000 from the bank to buy a painting, which he hopes to resell at a profit in 12 months’ time. The interest on the loan is 9.25% per annum, compounding monthly, and he makes monthly payments on the loan. How much will he need to sell the painting for in order not to lose money?

 18 Barbara wishes to start a scholarship which will reward the top mathematics student each year with a $500 prize. If the interest on the initial investment averages 2.7% per annum, compounded annually, how much should be invested? Give your answer to the nearest dollar.

 

19 Cathy wishes to maintain an ongoing donation of $5500 per year to the Collingwood Football Club. If the interest on the initial investment averages 2.75% per annum, compounded annually, how much should she invest?

 

20 Craig wins $1 000 000 in a lottery and decides to place it in a perpetuity which pays 5.75% per annum interest, compounding monthly. What monthly payment does he receive?

 

21 Suzie invests her inheritance of $642 000 in a per perpetuity which pays 6.1% per annum compounding quarterly. What quarterly payment does she receive?

  

                                                                                                                    ANSWERS

1 a $83 056.06

b $62 299.58

c $308 520 d $218 520

 

2 a $98 896.93 b $1451.48

 

3  a $132 119.82 b $1196.29 

 

4.  number of repayments: 55; interest (calculated 

for 55 months) charged: $4750 

 

 

6 a $3023.66 b 21 quarters 

 

8 a $81 939.67 b $67 141.09

 

 

9 a A: $30 057, B: $30 211

 b A: $5057; B: $5211 ∴ option A is best.

 

10 $5414.60

 

11 a $349.43 b $48 863.20 c $32 437.73 d $418.65 e $13 290.24

 

12 a $101 283.53

b $143 716.47

c $76 679.37

d $50 382.81

e paid off after 40 years and 8 months

f $671.64 5

 

13 a $156.84

b final payment = $54.58; saving$2.26

 

14 a $7082.15     b $520.83     c $426.67, $94.16       d $2259.92 7  

 

15 a i $10 950 ii $9392            b 6.3% (to 1 d.p.) 

 

16   $339.58

 

17 $32 775

 

18 $18 519

19 $200 000

 

20  $4791.67

 

21  $9790.50

 

 

 

 

 

   

The following powerpoint consists theory on Percentages, Simple Interest and Compound Interest 
 

<iframe src=”http://docs.google.com/present/embed?id=dg436rpv_103d2tjrpct” frameborder=”0″ width=”410″ height=”342″></iframe>

The related exercises will be found in the following links:  percentage-change , simple-interest, compound-interest1

Open the following link to access the examfurther-mathematics-exam-semester-1-20091 and the answerssolutions-for-trial-exam-1

Complete the following questions for revision, before the ‘Practice SAC’.revision-questions-on-geometry-and-trigonometry You will also be required to attend this week’s practice test.

Open the following document for practice SAC 2practice-sac-2-questions1

The diagram above is called contour map. 

 

The lines are called contour lines and they represent different heights above sea level.

The map gives the horizontal scaled distance between the lines and not the actual distance.

The real/cross-sectional representation of the contour map above is as follows:  

 

To find the distance between B and C, first determine from the diagram the horizontal distance B′C′. Suppose this distance is 80 m.Then triangle BCH in the first diagram can be used to find the distance between B and C and the average slope between B and C. 

 

A cross-sectional profile can be drawn from a contour map for a given cross-section AB. 

This is illustrated below. The horizontal distance that is represented on the cross-sectional profile is the real distance.If the distance AB on the contour map is 6cm then the distance on the cross-sectional profile will be 600cm which is 6m.

 scale 1:100

CONVERSION OF UNITS

When we are converting from smaller units to larger we divide.

When we are converting from larger to smaller units we multiply.

Conversion of units of length

a 1 cm (mm) = 1 × 10 = 10 mm

b 1 m (cm) = 1 × 100 = 100 cm

c 1 km (m) = 1 × 1000 = 1000km

d 1 mm(cm) = 1/10 = 0.1 cm

e 1 cm ( m) = 1/100 = 0.01 m

f  1m (km) = 1/1000 = 0.001 km 

Conversion of units of area 

 1 cm² (mm²) = 1 × 10² = 100 mm²

 1 m² (cm²) = 1 × 100² = 10000 cm² = 10⁴ cm²

 1 km² (m²) = 1 × 1000² = 1000000km² = 10⁶ km²

 1 mm²(cm²) = 1/10² = 0.01 cm² = 10^-2 cm²

 1 cm² (m²) = 1/100² = 0.0001 m² = 10^-4 m²

 1m² (km²) = 1/1000² = 0.000001 km²= 10^-6 

Example: 

a Convert 156000 m² to km²

Answer:

156000 m² / 1000² = 156000/1000000 = 0.156 km²

b Convert 20000mm² to cm² 

Answer:

