easter term 2021 grade 9 mathematics ... - education.gov.gy

Ministry of Education

MINISTRY OF EDUCATION

EASTER TERM 2021

GRADE 9

MATHEMATICS

WORKSHEETS


WEEK 1

LESSON 1

Topic: Consumer Arithmetic – Percentage Profit and Percentage Loss

𝑻𝒉𝒆 𝑷𝒓𝒐𝒇𝒊𝒕 = 𝑻𝒉𝒆 𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝑷𝒓𝒊𝒄𝒆 − 𝑻𝒉𝒆 𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆

𝑷 = 𝑺. 𝑷 − 𝑪. 𝑷

𝑻𝒉𝒆 𝑷𝒓𝒐𝒇𝒊𝒕 % =𝑻𝒉𝒆 𝑷𝒓𝒐𝒇𝒊𝒕

𝑻𝒉𝒆 𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆 × 𝟏𝟎𝟎%

𝑷% =𝑺.𝑷−𝑪.𝑷

𝑪.𝑷 × 𝟏𝟎𝟎%

𝑻𝒉𝒆 𝑳𝒐𝒔𝒔 = 𝑻𝒉𝒆 𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆 − 𝑻𝒉𝒆 𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝑷𝒓𝒊𝒄𝒆

L = 𝑪. 𝑷 − 𝑺. 𝑷

𝑻𝒉𝒆 𝑳𝒐𝒔𝒔 % =𝑻𝒉𝒆 𝑳𝒐𝒔𝒔

𝑻𝒉𝒆 𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆 × 𝟏𝟎𝟎%

𝑳% =𝑪.𝑷−𝑺.𝑷

𝑪.𝑷 × 𝟏𝟎𝟎%

Example 1.

A shop keeper bought 25 cricket balls at a total cost of $87 500.

a) He sold them at $4 200 each. What was his percentage profit?

b) He sold them at $2 700 each. What was his percentage loss?

Solution

a) C.P of the balls = $87 500

S.P of the balls = 25 × $4 200

= $105 000

P = S.P – C.P

= $105 000 – $87 500

= $ 17 500

𝑃% =𝑆.𝑃−𝐶.𝑃

𝐶.𝑃 × 100%

=$17 500

$87 500 × 100%

= 20%

b) C.P of balls = $87 500

S.P of balls = 25 × $2 700

= $67 500

L = C.P – S.P


= $87 500 – $67 500

= $ 20 000

𝐿% =𝐶.𝑃−𝑆.𝑃

𝐶.𝑃 × 100%

=$20 000

$87 500 × 100%

= 22.85%

Example 2.

A businesswoman bought a stove for $42 000.

a) Calculate the selling price of the stove if she makes a profit of 15%.

b) The stove was damaged in transporting it to the customer. Determine the selling price of

the stove if she incurred a loss of 8% on the cost price.

Solution

a) C.P = $42 000 Alternatively: CP represents 100%

P = 15% of C.P 100% = $42 000

= 15% of $42 000 1% = 42 000

100

=15

100 × $42 000 115% =

42 000

100 ×115

= $6 300 = $ 6 300

S.P = C.P + P

= $42 000 + $6 300

= $48 300

b) C.P = $42 000

L = 8% of C.P

= 8% of $42 000

=8

100 × $42 000

= $3 360

S.P = C.P – L

= $42 000 – $3 360

= $38 640


Exercises

1. Albert bought a bicycle for $27 500. He sold it for $35 500.

a) What was the amount of his profit?

b) What was his percentage profit?

2. Mrs. Jones bought 250 apples for $22 500. She sold 50 for $120 each, and the remainder

she sold 5 for $1 000 a parcel.

(a) State whether she made a profit or loss.

(b) Calculate her percentage profit or percentage loss.

3. A businesswoman bought a personal computer for $108 000.

a) Calculate her selling price on the personal computer if she wants to make a profit of

25%.

b) During transporting the personal computer to the customer, it was damaged. Calculate

her selling price if she incurred a loss of 5%.

4. A man bought a clock for $6 000 and sold it to make a profit of 15%. What was his

selling price?

5. A man offers to sell a boat for $240 000 in order to make a profit of 20%. Later he

changes his mind and decides to make a profit of 30%. What would be the selling price of

the boat?

6. A vender sells an umbrella for $1 800 in order to make a profit of 15%. What would be

the selling price if she decides to make a profit of 20%?

7. A businesswoman bought a refrigerator from a manufacturer for $68 000. Calculate:

a) the selling price if she makes a profit of 17.5%.

b) the selling price if she incurred a loss of 3.5%.

8. A salesman bought a computer from a manufacturer. The salesman the sold the computer

for $156 000, making a profit of 25%. What amount did the salesman pay the

manufacturer?


WEEK 1

LESSON 2

Topic: Consumer Arithmetic – Percentage Profit and Percentage Loss (continued)

Example 1.

A trader sold a vase for $7 600 and made a loss of 5% on what he paid for it. How much did he

pay for it?

Solution

S.P = $7 600

L = 5%

S.P in terms of % = 100% – 5%

= 95%

∴ 95% = $7 600

So 1% = $7 600

95

Then C.P (100%) = $7 600

95 × 100

= $8 000

Example 2.

A man sold a cabinet for $85 000 and made a profit of 25% on what he paid for it. How much

did he pay for it?

Solution

S.P = $85 000

P = 25%

S.P in terms of % = 100% + 25%

= 125%

∴ 125% = $85 000

So 1% = $85 000

125

Then C.P (100%) = $85 000

125 × 100

= $68 000


Exercises

1. A trader sold a toaster for $38 000 and loss 15%. How much did he pay for the stove?

2. When a radio was sold for $36 000, a profit of 12% was made. Calculate the cost

price.

3. A businessman purchased a car and sold it for $1 900 000 making a profit of 15%.

What was the cost price of the car?

4. Three motor bikes were sold for $1 575 000. If the seller made a profit of 20%

altogether, calculate how much he paid for the motor bikes.

5. Mark is an auto dealer. He sold a car and made a loss of 10%. If his loss was $150

000, calculate:

a) Calculate how much he sold the car for.

b) What was his cost price?.

6. An item is sold for $500 000. The salesman made a loss of 5%. Calculate how much

the item cost.


WEEK 1

LESSON 3

Topic: Consumer Arithmetic – Hire Purchase

Hire purchase is a type of credit system where you are allowed to purchase an item buy paying

some money down (down payment or deposit) then paying the remainder in equal monthly

instalments. The buyer is allowed to take the item home but cannot own it until all instalments

are paid.

Example 1.

A new freezer costs $200 000 but it may be bought on hire purchase by making a deposit of

$75 000 and 24 monthly instalments $7 292.

How much does the freezer cost by the hire purchase system?

Solution

Deposit = $75 000

1 instalment = $7 292

24 instalments = $7 292 × 24

= $175 008

Total cost of freezer = deposit + total instalments

= $75 000 + $175 008

= $250 008

Example 2.

Mrs. Sam checks on the new section suite she wants to buy. The cash price is $60 000 or she can

pay 40% deposit on the item and 12 monthly instalments of $4 000 each.


a) How much extra would she pay by the hire purchase method?

b) What rate of interest would she pay for the suite?

Solution

a) Deposit = 40% of $60 000

= 40

100 × $60 000

= $24 000

Instalments = $4 000 × 12

= $48 000

Total = $24 000 + $48 000

= $72 000

Amount extra paid = $72 000 – $60 000

= $12 000

b) Interest is paid on = $60 000 – $24 000

= $36 000

Rate of interest = $12 000

$36 000 × 100

= 33.33%


Exercises

1. A small freezer cost $125 000. It can be bought on hire purchase terms by making a

deposit of $65 000 and 24 monthly instalments of $4 200. How much does the freezer

cost by hire purchase system?

2. A dining set can be bought for $210 000 cash or by paying $85 000 deposit and 12

monthly instalments $13 500. How much more will the dining set cost by using the hire

purchase method of payment?

3. A vanity can be bought for $85 000 or by paying a deposit of $25 000 and 24 equal

monthly instalments each of $3 200.

a) How much more does the vanity cost by using the hire purchase method of payment?

b) Express the increase as a percentage of the cash price.

4. A tablet can be bought for $15 000 cash or by paying a 10% deposit and 12 monthly

instalments, each of $1 500. Find how much extra it costs to use the hire purchase plan

and express this extra cost as a percentage of the cash price.

5. A house wife wants to buy a fridge costing $163 000 from Singers. She can have it now

and pay for it by instalments during a period of one year, in which case the price is

increased by 10%. If she is required to pay a deposit and 12 monthly instalments of

$5 200, calculate the deposit to the nearest dollar.

6. The cash price of an article is $65 000. Work out a monthly plan if a deposit of 10% is

paid. The remaining amount must be paid off in 12 monthly instalments. The interest rate

is 15%.

7. The cash price of a smart TV is $250 000. Work out an instalment plan for 2 years’

monthly payments, if the deposit is 15% and interest is at the rate of 12%.

8. The retail price of a television set is $175 000. If the buyer pays cash, the price is 10%

below the retail price. If the set is bought on hire purchase, the buyer pays a down

payment $65 000 and 24 monthly instalments of $10 500.

a) Determine the difference between the hire purchase price and the cash price.

b) Calculate the difference as a percentage of the retail price.


9. The marked price of a used car is $450 000. A person can pay a deposit of 30% and

interest at 12% per annum is charged on the outstanding balance. The total amount

payable is to be paid in 21

2 years.

Calculate:

a) the amount of each monthly instalment.

b) the hire purchase price of the car.

10. A television set can be bought on hire purchase by making a deposit of $26 000 and 24

monthly instalments of $11 000 each. Calculate the hire purchase price.


WEEK 1

LESSON 4

Topic: Consumer Arithmetic – Mortgage

Some items such as cars, machinery, properties, etc are usually very expensive to purchase. These

can be bought under a special hire purchase agreement with a bank called mortgage.

The purchaser pays a deposit on the item and the bank lends him/her the remainder

(loan/mortgage) to pay for the item. The item is legally the property of the bank until the

mortgage is repaid.

The loan plus an interest is usually repaid in equal instalments over a long period.

Example 1.

A property is advertised for sale. The sale price is $4 800 000. The property can be bought with

a 20% deposit and a mortgage.

Calculate:

a) the deposit.

b) the mortgage.

c) the total amount to be repaid to the bank if monthly instalments of $27 733 are to be

made over a period of 15 years.

d) the interest on the mortgage.