20000mm² /10² = 20000/100 = 200 cm²

Conversion of units of volume

1000mm³ (cm³) = 1000/10³ = 1000/1000 = 1 cm³

1000000cm³(m³) = 1000000/100³ = 1000000/1000000 = 1 m³

1 litre = 1000cm³

1000 litres = 1m³

Open the following document for questions contour-maps

 

 

 

 

Example 1:ABCDEFGH is a cuboid. Find: 

 

a distance DB 

b distance HB 

c the magnitude of angle HBD 

d distance HA 

e the magnitude of angle HBA 

Solution: 

a.By selecting the appropriate triangle from the diagram and by using the Pythagoras Theorem:

 

 

 

 

The angle of elevation is the angle between the horizontal and a direction above the horizontal. 

The angle of depression is the angle between the horizontal and a direction below the horizontal. 

 

Example 1: The pilot of a helicopter flying at 400 m observes a small boat at an angle of depression of 1.2^◦ . Draw a diagram and calculate the horizontal distance of the boat to the helicopter, correct to the nearest 10 metres. 

 

Solution:  The horizontal distance of the boat to the helicopter is AB. The angle of depression H = B ( alternate).  

 tan(1.2^◦) = 400

                     AB

AB =   400 

        tan(1.2^◦)

AB = 19 095.80 m

 

Example 2:  The light on a cliff-top lighthouse, known to be 75 m above sea level, is observed from a boat at an angle of elevation of 7.1^◦ . Draw a diagram and calculate the distance of the boat from the lighthouse, to the nearest metre.

Solution:  tan (7.1^◦) = 75

                                     AB

                        AB = 75

                               tan (7.1^◦)

                        AB = 602 m

 

Example 3: From a point A, a man observes that the angle of elevation of the summit of a hill is 10^◦ . He then walks towards the hill for 500 m along flat ground. The summit of the hill is now at an angle of elevation of 14^◦ . Draw a diagram and find the height of the hill above the level of A, to the nearest metre. 

Solution

 

 

 

 We need to find HC

To find HC we need to find HB

3 angles and a side are given- sine rule  

  HB       =     500

sin(10^◦)     sin(4^◦)

HB = 500 sin(10^◦) =  1244.67 m

             sin(4^◦)

sin(14^◦) =     HC

               1244.67

HC = 1244.67 sin(14^◦) =  301.11m

 

                                                                         BEARINGS

Bearings are used to indicate direction.

The three-figure bearing, the True Bearing is the direction measured clockwise from north and starts from 0° to 360°. 

For the Compass Bearing we seperate the plane in 4 sections of 90°.   

 

Example: 

A  True Bearing is 30°.

      Compass Bearing = N 30°E

B  True Bearing = 150°

     Compass Bearing = E 30°B

C  True Bearing = 210°

     Compass Bearing = S 30°W

D  True Bearing = 330°

     Compass Bearing = W60°N

 

 

Example 1:The road from town A runs due west for 14 km to town B. A television mast is located due south of B at a distance of 23 km. Draw a diagram and calculate the distance of the mast from the centre of town A, to the nearest kilometre. Find the bearing of the mast from the centre of the town.

 

Solution:  AT² = 14² + 23²

                     AT = √( 14² + 23² ) = 26.93 km

For Bearing find θ

                     tan θ = 23

                            14

                θ = tan^-1(23/14) = 58.67^°

     Bearing = 270^°- 58.67^°= 211.33^°

The bearing of the mast from A is 211.33^°T

 

 

 

Example 2: A yacht starts from a point A and sails on a bearing of 038^◦ for 3000m. It then alters its course to a bearing of 318^◦, and after sailing for 3300 m it reaches a point B. 

a Find the distance AB,correct to the nearest metre. 

b Find the bearing of B from A, correct to the nearest degree. 

Solution

 

 a 2 sides and the included angle are given- cosine rule

AB = √(3300² +3000² - 2 × 3300 × 3000 cos 100^°)

AB =  4830 m

b To find the bearing of B from A we need to find angle A

   We may use sine and cosine rule

   3300 = 4830

  sinA°    sin100°

  A°= sin^-1 (3300sin100°  ) =  42.28°

                       4830

The bearing of B from A = 360^◦ −(42.29^◦- 38^◦ ) = 355.71^◦

The bearing of B from A is 356^◦T , to the nearest degree.

 

 

Example 3 : Two points, A and B, are on opposite sides of a lake so that the distance between them cannot be measured directly. A third point, C, is chosen at a distance of 100 m from A and with angles BAC and BCA of 65^◦ and 55^◦ , respectively. Calculate the distance between A and B, correct to two decimal places. 

 

Solution: 2 angles and a side- sine rule

AB   =  100

sin55°  sin60°

AB = 100sin55°

           sin60°

AB = 94.59m

Open the following document for questionsapplications-of-geometry-and-trigonometry