Solution

a) deposit = 20% of sales

= 20% 0f $4 800 000

= 20

100 × $4 800 000

= $960 000


b) mortgage = sales price – deposit

= $4 800 000 – $960 000

= $3 840 000

c) total amount to be repaid = monthly installments × 12 × 15

= $27 733 × 12 × 15

= $4 991 900

d) interest = total amount to be repaid – mortgage

= $4 991 900 – $3 840 000

= $1 151 900

Example 2.

A man borrows $300 000 from the bank. The bank charges 18% interest for the entire period of

the loan. If repayments are 36 monthly instalments, calculate the amount of each instalment.

Solution

Interest = 18% of $300 000

= 18

100 × $300 000

= $54 000

The amount to be repaid = $300 000 + $54 000

= $354 000

Amount of each instalment = $354 000

36

= $9 833


Exercises

1. A property is advertised for sale. The sale price is $5 000 000. The property can be

bought with a deposit of 10% and a mortgage.

Calculate:

a) the deposit.

b) the mortgage.

2. The sale price of a property is $10 000 000. It can be bought with a deposit of 15% and a

mortgage.

Calculate:

a) the deposit

b) the mortgage

3. The selling price of an outboard engine is $2 700 000. A deposit of 25% of the sale price

must be made before obtaining the mortgage.

Calculate the:

a) deposit.

b) Mortgage.

c) total amount that will be due to the bank if monthly instalments of $13 889 will be

made over a period of three years.

d) Interest on the mortgage.

4. A businessman borrows $240 000 from a bank. The bank charges 15% interest for the

entire period of the loan. If repayments are 12 monthly instalments, calculate the amount

of each monthly instalment to the nearest dollar.

5. Mr. Melville borrows $3 200 000 from a bank at 15% interest over the entire period of

the loan. If the loan plus the interest is to be repaid in 24 monthly instalments, calculate

the amount of each instalment to the nearest dollar.

6. A condominium is on sale for $2 750 000. It is possible to buy the condominium by

making a 10% deposit and taking a bank mortgage.

Calculate:

a) the deposit.

b) the amount borrowed.


c) The total amount to be paid to the bank, if monthly payments of $28 000 are made

over a 10-year period.

7. A bungalow is priced at $950 000. A 90% mortgage can be obtained over a 3-years

period.

Determine:

a) the deposit.

b) the amount of the loan required.

c) the total amount paid to the bank if each monthly instalment was $29 000.

d) the actual amount paid for the house.

8. A luxury apartment is priced at $2 350 000. An 85% mortgage can be obtained over a 20-

year period.

Calculate:

a) the deposit payable.

b) the loan amount needed.

c) the total amount of money paid to the bank if each monthly payment was $28 290.

d) the total amount paid for the house


WEEK 2

LESSON 1

Topic: Consumer Arithmetic – Simple Interest

The interest paid to a bank is called simple interest, providing the principal (amount borrowed)

in calculating the interest remains the same during the period of the loan.

The borrower will have to repay the interest plus the principal.

Principal, P – amount of money borrowed

Rate, R – percentage rate charged by the bank

Time, T – period for the loan to be repaid

𝑆𝑖𝑚𝑝𝑙𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 ×𝑅𝑎𝑡𝑒 ×𝑇𝑖𝑚𝑒

100

𝐼 = 𝑃𝑅𝑇

100

Example 1.

Mrs. Lord borrowed $100 000 from a bank at a simple interest rate of 8% per annum for 3

years.

a) What amount is the simple interest payable?

b) Calculate the amount accruing for the loan.

c) Determine the amount of each monthly instalment.

Solution

a) The simple interest payable, 𝐼 = 𝑃𝑅𝑇

100

= $100 000 ×8 ×3

100

=$24 000

b) Amount accruing, 𝐴 = 𝑃 + 𝐼

= $100 000 + $24 000

= $124 000


c) The amount of each instalment = 𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑎𝑐𝑐𝑟𝑢𝑖𝑛𝑔

𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑛𝑡ℎ𝑠

= $124 000

36

= $3 444.44

Exercises

1. $960 000 is invested 8% per annum simple interest for 5 years.

a) What is the amount of simple interest payable?

b) Calculate the amount accruing for the investment.

2. Mr. Ford invested $1 240 000 in a bank at 7.25% per annum simple interest for 6 years.

Calculate:

a) the interest he paid.

b) the total of amount of money he would have received at the end of the period of the

investment.

3. Mrs. Ricky borrowed $534 000 from a bank at 9.5% per annum simple interest for 5

years.

a) the sum of money paid in interest to the bank.

b) the total amount of money repaid to the bank.

c) the value of each monthly instalment.

4. Mr. Isaacs borrowed $368 000 from the bank for 6 months at 8.75% per annum.

Calculate:

a) the sum of money Mr. Isaacs had to pay the bank as simple interest

b) the total amount of money repaid to the bank

c) the amount of money paid monthly

5. A television set can be bought on hire purchase by paying down a down payment of

$85 000 and 18 monthly instalments of $10 900 each. Calculate the hire purchase price.

The television set can be bought for cash by taking a loan for 12 months at 11% per

annum simple interest. If the cash price is 90% of the hire purchase price, calculate the

amount payable to the bank each month.


WEEK 2

LESSON 2

Topic: Consumer Arithmetic – Simple Interest (continued)

Example 1.

a) The simple interest on a sum of money invested for 6 months at 5% per annum is $16

800. Determine the amount of money invested.

b) The simple interest on $850 000 invested for 31

2 years is $38 675. Calculate the rate

percent per annum.

c) The simple interest on $540 000 invested at 8.75% per annum is $189 000. Determine the

period of the investment.

Solution

a) 𝐼 = 𝑃𝑅𝑇

100

$16 800 = 𝑃 ×0.5 ×5

100

$16 800 ×100

0.5 ×5= 𝑃

𝑃 = $1 680 000

2.5

= $672 000

b) 𝐼 = 𝑃𝑅𝑇

100

$38 675 = $850 000 × 3.5 ×𝑅

100

$38 675×100

$850 000 × 3.5= 𝑅

𝑃 = $3 867 500

$2 975 000

= $672 000

c) 𝐼 = 𝑃𝑅𝑇

100


$189 000 = $540 000 × 𝑇 ×8.75

100

$189 000×100

$540 000 × 8.75= 𝑇

𝑇 = $18 900 000

$4 725 000

= 4 year

Exercises

1. Mr. Jerome borrowed $126 000 from a bank for 5 years and paid $17 000 simple interest.

Calculate the rate percent per annum that he had to pay the bank.

2. Calculate the number of years in which $560 000 will earn $55 000 interest at 4% per

annum.

3. The simple interest on $200 000 invested at 8% per annum is $21 500. Calculate the

number of years for which the sum was invested.

4. Mr. Kallicharan took a loan from the bank at 11.25% per annum for 9 months and paid

$270 000 simple interest. Calculate the amount of money Mr. Kallicharan borrowed from

the bank.

5. Calculate the rate percent per annum if $576 000 simple interest is paid when $1 280 000

is invested for 6 years.

6. $855 000 simple interest was paid when $4 500 000 was invested at 4.75% per annum.

Determine the period of investment.

7. Calculate the number of years in which $560 000 will earn $112 000 simple interest at

4% per annum.

8. Mrs. Kanhai borrowed a sum of money from the bank for 9 months at 13.5% per annum

and paid $88 600 simple interest. Determine the sum of money Mrs. Kanhai borrowed

from the bank.


WEEK 2

LESSON 3

Topic: Consumer Arithmetic – Compound Interest

When money is invested, it is called the principal. Each year the principal achieves an

interest. If the interest payable is reinvested at the end of each year in the same fixed deposit,

then the principal at the beginning of each New Year is greater than the principal of the

previous year. Thus, the interest payable at the end of each New Year is greater than the

interest paid at the end of the previous year. Money invested in this way is set to attract

compound interest. (interest attracts interest)

The compound interest formula:

Compound Interest, C. I = A – P

Where:

A = the amount of money accruing after n years

P = the principal

R = the rate percent per annum

n = the number of years for which the money was invested

Example 1.

Calculate the compound interest earned and the amount accruing when $100 000 is invested at

8% per annum for 3 years.

Solution

𝐴 = 𝑃(1 + 𝑅

100)𝑛

= $100 000(1 + 8

100)3

= $100 000(1 + 0.08)3

= $100 000( 1.08)3

𝐴 = 𝑃(1 + 𝑅

100)𝑛


= $125 971.20

The amount accruing after 3 years is $125 971.20

The compound interest

C. I = A – P

= $125 971.20 – $100 000

= $25 971.20

Exercises

1. Calculate the compound interest on investing $60000 for 2 years at 7% per annum.

2. Calculate the compound interest on investing $850 000 for 3 years at 5 per annum.

3. Which is a better investment?

$1 200 000 at 9% simple interest for 2 years

Or $1 200 000 at 8% compound interest for 2 years.

4. $800 000 was put in a fixed deposit account on 1st January, 2019 for 1 year. The rate of

interest was 7.5% per annum. On 1st January, 2020, the total amount was reinvested for a

further 1 year at 7% per annum. Calculate the final amount received at the end of the

year.

5. A man wishes to invest $350 000. He can buy saving bonds which pay simple interest at

the rate of 12% per annum or he can start a savings account which pays compound

interest at the same rate.

Calculate the difference in the amounts of the two investments at the end of three years.

6. A man placed $115 000 in a fixed deposit account for 5 years at 8% per annum.

a) Calculate the total amount received at the end of the period under compound interest.

b) Determine the total amount received at the end of the period under simple interest.

c) State the difference in the interest received.

7. A woman wishes to invest $2 500 000. She can purchase savings bonds which pays

simple interest at the rate of 7% per annum or she can start a savings account which pays

compound interest at the same rate. Calculate the difference between the amounts of the

two investments at the end of 6 years.

8. Bill wishes to invest $5 700 000. He can buy savings bonds which pays simple interest at

a rate of 8.5% per annum or he can start savings account which pays compound interest

at the same rate.

Calculate the difference in amounts of the two investments at the end of 5 years.

9. Mr. Roland invests $950 000 in a bank for 3 years and receives simple interest at 7.5%

per annum. If he invests his money in bonds for the same period, he will receive

compound interest at 4.5% per annum. Calculate the interest Mr. Roland received in both

cases. State which investment is the better one and the difference in interest received.

10. Calculate the simple interest and compound interest earned when $140 000 is invested for

5 years at 7% per annum. State which investment is better and give a reason for youe

answer.


WEEK 2

LESSON 4

Topic: Consumer Arithmetic – Depreciation

Many assets such as cars, mini bus, speed boat engine, furniture, etc. tend to decrease in value

over time as a result of wear and tear due to constant use. It is a common practice to reassess

such values annually. The loss in value is referred to as depreciation.

Depreciation is calculated as a percentage of the value of the article at the beginning of each

year.

The depreciation formula:

Compound Interest, D = P – A

Where:

A = the book value after n years

P = the initial cost of the asset

R = the rate of depreciation per annum

n = the number of years for which the asset was depreciated

Example 1.

A small vehicle is bought for $2 500 000. The insurance company calculates the depreciation of

the vehicle at a rate of 25% per annum. What will be the value, to the nearest dollar, of the

vehicle at the end of 3 years?

Solution

Cost of vehicle = $2 500 000

Depreciation during the 1st year = 25% of cost

= 25

100 × $2 500 000

𝐴 = 𝑃(1 − 𝑅

100)𝑛


= $625 000

Value of vehicle at the beginning of 2nd year = cost price – depreciation

= $2 500 000 – $625 000

= $1 875 000

Depreciation during the 2nd year = 25

100 × $1 875 000

= $468 750

Value of vehicle at the beginning of 3rd year = $1 875 000 – $468 750

= $1 406 250

Depreciation during the 3rd year = 25

100 × $1 406 250

= $351 562. 50

Value of vehicle at the beginning of 4th year = value of vehicle after 3 years

= $1 406 250 – $351 562.50

= $1 054 687.50

Using the formula

The value after 3 years, 𝐴 = 𝑃(1 − 𝑅

100)𝑛

= $2 500 000(1 − 25

100)3

= $2 500 000(1 − 0.25)3

= $2 500 000(0.75)3

= $1 054 686.50

Exercises

1. A motor cycle is bought for $400 000. The insurance company calculates the depreciation

of the motor cycle at 15% per annum. What will be the value of the motor cycle at the

end of 3 years?

2. Mr. Bacchus buys a suite for $260 000. If its value depreciates at the rate of 10% per

annum, what will be the value of the suite at the end of 3 years?

3. How much will a car, which cost $4 250 000, depreciate after the first year if the

insurance company calculates the depreciation at a rate of 12%?


4. The value of a computer depreciates each year by 20% of its value at the beginning of the

year. If the purchase price was $230 000, what will be its value at the end of 4 years?

5. A property is bought for $8 000 000.it depreciates in value at a rate of 10% each year.

What will be the value of the property at the end of 2 years?

6. A man buys a portion of land for $500 000. The land depreciates in value at the rate of 2

% per annum.

a) What will be the value of the land at the end of 4 years?

b) What was the total amount depreciated over the period?

7. A lathe was bought for $98 000. The insurance company decides to calculate depreciation

each year at 5.5% of the book value at the beginning of the year. Calculate the book value

at the end of 10 years.

8. A man bought a car for $6 000 000. After using it for two years, he decides to trade-in the

car. The company estimates a depreciation of 15% for the first year of its use and a

further 15% on its reduced value for the second year.

a) Calculate the value of the car after 2 years.

b) Express the value of the car after 2 years as a percentage of the original value.

c) Express the depreciation after the 2-years period as a percentage of the original value.


WEEK 3

LESSON 1

Topic: Algebra 2 - Factorization of Quadratic expressions

𝑻𝒉𝒆 𝑮𝒆𝒏𝒆𝒓𝒂𝒍 𝒇𝒐𝒓𝒎 𝒐𝒇 𝒂 𝒒𝒖𝒂𝒅𝒓𝒂𝒕𝒊𝒄 𝒆𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒊𝒔 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄

Examples of quadratic expressions:

1. 𝑥2 + 7𝑥 + 6

Here: a = 1, b = 7 and c = 6

2. 3𝑥2 + 5𝑥 + 2

a = 3, b = 5 and c = 2

3. 𝑥2 − 9

a = 1, b = 0 and c = -9

Example 1.

Factorize completely 𝑥2 + 5𝑥 + 6

Solution

We multiply the coefficient of 𝑥2 ‘a’ by the constant ‘c’

I.e. ac = 1 × 6 = 6

We write out the factors of 6 and find the sum of each pair

Factors of 6 Sum of factors

6 and 1 7

-6 and -1 -7

3 and 2 5

-3 and -2 -5

Choose the pair of factors whose sum is equal to the coefficient of x. In this case the coefficient

of x is 5 so we choose the factors 3 and 2.

We rewrite our algebraic expression using these two terms.

𝑥2 + 5𝑥 + 6


= 𝑥2 + 3𝑥 + 2𝑥 + 6

= (𝑥2 + 3𝑥) + (2𝑥 + 6)

= 𝑥(𝑥 + 𝑥) + 2(𝑥 + 3)

= (𝑥 + 3)(𝑥 + 2)

Example 2.

Factorize the expression 𝑚2 − 2𝑚 − 8

𝑚2 − 2𝑚 − 8 𝑎𝑐 = 1 × −8 = −8

= 𝑚2 − 4𝑚 + 2𝑚 − 8

= (𝑚2 − 4𝑚) + (2𝑚 − 8)

= 𝑚(𝑚 − 4) + 2(𝑚 − 4)

= (𝑚 − 4)(𝑚 + 2)

Example 3.

Factorize the expression 6𝑡2 ∓ 7𝑡 − 3

6𝑡2 + 7𝑡 − 3 𝑎𝑐 = 6 × −3 = −18

= 6𝑡2 + 9𝑡 − 2𝑡 − 3

= (6𝑡2 + 9𝑡) − (2𝑡 + 3)

Leave the negative sign outside the brackets and

change the signs inside

= 3𝑡(2𝑡 + 3) − 1(2𝑡 + 3)

= (2𝑡 + 3)(3𝑡 − 1)

Factor of ac (-8) sum of factors

Must be = b (-2)

-8 and 1 -7

8 and -1 7

-4 and 2 -2

4 and -2 2

Factor of ac (-8) sum of factors

Must be = b (-7)

-18 and 1 -17

18 and -1 17

-9 and 2 -7

9 and -2 7

-6 and 3 -3

6 and -3 3


Exercises

Factorize the following quadratic expressions completely:

1. 𝑦2 + 6𝑦 + 5

2. 𝑎2 + 14𝑎 + 13

3. 7𝑥2 + 15𝑥 + 2

4. 12 + 8𝑦 + 𝑦2

5. 4𝑦2 + 10𝑦 − 6

6. 8𝑥2 + 13𝑥 − 6

7. 𝑥2 + 11𝑥 − 60

8. 𝑥2 + 𝑥 − 6

9. 6𝑥2 + 2𝑥 − 4

10. 𝑥2 − 5𝑥 + 6

11. 11𝑏2 + 14𝑏 + 3

12. 𝑥2 − 11𝑥 + 30

13. 𝑡2 + 9𝑡 − 10

14. 𝑝2 − 2𝑝 − 15

15. 8𝑥2 − 3𝑥 − 5


WEEK 3

LESSON 2

Topic: Algebra 2 - Factorization of Quadratic expressions that can be expressed as perfect

squares

𝑇ℎ𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 (𝑎 + 𝑏)2 𝑖𝑠 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒

When expanded, we get 𝒂𝟐 + 𝟐𝒂𝒃 + 𝒃𝟐

Similarly (𝑎 − 𝑏)2 𝑖𝑠 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒

When expanded, we get 𝒂𝟐 − 𝟐𝒂𝒃 + 𝒃𝟐

To factorize a perfect square we follow the same procedure as previous lesson, then we write our

factors of the expression like (𝑎 + 𝑏)2 𝑜𝑟 (𝑎 − 𝑏)2.

Example 1.

Factorize each quadratic expression below completely.

a) 𝑥2 + 2𝑥 + 1

b) 𝑥2 − 6𝑥 + 9

c) 9𝑥2 + 30𝑥 + 25

Solutions

𝑎) 𝑥2 + 2𝑥 + 1 𝑎𝑐 = 1 × 1 = 1

= 𝑥2 + 𝑥 + 𝑥 + 1

= (𝑥2 + 𝑥) + (𝑥 + 1)

= 𝑥(𝑥 + 1) + 1(𝑥 + 1)

= (𝑥 + 1)(𝑥 + 1)

= (𝑥 + 1)2

𝑏) 𝑥2 − 6𝑥 + 9 𝑎𝑐 = 1 × 9 = 9

= 𝑥2 − 3𝑥 − 3𝑥 + 9

= (𝑥2 − 3𝑥) − (3𝑥 − 9)

= 𝑥(𝑥 − 3) − 3(𝑥 − 3)

= (𝑥 − 3)(𝑥 − 3)

= (𝑥 − 3)2

Factor of ac (1) sum of factors

Must be = b (2)

1 and 1 2


Must be = b (-6)

9 and 1 10

-9 and -1 -10

3 and 3 6

-3 and -3 -6


𝑐) 9𝑥2 + 30𝑥 + 25 𝑎𝑐 = 9 × 25 = 225

= 9𝑥2 + 15𝑥 + 15𝑥 + 25

= (9𝑥2 + 15𝑥) + (15𝑥 + 25)

= 3𝑥(3𝑥 + 5) + 5(3𝑥 + 5)

= (3𝑥 + 5)(3𝑥 + 5)

= (3𝑥 + 5)2

Exercises

Factorize the following quadratic expressions completely:

1. 𝑥2 + 10𝑥 + 25

2. 𝑎2 + 8𝑎 + 16

3. 𝑥2 + 2𝑥 + 1

4. 64 + 16𝑦 + 𝑦2

5. 4𝑦2 + 4𝑦 + 1

6. 4𝑥2 + 12𝑥 + 9

7. 𝑥2 − 8𝑥 + 16

8. 𝑥2 + 2𝑥𝑦 + 𝑦2

9. 25𝑥2 + 40𝑥 + 16

10. 𝑥2 + 12𝑥 + 36

11. 64𝑏2 + 80𝑏 + 25

12. 81𝑥2 + 126𝑥 + 49

13. 𝑡2 + 18𝑡 + 81

14. 64𝑝2 + 144𝑝 + 81


Must be = b (30)

After writing all the factors we realize:

15 and 15 30


WEEK 3

LESSON 3

Topic: Algebra 2 - Factorization of Quadratic expressions that can be expressed as

difference of squares

The product of (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏2

Hence 𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)

Example 1.

𝑥2 − 36

Solution

𝑥2 − 36 = 𝑥2 − 62

= (𝑥 + 6)(𝑥 − 6)

Example 2.

Factorize the expression completely:

144 − 25𝑦2

Solution

144 − 25𝑦2 = 122 − (5𝑦)2

= (12 + 5𝑦)(12 − 5𝑦)

Example 3.

Factorize completely:

36𝑥2 − 1

Solution

36𝑥2 − 1 = (6𝑥)2 − 12

= (6𝑥 + 1)(6𝑥 − 1)


Exercises

Factorize each completely

1. 𝑦2 − 1

2. 𝑥2 − 22

3. 𝑥2 − 4𝑦2

4. 𝑥2 − 25𝑏2

5. 36 − 9𝑚2

6. 36𝑎2 − 169𝑥2

7. 81𝑥2 − 49𝑦2

8. 49𝑟2 − 16𝑠2

9. 144𝑥2 − 49

10. 64𝑥2 − 25

11. 81𝑎2 − 64𝑏2

12. 1 − 16𝑥2

13. 49𝑥2 − 1

14. (𝑎 + 𝑏)2 − 22

15. (𝑎 − 𝑏)2 − 1


WEEK 3

LESSON 4

Topic: Algebra 2 – Changing the subject of the formula

A formula is really an equation that tells the relationship between two quantities. For instance,

𝑨 = 𝒍𝒃 , A is the subject of the formula because it is written in front of the formula.

This formula can be rearranged to make either l or b the subject.

Example 1.

In 𝑨 = 𝒍𝒃, make l the formula

𝑨 = 𝒍𝒃

𝑨

𝒃=

𝒍𝒃

𝒃

𝑨

𝒃= 𝒍

𝒍 =𝑨

𝒃

Example 2.

Given 𝑣 = 𝜋𝑟2ℎ, make:

a) h the subject of the formula

b) r the subject of the formula

Solution

a) 𝑣 = 𝜋𝑟2ℎ

𝑣

𝜋𝑟2 =𝜋𝑟2ℎ

𝜋𝑟2

𝑣

𝜋𝑟2 = ℎ

h =𝑣

𝜋𝑟2

b) 𝑣 = 𝜋𝑟2ℎ

𝑣

𝜋ℎ=

𝜋𝑟2ℎ

𝜋ℎ

𝑣

𝜋ℎ= 𝑟2

√𝑟2 = √𝑣

𝜋𝑟2


𝑟 = √𝑣

𝜋𝑟2

Example 3.

Given = 𝑡 +𝑥

𝑦 , make x the subject of this formula.

Solution

𝑀 = 𝑡 +𝑥

𝑦

𝑀 − 𝑡 = 𝑡 +𝑥

𝑦− 𝑡

(𝑀 − 𝑡) × 𝑦 =𝑥

𝑦× 𝑦

(𝑀 − 𝑡)𝑦 = 𝑥

𝑥 = 𝑦(𝑀 − 𝑡)

Example 4.

Make c the subject of the formula in the relation 𝑏 =7

2𝑐+5

Solution

𝑏 =7

2𝑐+5

𝑏(2𝑐 + 5) =7

2𝑐+5× 2𝑐 + 5

𝑏(2𝑐 + 5) = 7

2𝑏𝑐 + 5𝑏 = 7

2𝑏𝑐 + 5𝑏 − 5𝑏 = 7 − 5𝑏

2𝑏𝑐

2𝑏=

7−5𝑏

2𝑏

𝑐 =7−5𝑏

2𝑏


Exercises

1. Given 𝐴 = 𝜋𝑟𝑙, 𝑚𝑎𝑘𝑒 𝑙 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

2. Given 𝑃 =𝑅𝑇

𝑉, 𝑚𝑎𝑘𝑒 𝑇 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

3. Given 𝐼 =𝑃𝑅𝑇

100, 𝑚𝑎𝑘𝑒 𝑇 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

4. Given 𝑉 =𝜋𝑑2ℎ

4, 𝑚𝑎𝑘𝑒 𝑑 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

5. Given 𝑉 = 𝑢 + 𝑎𝑡, 𝑚𝑎𝑘𝑒 𝑡 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

6. Given 𝑇 =𝐴

𝑅−𝑟, 𝑚𝑎𝑘𝑒 𝑟 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

7. Given 𝑄 =𝑅(𝐵−𝑇)

𝐵, 𝑚𝑎𝑘𝑒 𝐵 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

8. Given 𝑃 − 𝑚𝑔 =𝑀𝑣2

𝑟, 𝑚𝑎𝑘𝑒:

a) v the subject of the formula.

b) m the subject of the formula.

9. Given 𝑚2 =𝑎+𝑏

𝑎−𝑏, 𝑚𝑎𝑘𝑒:

c) b the subject of the formula.

d) a the subject of the formula.

10. Given 𝐴 =ℎ(𝑎+𝑏)

2, 𝑒𝑥𝑝𝑟𝑒𝑠𝑠 𝑏 𝑎𝑠 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

11. 𝑅 = √𝑝𝑚−𝑞

𝑞+𝑐𝑚, make m the subject of the formula.

12. Given 𝑛 =1

𝑑2 ,make d the subject of the formula.

13. 𝑉 = √𝑎2−𝑥2, make x the subject of the formula.

14. Given that 𝑃 = 𝑟2ℎ, make h the subject of the formula.

15. If =𝑏𝑟𝑡

𝑣−𝑏 , make r the subject of the formula.


WEEK 4

LESSON 1

Topic: Algebra 2 – Solution of Simultaneous Equations by the method of Substitution

The method of Substitution

Example 1.

Solve the equations 𝑥 + 2𝑦 = 4 𝑎𝑛𝑑 2𝑥 + 3𝑦 = 7

Solution

Step 1. Rewirte the equations

𝑥 + 2𝑦 = 4 …………eq(1)

2𝑥 + 3𝑦 = 7 …….....eq(2)

Step 2. Since the coefficient of one of the variables in the equations is 1 express one unknown in

terms of the other

Using eq 1

𝑥 + 2𝑦 = 4

𝑥 = 4 − 2𝑦 …………eq(3)

Step 3. Substitute for x in eq(2) 4-2y and solve for y

2𝑥 + 3𝑦 = 7

2(4 − 2𝑦) + 3𝑦 = 7

8 − 4𝑦 + 3𝑦 = 7

−𝑦 = 7 − 8

−𝒚 = −𝟏

𝒚 = 𝟏

Step 4. Substitute for y in eq(3)

𝑥 = 4 − 2𝑦

𝑥 = 4 − 2(1)

𝑥 = 2

Hence 𝑥 = 2, 𝑦 = 1

Note: It is better to use this method when the coefficient of one of the variable is equal to 1.


Exercises

Solve the following pairs of simultaneous equations by method of substitution.

1. 𝑥 − 𝑦 = 1

𝑥 + 𝑦 = 3

2. 𝑦 − 𝑥 = 1

𝑥 + 𝑦 = 5

3. 𝑦 = 4𝑥 + 4

𝑥 + 𝑦 = 14

4. 𝑎 = 4 − 𝑏

2𝑎 + 𝑏 = 1

5. 5𝑥 − 2𝑏 = 1

𝑥 = 5 − 2𝑏

6. 𝑥 − 𝑦 = 4

𝑥 + 2𝑦 = 10

7. 𝑎 − 2𝑏 = 1

3𝑎 − 4𝑏 = 1

8. 4𝑥 − 3𝑦 = 1

𝑥 − 2𝑦 = 4

9. 𝑎 + 2𝑝 = 4

𝑥 − 2𝑦 = 6

10. 2𝑥 + 3𝑦 = 5

−𝑥 + 2𝑦 = 8

11. 𝑦 − 𝑥 = 2

2𝑦 − 4𝑥 = 10

12. 𝑥 + 6𝑦 = 22

2𝑥 + 10𝑦 = 38

13. 2𝑥 − 𝑦 = 5

𝑥 + 𝑦 = 4

14. 𝑥 + 𝑦 = 7

2𝑥 + 𝑦 = 10

15. 𝑥 + 𝑦 = 144

2𝑥 − 3𝑦 = 63


WEEK 4

LESSON 2

Topic: Algebra 2 – Solution of Simultaneous Equations by the method of Substitution

The method of Substitution (where none of the coefficients of the variable is equal to 1)

Example 1.

Solve the equations 5𝑥 + 2𝑦 = 16 𝑎𝑛𝑑 − 3𝑥 + 4𝑦 = 6

Solution


5𝑥 + 2𝑦 = 16 …………eq(1)

−3𝑥 + 4𝑦 = 6 …….....eq(2)

Step 2. Make x the subject of eq(1)

Using 5𝑥 + 2𝑦 = 16 ……..eq (1)

𝑥 =16−2𝑦

5… … … … . . 𝑒𝑞(3)

Step 3. Substitute for x in eq(2) 4-2y and solve for y

−3𝑥 + 4𝑦 = 6

−3(16−2𝑦

5) + 4𝑦 = 6

(−48+6𝑦

5) + 4𝑦 = 6

(−48+6𝑦

5) × 5 + 4𝑦 × 5 = 6 × 5

−48 + 6𝑦 + 20𝑦 = 30

26𝑦 = 78

𝑦 =78

26

𝒚 = 𝟑

Step 4. Substitute for y in eq(3)

𝑥 =16−2𝑦

5

𝑥 =16−2(3)

5

𝑥 =16−6

5


𝒙 = 𝟐

Hence 𝑥 = 2, 𝑦 = 3

Exercises

Solve the following pairs of simultaneous equations by method of substitution.

1. 4𝑥 + 3𝑦 = 17

5𝑥 − 2𝑦 = 4

2. 3𝑥 − 5𝑦 = −13

−2𝑥 + 3𝑦 = 8

3. 5𝑥 + 3𝑦 = 27

2𝑥 + 5𝑦 = 26

4. 2𝑥 + 3𝑦 = −8

5𝑥 − 2𝑦 = 18

5. 9𝑥 + 5𝑏 = 15

3𝑥 − 2𝑏 = −6

6. 3𝑥 + 2𝑦 = 21

2𝑥 + 3𝑦 = 13

7. 7𝑥 − 2𝑦 = 19

3𝑥 + 5𝑦 = 14

8. 5𝑥 + 2𝑦 = 16

−3𝑥 + 4𝑦 = 6

9. 7𝑥 + 2𝑦 = 17

2𝑥 − 2𝑦 = 1

10. 3𝑥 − 2𝑦 = −1

4𝑥 + 7𝑦 = 18

11. 3𝑥 + 4𝑦 = 27

5𝑥 − 2𝑦 = 19

12. 7𝑥 + 5𝑏 = 20.15

5𝑥 + 7𝑏 = 18.85


WEEK 4

LESSON 3

Topic: Algebra 2 – Solution of Simultaneous Equations by the method of Elimination

Procedure 1.

- Check to see which of the variables has the same coefficient

- If the signs of that variable are the same, subtract the equations; if the signs of that

variable are not the same, add the equations.

Example 1.

Solve the equations 3𝑥 + 𝑦 = 18 𝑎𝑛𝑑 2𝑥 − 𝑦 = 7

Solution 2.


3𝑥 + 𝑦 = 18 …………eq(1)

2𝑥 − 𝑦 = 7 …….....eq(2)

Step 2. Coefficients of y are the same; signs are different so we will add the equations

3𝑥 + 𝑦 = 18

2𝑥 − 𝑦 = 7

5𝑥 = 25

Then: 𝑥 =25

5

𝑥 = 5

Step 3. Substitute for x in eq(1) and solve for y (any equation could have worked, it’s your

choice)

3𝑥 + 𝑦 = 18

3(5) + 𝑦 = 18

15 + 𝑦 = 18

𝑦 = 18 − 15

𝒚 = 𝟑

Hence 𝑥 = 5, 𝑦 = 3


Exercises

Solve the following pairs of simultaneous equations by method of elimination.

1. 𝑥 − 𝑦 = 1

𝑥 + 𝑦 = 3

2. 𝑦 − 𝑥 = 1

𝑥 + 𝑦 = 5

3. 𝑦 = 4𝑥 + 4

𝑥 + 𝑦 = 14

4. 𝑎 = 4 − 𝑏

2𝑎 + 𝑏 = 1

5. 5𝑥 − 2𝑏 = 1

𝑥 = 5 − 2𝑏

6. 𝑥 − 𝑦 = 4

𝑥 + 2𝑦 = 10

7. −4𝑥 − 6𝑦 = 7

4𝑥 + 𝑦 = −2

8. 4𝑥 − 2𝑦 = 6

𝑥 + 2𝑦 = 4

9. 𝑥 + 2𝑦 = 3

2𝑥 − 2𝑦 = 6

10. 2𝑥 − 2𝑦 = 8

2𝑥 + 2𝑦 = 20

11. 4𝑥 − 3𝑦 = −4

2𝑥 − 3𝑦 = 4

12. 2𝑥 + 4𝑦 = 5

10𝑥 + 4𝑦 = 52


WEEK 4

LESSON 4

Topic: Algebra 2 – Solution of Simultaneous Equations by the method of Elimination

Procedure 1.

- If the coefficients of the variables are not the same, choose which variable you want to

eliminate.

- Make the coefficients of that variable the same by multiplying the first equation by the

coefficient of that variable in the second equation then multiply the second equation by

the coefficient of that variable in the first equation.

- If the signs of that variable are the same, subtract the equations; if the signs of that

variable are not the same, add the equations.

Example 1.

Solve the equations 5𝑥 + 3𝑦 = 21 𝑎𝑛𝑑 2𝑥 + 7𝑦 = 20

Solution 2.


5𝑥 + 3𝑦 = 21 …………eq(1)

2𝑥 + 7𝑦 = 20 …….....eq(2)

Step 2. Getting rid of x

Multiply eq(1) by coefficient of x in eq(2) and multiply eq(2) by coefficient of x in eq(1)

2(5𝑥 + 3𝑦 = 21) …………eq(3)

5(2𝑥 + 7𝑦 = 20) ………….eq(4)

So we get

10𝑥 + 6𝑦 = 42 …………eq(3)

10𝑥 + 35𝑦 = 100………eq(4)

Step 3. Signs are the same so we will subtract the equations

10𝑥 + 6𝑦 = 42

- (10𝑥 + 35𝑦 = 100)

So we get −29𝑦 = −58

𝑦 =−58

−29


𝑦 = 2

Step 3. Substitute for y in eq(1) and solve for x (any equation could have worked, it’s your

choice)

5𝑥 + 3𝑦 = 21

5𝑥 + 3(2) = 21

5𝑥 + 6 = 21

5𝑥 = 21 − 6

5𝑥 = 15

𝑥 =15

5

𝒙 = 𝟑

Hence 𝑥 = 3, 𝑦 = 2

Exercises

Solve the following pairs of simultaneous equations by method of elimination.

1. 4𝑥 + 3𝑦 = 17

5𝑥 − 2𝑦 = 4

2. 3𝑥 − 5𝑦 = −13

−2𝑥 + 3𝑦 = 8

3. 5𝑥 + 3𝑦 = 27

2𝑥 + 5𝑦 = 26

4. 2𝑥 + 3𝑦 = −8

5𝑥 − 2𝑦 = 18

5. 9𝑥 + 5𝑏 = 15

3𝑥 − 2𝑏 = −6

6. 3𝑥 + 2𝑦 = 21

2𝑥 + 3𝑦 = 13

7. 7𝑥 − 2𝑦 = 19

3𝑥 + 5𝑦 = 14

8. 5𝑥 + 2𝑦 = 16

−3𝑥 + 4𝑦 = 6

9. 7𝑥 + 2𝑦 = 17

2𝑥 − 2𝑦 = 1

Mathematics Grade 9

WEEK 5

LESSON 1

Topic: Algebra 2 – Solution of Simultaneous Equations – word problems

Example 1.

The sum of the digits of a two-digit number is 12. The number is 13 times the tens digit. Find the

number.

Solution

in any two-digit number, we will have a numeral in the singles column and another in the tens

column giving rise to the number represented by 10t + s

Let t = the tens digit

s = the singles digit

Since the sum of digits is 12

That give the equation 𝒕 + 𝒔 = 𝟏𝟐 …………….eq(1)

So 𝟏𝟎𝒕 + 𝒔 = 𝟏𝟑 …………….eq(2)

Using the substitution method

Finding s in eq(1) gives

𝒕 = 𝟏𝟐 − 𝒔……………….eq(3)

Sub. for t in eq(2)

𝟏𝟎𝒕 + 𝒔 = 𝟏𝟑𝒕

𝟏𝟎(𝟏𝟐 − 𝒔) + 𝒔 = 𝟏𝟑(𝟏𝟐 − 𝒔)

𝟏𝟐𝟎 − 𝟏𝟎𝒔 + 𝒔 = 𝟏𝟓𝟔 − 𝟏𝟑𝒔

−𝟗𝒔 + 𝟏𝟑𝒔 = 𝟏𝟓𝟔 − 𝟏𝟐𝟎

𝟒𝒔 = 𝟑𝟔

𝒔 =𝟑𝟔

𝟒

𝒔 = 𝟗 Sub for s in eq(3)

𝒕 = 𝟏𝟐 − 𝒔

𝒕 = 𝟏𝟐 − 𝟗

𝒕 = 𝟑

Hence the number is 39.

Exercises

1. A two-digit number is 6 times its singles digit. The sum of the digits is 6. Find the

number.

2. A boy has five-dollar coins and y ten-dollar coins. There are 8 coins altogether and their

total value is $55. How many of each coins do he have?

3. A bill of $3 400 was paid with $20 notes and $100 notes. Ninety notes were used. How

many $20 notes and $100 notes were used?

4. The sum of two numbers is 38 and their difference is 10. Find the numbers.

5. The sum of two numbers is 11. Twice the first number minus the second number is 7.

Find the numbers.

6. The difference between two numbers is 4. Twice the first number plus three times the

second number is 28. Find the two numbers.

7. Find the cost of a pencil and eraser if two pencils and three erasers cost $140, and five

pencils and four erasers cost $280.

8. A pencil and pen cost $200. Six pencils cost the same as two pens. Find the cost of a

pencils and a pen.

9. Karen and Kate have 50 CDs between them. Karen has 6 less than Kate. How many each

has?

10. Ben is 7 years older than his sister. The sum of their ages is 31 years. What are their

ages?

11. A store clerk sold 25 Mathematics books and 10 English books for a total of $85 500. If

she sold 10 Mathematics books and 40 English books, she would have gotten $13 500

more. Calculate the price of each type of book.

12. The cost of two juice and three patties is $1 750. While the cost of four juice and three

patties is $3 050. Form a pair of simultaneous equations and solve them to determine:

a) the cost of a juice.

b) the cost of a patty.

WEEK 5

LESSON 2

Topic: Relation and Function – Identifying a Function

Relation is a set of ordered pairs (x, y) and a function is a relation in which no two pairs have

the same x-value (domain). It is a rule that assigns a single x-value to each y-value.

What is a function?

1. Arrow diagram – one and only one arrow must leave each element of the domain.

1 3

2 5

3 7

2. A set of ordered pairs – no two ordered pairs must have the same first element.

e.g. a) (2,3), (4,1), (7,1)

b) (1,2), (1,3), (5,2), (6,4) is not a function. The ordered pairs (1,2) and (1,3)

have the same first element ‘1’.

3. A graph – if a vertical line passes through two or more points of the graph, then the

relation is not a function. In a function, vertical lines pass through only one point of the

graph.

i.e.

-4

-2

0

2

4

6

8

10

-3 -2 -1 0 1 2 3 4 5

y=2x+1

Observe that each line only passes through one point on the graph.

𝑦 = 𝑥2 − 3

Again the vertical line only passes through one point on the curve.

Exercise

Using a scale of 1cm to represent 2units on the x-axis and y-axis, draw the graph of each of the

following sets of ordered pairs. Say which is a function.

a) (0,1),(2,2), (4,3), (6,4)

b) (1,3), (3,3), (5,3)

c) (1,2), (2,3), (3,4), (4,5)

d) (-2.-2). (-2,1), (-2,1), (-2,5)

e) (0,3), (1,4), (2,5), (4,7)

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

WEEK 5

LESSON 3

Topic: Relation and Function – Linear Functions (identifying a linear function)

All the points of a linear function lie of the same straight line. (a linear function is a straight

line)

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑦 = 2𝑥 + 1,

Notice that all the points { (-2, -3), (-1,-1), (0,1), (1,3), (2,5), (3,7) (4,9)} are lying on the straight

line.

Exercises

State which is a linear function.

a)

-4

-2

0

2

4

6

8

10

-3 -2 -1 0 1 2 3 4 5

y=2x+1

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6

b)

c)

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6

0

5

10

15

-4 -3 -2 -1 0 1 2 3 4

Chart Title

WEEK 5

LESSON 4

Topic: Relation and Function – Use of functional notation

We use symbols to describe a function. For example, 𝒇: 𝒙 → 𝟑𝒙. This states that ‘𝑓 is a function

such that x is mapped onto 3𝑥’. Here we have been supplied with the rule which must be used in

order to determine the image of any given value for ‘ 𝑥’.

The function 𝒇: 𝒙 → 𝟑𝒙 could also be expressed as 𝒇(𝒙) = 𝟑𝒙 or 𝒚 = 𝟑𝒙 which means the same.

Example 1.

Given 𝒇: 𝒙 → 𝟑𝒙 + 𝟐 find the image of 3

Solution

𝑓(𝑥) = 3𝑥 + 2

𝑓(2) = 3(2) + 2

= 8

Example 2.

Given

x f(x)

0 1

1 2

2 3

3 4

Use functional notation to state the rule

Solution

𝑥 𝑥 + 1 𝑓(𝑥) 0 0+1 1

1 1+1 2

2 2+2 3

3 3+1 4 𝑓: → 𝑥 + 1

or

𝑓(𝑥) = 𝑥 + 1

or

𝑦 = 𝑥 + 1

Exercises

1. Given 𝑓: → 𝑥 + 1, find the image of 2

2. Given 𝑓: → 3𝑥 − 1, find the images for the domain {1,2,3,4,5}

3. Given 𝑓(𝑥) = 𝑥2, find:

a) 𝑓(−1)

b) 𝑓(3)

4. Copy and complete the table

𝑥 1 2 3 4 5 6 7 8

𝑓(𝑥) -2 0 3 4 5

From the table write down

a) The rule of 𝑓(𝑥)

b) The values 𝑓(7)

c) The image of 5

d) The first element whose image is 5

5. a) Given 𝑓(𝑥) = 𝑥 − 1, for what value of 𝑥 will 𝑓(𝑥) = 10

6. Given

x f(x)

2 8

4 14

6 20

8 26

10 32

Use functional notation to state the rule.

WEEK 6

LESSON 1

Topic: Relation and Function – Graph of a linear function

When plotting the graph of a linear function, we should find at least 3 ordered pairs which satisfy the rule

of the function.

If the elements of the domain are not given, we can use any 3 values for ‘x’.

Example 1.

Using the domain x={-2, 0, 2} plot the graph of the function 𝑦 = 2𝑥 − 1

Solution

First, we find the corresponding y value for the domain

We can do this on a table as shown below

X 2x-1 y (x, y)

-2 2(-2)-1 -5 (-2, -5)

0 2(0)-1 -1 (0, -1)

2 2(2)-1 3 (2, 3)

𝑦 = 2𝑥 − 1

Exercises

Draw the graph for each function, using a scale of 1cm to represent 1 unit on both axis.

a) 𝑦 = 𝑥 − 5

b) 𝑓(𝑥) = 2𝑥 − 1

c) 𝑓: 𝑥 → 2𝑥 − 1

d) 𝑓: 𝑥 → 5𝑥 + 2

e) 𝑦 = −2𝑥 + 3

-6

-5

-4

-3

-2

-1

0

1

2

3

4

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

-6

-5

-4

-3

-2

-1

0

1

2

3

4

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

WEEK 6

LESSON 2

Topic: Relation and Function – Finding the gradient of a straight line

Between two points, a line rises a vertical distance and covers a horizontal distance. This is given as the

slope or gradient of the line. The gradient of a line is the ratio of the increase in the vertical rise to the

increase horizontal run.

i.e. 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑟 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒, 𝑚 = 𝑟𝑖𝑠𝑒

𝑟𝑢𝑛

Example 1.

Run=8 units (2, 3)

Rise= 8 units

(-2, -5)

𝑚 = 8

8= 1

Using the formula:

𝑻𝒉𝒆 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕, 𝒎 = 𝒚𝟐−𝒚𝟏

𝒙𝟐−𝒙𝟏

Considering (-2, -5) and (2, 3)

𝑚 = 𝑦2−𝑦1

𝑥2−𝑥1

𝑚 = 3−−5

2−−2

𝑚 = 8

8= 1

Exercises

1. Identify 2 points on each of the following lines and calculate the gradient:

a)

b)

2. Using a scale of 1cm to represent I unit on both axes, plot each pair of points on graph paper and

join the points. Determine the gradient of each line.

a) (4, 6) and (7, 10)

b) (-5, 2) and (4, 4)

3. Determine the gradient of the line passing through the points (2, 9) and (-7, 4).

4. Determine the value of q so that the points (q,3) and (5,9) passing through the line gives the

gradient 2.

-4

-3

-2

-1

0

1

2

3

4

5

6

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

-8

-6

-4

-2

0

2

4

6

8

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

WEEK 6

LESSON 3

Topic: Relation and Function – Equation of a straight line

𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑓𝑜𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦: 𝒚 = 𝒎𝒙 + 𝒄

Where: ‘m’ is the gradient

And ‘c’ is the y-intercept (where the line cuts the y-axis)

Take, for example:

The line above cuts the y-axis at 1, then the y-intercept c = 1

To find the equation of a specific line we need two points that passes through the line.

Procedure:

- Find the gradient

- Use one set of coordinates to find c using the general form of the equation for a straight line

𝒚 = 𝒎𝒙 + 𝒄 ; substitute the x and y values in the equation

- Write the equation replacing m and c

Example 1.

Given the gradient of a line is 3 and the y-intercept is 4, write down the equation of this line.

Solution

𝑦 = 𝑚𝑥 + 𝑐

m = 3 and c = 4

Then the equation of this line is

𝑦 = 3𝑥 +4

-4

-3

-2

-1

0

1

2

3

4

5

6

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

Example 2.

Given the line 𝒚 = 𝟐𝒙 − 𝟏, write down the gradient of this line and the y-intercept.

Solution

𝒚 = 𝒎𝒙 + 𝒄

Since the equation is given as: 𝒚 = 𝟐𝒙 − 𝟏

Then:

m =2

c = -1

Exercises

1. State the gradient and the y-intercept for each of the following equations:

a) 𝒚 = 𝟒𝒙 + 𝟏

b) 𝒚 = −𝟐𝒙 − 𝟗

c) 𝒚 = 𝒙 +𝟏

𝟐

d) 𝒚 = −𝒙 − 𝟓

e) 𝒚 = 𝟐𝒙

f) 𝒚 = 𝒙

g) 𝒚 = 𝟓

h) 𝒚 = −𝟐

𝟑𝒙 +

𝟏

𝟒

2. Write the specific equation for each straight line given:

a) 𝑚 = 4 𝑎𝑛𝑑 𝑐 = 2

b) 𝑚 = −2 𝑎𝑛𝑑 𝑐 = −1

c) 𝑚 = 1 𝑎𝑛𝑑 𝑐 =2

3

d) 𝑚 = −9 𝑎𝑛𝑑 𝑐 = 0

e) 𝑚 = 0 𝑎𝑛𝑑 𝑐 = 2

f) The gradient is 3 and the y-intercept is 5.

3. Find the equation of the line that passes through the point (1, 2) with gradient 3.

WEEK 6

LESSON 4

Topic: Relation and Function – Finding the equation of a straight line using two given

points

From our previous lesson:

To find the equation of a specific line we need two points that passes through the line.

Procedure:

- Find the gradient

- Use one set of coordinates to find c using the general form of the equation for a straight line

𝒚 = 𝒎𝒙 + 𝒄 ; substitute the x and y values in the equation

- Write the equation replacing m and c

Example 1.

Find the equation of the line that passes through the points (4, 3) and (6, 7)

Solution

Gradient, 𝒎 = 𝒚𝟐−𝒚𝟏

𝒙𝟐−𝒙𝟏

𝒎 = 𝟕−𝟑

𝟔−𝟒

𝒎 = 𝟒

𝟐

m = 2

now finding c, we apply the second procedure given above

using (4, 3)

we get; 3 = 2(4) + 𝑐

3 = 8 + 𝑐

3 − 8 = 𝑐

𝑐 = −5

Finally, we rewrite the equation

The general form of the equation is 𝒚 = 𝒎𝒙 + 𝒄

Then the specific equation is given by 𝒚 = 𝟐𝒙 − 𝟓

Exercises

1. Determine the equation of the line that passes through each pair of points:

a) (5, 1) and (6, 0)

b) (3, 8) and (-6, 6)

c) (-2, 4) and (2, -4)

d) (3, -2) and (-2, 9)

2. The straight line 𝒚 = 𝒎𝒙 + 𝒄 passes through the points (-2, -3) and (4,9). Determine the values of

m and c and hence write the particular equation that represents the straight line.

3. Calculate the values of m and c if the straight line 𝒚 = 𝒎𝒙 + 𝒄 passes through the points (-3, -2)

and (1, 6)

4. The coordinates of A and B are (4, 7) and (6, 3) respectively. Determine the particular equation of

the straight line.

5. Plot the points L(3, 6) and M(9, 8) on graph paper. From the graph, determine:

a) the gradient of LM

b) the intercept of LM produced on the y-axis

c) Hence, state the particular equation of the straight line passing through the given points

6. Using a scale of 1cm to represent 1 unit on each axis, plot on graph paper the points A(-3, 2) and

B(3, -2).

From your graph,

a) calculate the gradient of AB

b) state the point where AB meets the y-axis.

c) Write down the equation of AB in the form 𝒚 = 𝒎𝒙 + 𝒄

WEEK 7

LESSON 1

Topic: Relation and Function – Using graphs to solve linear simultaneous equations

Example 1.

Solve graphically, the pair of simultaneous equations:

2𝑥 + 𝑦 = 8 and 3𝑥 − 𝑦 = 7

Solution

Step 1: Rewrite the equations

2𝑥 + 𝑦 = 8 ……eq(1)

3𝑥 − 𝑦 = 7 ……eq(2)

Step 2: make y the subject of the formula in both equations (i.e. rewrite the equations in the form

𝑦 = 𝑚𝑥 + 𝑐)

We get: 𝑦 = −2𝑥 + 8

𝑦 = 3𝑥 − 7

Step 3: construct a table of values for each equation (Assume values for x).

𝑦 = −2𝑥 + 8

x -2x+8 y (x,y)

-2 -2(-2) +8 12 (-2,12)

0 -2(0) +8 8 (0,8)

2 -2(2) +8 4 (2,4)

𝑦 = 3𝑥 − 7

x 3x-7 y (x,y)

-2 3(-2) -7 -13 (-2, -13)

0 3(0) -7 -7 (0, -7)

2 3(2) -7 -1 (2, -1)

Step 4:

- Draw axes and choose suitable scale

- Plot points for both equations on same graph and draw lines, extending until they meet

- Name the coordinates of the point of intersection.

The graph is shown below

The coordinates are (3, 2)

So, the solution of the equations is x=3, y=2

Exercises

Solve the following pairs of simultaneous equations graphically.

1. 𝑥 − 𝑦 = 1

𝑥 + 𝑦 = 3

2. 𝑦 − 𝑥 = 1

𝑥 + 𝑦 = 5

3. 𝑦 = 4𝑥 + 4

𝑥 + 𝑦 = 14

4. 𝑦 = 4𝑥 𝑎𝑛𝑑 𝑥 + 𝑦 = 5

5. 2𝑥 + 𝑦 − 5 = 0 𝑎𝑛𝑑 𝑦 = 𝑥 − 4

6. 3𝑥 + 𝑦 = 0 𝑎𝑛𝑑 4𝑥 + 𝑦 = 2

7. 4𝑥 = 𝑦 − 2 𝑎𝑛𝑑 3𝑥 − 𝑦 = 8

8. 𝑥 + 2𝑦 = 7 𝑎𝑛𝑑 𝑦 − 2𝑥 − 1 = 0

-15

-10

-5

0

5

10

15

-3 -2 -1 0 1 2 3 4 5

WEEK 7

LESSON 2

Topic: Relation and Function – Graphs of linear inequalities

Example 1.

Draw the graph of 𝑦 < 𝑥 + 2

Solution

Write equation as 𝑦 = 𝑥 + 2, assume points for x in order to find their corresponding y values.

x x+2 y (x,y)

-2 -2+ -2 -4 (-2, -4)

0 -2+0 -2 (0, -2)

2 -2+2 0 (2, -0)

Plot points

Notice that the line is drawn broken, the line is drawn bold when 𝒚 ≤ 𝒐𝒓 ≥.

Then we shade the region (the arrows indicate the shaded part)

Notice also the arrows are not touching the broken lines that represents 𝑦 < 𝑥 + 2

Example 2.

Draw the graph of 2𝑥 + 𝑦 ≤ 7

0

1

2

3

4

5

6

7

-3 -2 -1 0 1 2 3 4 5

Solution

Write equation as 𝑦 = −2𝑥 + 7, assume points for x in order to find their corresponding y values.

x -2x+7 y (x,y)

-2 -2(-2) +7 11 (-2, 11)

0 -2(0) +7 7 (0, 7)

2 -2(2) +7 3 (2, 3)

Plot points

Notice that the line is drawn bold.

Then we shade the region (the arrows indicate the shaded part)

Notice also the arrows are touching the line that represents 𝑦 ≤ −2 + 7

Exercises

Draw graphs to represent each of the following inequations.

a) 𝑥 + 𝑦 > 5

b) 𝑥 + 𝑦 < 5

c) 𝑦 > 𝑥 + 4

d) 𝑦 < 𝑥 + 4

e) 𝑥 + 2𝑦 ≥ 8

f) 2𝑥 + 𝑦 ≤ 5

g) 𝑦 ≤ 5

-2

0

2

4

6

8

10

12

-3 -2 -1 0 1 2 3 4 5

WEEK 7

LESSON 3

Topic: Geometry – Pythagoras’ Theorem

Pythagoras’ Theorem is used to find for the missing side of a right-angled triangle.

Consider the right-angled triangle

A

Hypotenuse (the side opposite the right angle)

b c

C a B

The Pythagoras’ Theorem states that the square of the hypotenuse of any right-angled triangle is equal to

the sum of the squares of the other two sides.

i.e. 𝑐2 = 𝑎2 + 𝑏2

As demonstrated below:

Example 1.

Find the length of the missing side.

C

6cm

A 8cm B

Note: Here we see that the side BC is the hypotenuse.

Then:

𝑩𝑪𝟐 = 𝑨𝑩𝟐 + 𝑨𝑪𝟐

Which is the same as 𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐

Solution

𝑎2 = 𝑏2 + 𝑐2

𝑎2 = (6𝑐𝑚)2 + (8𝑐𝑚)2

𝑎2 = 36𝑐𝑚2 + 64𝑐𝑚2

𝑎2 = 100𝑐𝑚2

𝑎 = √100𝑐𝑚2

𝑎 = 10𝑐𝑚

Exercises

1. Find the missing side for each triangle below correct to the nearest whole number.

A P 12.5cm Q

2m 4cm M 3.2cm N

C 4m R

11cm

X

Q

4cm

R S 3.6cm

11.25cm

Y Z

2.5cm

WEEK 7

LESSON 4

Topic: Geometry – Pythagoras’ Theorem (continued)

Example 1.

Find the length of the missing side.

C

x cm 10cm

A 8cm B

Note:

Here we see that the side BC is the hypotenuse.

Then:

𝑩𝑪𝟐 = 𝑨𝑩𝟐 + 𝑨𝑪𝟐

Which is the same as 𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐

Solution

𝑎2 = 𝑏2 + 𝑐2

(10𝑐𝑚)2 = 𝑏2 + (8𝑐𝑚)2

100𝑐𝑚2 = 𝑏2 + 64𝑐𝑚2

100𝑐𝑚2 − 64𝑐𝑚2 = 𝑏2

𝑏2 = 36𝑐𝑚2

𝑏 = √36𝑐𝑚2

𝑏 = 6𝑐𝑚

Exercises

1. Find the missing side for each triangle below correct to the nearest whole number.

A P 12.5cm Q

5m 4cm M 13.2cm N

C 4m R

11.1cm O

Q X

4cm 10cm

R S 6.6cm

Y Z

2.5cm

Mixed problems

2. A vertical tower AB which is 15.6m high was built on level ground. The distance of a point C

on the ground from the base of the tower is 9.5m. Calculate the distance from the top of the

tower to the point C.

3. In the triangle below, AB = 5cm, angle A = angle B = 45˚ and the altitude is h.

A 5cm B

h

C

a) Determine the value of h.

b) Hence, calculate the length of AC.

4. The height of a vertical lamp post XY which was placed in level ground is 10.5m. the

distance from the top of the lamp post X to a point Z on the ground is 15.4m. calculate the

distance of the point Z from the base of the lamp post.

5. L

13cm

M P N

18cm

In triangle LMN above, MN = 18cm, LP is perpendicular to MN. LP = 13cm and angle

PLN = angle PNL

a) State the length of pn. Give a reason for your answer.

b) Calculate the length of

(i) LM

(ii) LN

WEEK 8

LESSON 1

Topic: Geometry – Similar triangles

Equi-angular triangles are said to be similar. Similar triangles are said to have the same shape.

To identify similar triangles, there are some properties we look for. These are below.

The knowledge of the properties of similar triangles can help you to solve problems involving triangles.

Example 1.

In the figure below, MN is parallel to QR, PM = 5cm, MQ = 2cm and QR = 6 cm.

a) Prove that the triangles PMN and PQR are similar.

b) Calculate the length of MN

The figure PQR.

P

5cm

M N

2cm

Q 6cm R

Solution

a) In the triangles PMN and PQR, angle PMN = angle PQR (corresponding angles, MN is parallel

to QR)

Angle PNM = angle PRQ (corresponding angles, MN is parallel to QR)

P is common

The triangles are equi-angular; therefore, they are similar.

b) 𝑃𝑀

𝑃𝑄 =

𝑃𝑁

𝑃𝑅=

𝑀𝑁

𝑄𝑅

5𝑐𝑚

7𝑐𝑚=

𝑀𝑁

6𝑐𝑚

𝑀𝑁 = 30

7

𝑀𝑁 = 4.3𝑐𝑚

Exercises

1. P W

15cm p cm

X 9cm Y

Q 12cm R

The triangles above, not drawn to scale, are similar. Calculate the length of the side marked p cm.

2. In the figure below OMN and OAB are similar. MN = 4cm OM = 4.5cm and AB =6cm. calculate

the length of OA.

O

4.5cm

M 4cm N

A 6cm B

3. E 3cm F

2cm

O

G 8cm H

In the figure above, not drawn to scale, EOF and GOH are similar triangles. EF = 3cm, GH = 8cm and

FO = 2cm. Calculate the length of OG.

WEEK 8

LESSON 2

Topic: Geometry – Congruent triangles

Sometimes, two or more triangles are exactly the same. When this occur, these triangles are said to be

congruent triangles.

Below are the properties of congruent triangles.

Example 1.

B Z 9cm X

3cm 5cm 5cm 3cm

A 9cm C Y

Given the triangles ABC and XYZ above, prove whether or not the two triangles are congruent.

Solution

AB =XY =3cm (corresponding sides equal)

BC = YZ =5cm (corresponding sides equal)

AC = XZ = 9cm (corresponding sides equal)

Hence, triangle ∆𝐴𝐵𝐶 ≡ ∆𝑋𝑌𝑍

Exercises

Prove whether or not each pair of triangles is congruent.

1. A M

10cm 6cm

N 10cm O

B 6cm C

2. X

P 4cm

4cm 8cm 15cm

Q R Y

15cm

8cm

Z

3. A

B D C

In the figure above, ∆𝐴𝐵𝐶 is such that BD = CD. Prove that AB = AC.

WEEK 8

LESSON 3

Topic: Geometry – Angles in a circle

Before we can go on to the above topic, we must remember the parts of a circle.

Theorem 1

The angle at the centre of a circle is twice the angle at the circumference standing on the same

arc (or chord).

Here we notice that the angle at the centre is 180° so the angle at the centre is half of that.

Example 1.

O

C B

A

If angle AOB = 70°, then find angle ACB.

Solution

Since angle AOB is subtended by the arc AB at the centre O and angle ACB is the angle

subtended by AB at the circumference,

Then angle AOB = 2 angle ACB

Since angle AOB = 70°

Then angle ACB = 1

2(70°)

= 35°

Exercises

1.

O

C 20˚ B

A

Find the size of the angle AOB.

2.

M

O

105˚ N

P

Find the size of the angle PMN

3.

X Y

Z

Justify that angle XZY is a right angle.

WEEK 8

LESSON 4

Topic: Geometry – Angles in a circle (continued)

Theorem 2

Angles in the same segment are equal

Example 1.

C

D

O

A

B

If angle ACB = 59˚, determine the size of the angle ADB. State a reason for your answer.

Solution

Given that angle ACB =59˚

Then angle ADB = angle ACB = 59˚

(angles in the same segment)

Exercises

1. Calculate the size of the angles x and y, giving reasons for your answer.

58˚ x

y

37˚

2. If angle AQB = 21˚ , determine the magnitude of angle APB, stating a reason for your

answer.

A

B

Q

P

3.

44˚

20˚

Given the two angles in the figure above, determine the remaining four angles stating a

reason for each.

WEEK 9

LESSON 1

Topic: Trigonometry – Trigonometric Ratios

Trigonometric Ratios (Trig. Ratios) are used to solve for missing angles or missing sides of a

right-angled triangle.

Trig. Ratios are comparison of two sides of a right-angled triangle to its angle.

A

Adjacent to angle A α

b c

Opposite to angle B Hypotenuse

β

C a B

Opposite to side A

Adjacent to side B

Sine of an angle

𝑻𝒉𝒆 𝒔𝒊𝒏𝒆 𝒐𝒇 𝒂𝒏 𝒂𝒏𝒈𝒍𝒆 =𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆

𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆

𝒔𝒊𝒏 𝜽 = 𝒐𝒑𝒑

𝒉𝒚𝒑

i.e. From triangle above

𝒔𝒊𝒏 𝑨 =𝒂

𝒄 Or 𝒔𝒊𝒏 𝑩 =

𝒃

𝒄

Example 1.

Using a scientific calculator, find:

a) sin 30˚ b) sin 45˚ c) sin 60˚ d) sin 75˚

Solution

a) sin 30˚ = 0.5

b) sin 45˚ = 0.7071 067811865 = .07071 (4 s.f)

c) sin 60˚ = 0.8660 (4 s.f)

d) sin 75˚ = 0.9659 (4 s.f)

The inverse sine of an angle

𝒔𝒊𝒏−𝟏𝒙 𝒐𝒓 𝒂𝒓𝒄 𝒔𝒊𝒏𝒙

Example 2.

Using a scientific calculator, evaluate the following to determine the angle:

a) sin A = 0.429 b) B = arc sin 0.905

Solution

a) sin A = 0.429

𝐴 = 𝑠𝑖𝑛−10.429

A = 25.4˚

b) B = arc sin 0.905

B = 64.82˚

Exercises

1. Using a scientific calculator, find the sine for each of the following:

a) sin 8˚ c) sin 24˚

b) sin 16˚ d) sin 48˚


a) sin 53˚ c) sin 63˚

b) sin 74˚ d) sin 87˚


a) sin 23.2˚ c) sin 67.6˚

b) sin 7.4˚ d) sin 81.9˚


a) sin−1 0.235 c) sin−1 0.799

b) arc sin 0.435 d) arc sin 0.9134

5. Using a scientific calculator, find the sine for each of the following unknown angles,

given that:

a) sin a = 0.602 c) sin β = 0.883

b) sin A = 0.786 d) sin C = 0.7387

WEEK 9

LESSON 2

Topic: Trigonometry – Trigonometric Ratios (Sine Ratio)

Calculating an unknown angle using Sine Ratio

Example

b 15cm

8cm

Find the magnitude of the angle marked b.

Solution

𝑠𝑖𝑛 𝜃 = 𝑜𝑝𝑝

ℎ𝑦𝑝

𝑠𝑖𝑛 𝑏 = 8𝑐𝑚

15𝑐𝑚

𝑠𝑖𝑛 𝑏 = 0.5333

angle b = sin−1 0.5333

angle b = 32.23˚

Exercises

Determine the size of the marked angle

y

4cm 6.2cm

8cm 15.5cm

x

b

13cm 6cm 12cm

7cm

a

11.5cm

WEEK 9

LESSON 3

Topic: Trigonometry – Trigonometric Ratios (Sine Ratio)

Finding a missing side using the Sine Ratio

Example

40˚ 15cm

x cm

Find the length of the side marked x.

Solution


ℎ𝑦𝑝

𝑠𝑖𝑛 40° = 𝑥 𝑐𝑚

15𝑐𝑚

sin 40° × 15𝑐𝑚 = 𝑥𝑐𝑚

x cm = 0.64278 × 15cm

x cm = 9.63cm (3 s.f)

Exercises


29.2˚

b 6.2cm

8cm y

52˚

m 6cm 12cm

7cm

50˚ 9˚

WEEK 9

LESSON 4

Topic: Trigonometry – Trigonometric Ratios (Cosine Ratio)

A


b c


β

C a B

Opposite to side A

Adjacent to side B

Cosine of an angle

𝑻𝒉𝒆 𝒄𝒐𝒔𝒊𝒏𝒆 𝒐𝒇 𝒂𝒏 𝒂𝒏𝒈𝒍𝒆 =𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆

𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆

𝒔𝒊𝒏 𝜽 = 𝒂𝒅𝒋

𝒉𝒚𝒑


𝒄𝒐𝒔 𝑨 =𝒃

𝒄 Or 𝒄𝒐𝒔 𝑩 =

𝒂

𝒄

Example 1.


a) cos 30˚ b) cos 45˚ c) cos 60˚ d) cos 75˚

Solution

a) cos 30˚ = 0.8660 (4s.f)

b) cos 45˚ = 0.7071 (4 s.f)

c) cos 60˚ = 0.5

d) cos 75˚ = 0.9659 (4 s.f)

The inverse cosine of an angle

𝒄𝒐𝒔−𝟏𝒙 𝒐𝒓 𝒂𝒓𝒄 𝒄𝒐𝒔𝒙

Example 2.


a) cos A = 0.429 b) B = arc cos 0.905

Solution

a) cos A = 0.429

𝐴 = 𝑐𝑜𝑠−10.429

A = 64.6˚

b) B = arc cos 0.905

B = 25.18˚

Exercises


a) cos 8˚ c) cos 24˚

b) cos 16˚ d) cos 48˚


a) cos 53˚ c) cos 63˚

b) cos 74˚ d) cos 87˚


a) cos 23.2˚ c) cos 67.6˚

b) cos 7.4˚ d) cos 81.9˚


a) cos−1 0.235 c) cos−1 0.799

b) Arc cos 0.435 d) arc cos 0.9134


given that:

a) cos a = 0.602 c) cos β = 0.883

b) cos A = 0.786 d) cos C = 0.7387

WEEK 10

LESSON 1


Calculating an unknown angle using Cosine Ratio

Example

b 15cm

6cm


Solution

𝑐𝑜𝑠 𝜃 = 𝑎𝑑𝑗

ℎ𝑦𝑝

𝑐𝑜𝑠 𝑏 = 6𝑐𝑚

15𝑐𝑚

𝑐𝑜𝑠 𝑏 = 0.4

angle b = cos−1 0.4

angle b = 66.42˚

Exercises


6.2cm

8cm y 15.5cm

x

3cm

a b 4.1cm

13cm 6cm 12cm

WEEK 10

LESSON 2


Finding a missing side using the Cosine Ratio

Example

40˚ 15cm

x cm


Solution


ℎ𝑦𝑝

𝑐𝑜𝑠 40° = 𝑥 𝑐𝑚

15𝑐𝑚

cos 40° × 15𝑐𝑚 = 𝑥𝑐𝑚

x cm = 0.7660 × 15cm

x cm = 11.49cm (4 s.f)

Exercises


32.1˚

8cm 6.2cm

52˚ y

b

29˚

m 6cm c

7cm

9˚

WEEK 10

LESSON 3

Topic: Trigonometry – Trigonometric Ratios (Tangent Ratio)

A


b c


β

C a B

Opposite to side A

Adjacent to side B

Tangent of an angle

𝑻𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒐𝒇 𝒂𝒏 𝒂𝒏𝒈𝒍𝒆 =𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒔𝒊𝒅𝒆

𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆

𝒕𝒂𝒏 𝜽 = 𝒐𝒑𝒑

𝒂𝒅𝒋


𝒕𝒂𝒏 𝑨 =𝒂

𝒃 Or 𝒕𝒂𝒏 𝑩 =

𝒃

𝒂

Example 1.


a) tan 30˚ b) tan 45˚ c) tan 60˚ d) tan 75˚

Solution

a) tan 30˚ = 0.5774 (4s.f)

b) tan 45˚ = 1

c) tan 60˚ = 1.732(4 s.f)

d) tan 75˚ = 3.732 (4 s.f)

The inverse tangent of an angle

𝒕𝒂𝒏−𝟏𝒙 𝒐𝒓 𝒂𝒓𝒄 𝒕𝒂𝒏 𝒙

Example 2.


a) tan A = 0.429 b) B = arc tan 2.905

Solution

a) tan A = 0.429

𝐴 = 𝑡𝑎𝑛−10.429

A = 23.22˚

b) B = arc cos 2.905

B = 71˚

Exercises


a) tan 8˚ c) tan 24˚

b) tan 16˚ d) tan 48˚


a) tan 53˚ c) tan 63˚

b) tan 74˚ d) tan 87˚


a) tan 23.2˚ c) tan 67.6˚

b) tan 7.4˚ d) tan 81.9˚


a) tan−1 0.235 c) tan−1 1.799

b) arc tan 0.435 d) arc tan 2.434


given that:

a) tan a = 0.602 c) tan β = 3.803

b) tan A = 0.786 d) tan C = 1.2367

WEEK 10

LESSON 4


Calculating an unknown angle using Tangent Ratio

Example

b 15cm

6cm

13.75cm


Solution

𝑡𝑎𝑛 𝜃 = 𝑜𝑝𝑝

𝑎𝑑𝑗

𝑡𝑎𝑛 𝑏 = 13.75𝑐𝑚

6𝑐𝑚

𝑡𝑎𝑛 𝑏 = 2.292

angle b = tan−1 2.292

angle b = 66.43˚

Exercises


6.2cm y 4.3cm

8cm

x

3cm 6.3cm

a c

6cm 3.1cm

9cm


MINISTRY OF EDUCATION

EASTER TERM 2021

GRADE 9

MATHEMATICS

WORKSHEETS


WEEK 11

LESSON 1


Finding a missing side using the Tangent Ratio

Example

41.4˚

x cm

13.2cm


Solution


𝑎𝑑𝑗

𝑡𝑎𝑛 40° = 𝑥 𝑐𝑚

13.2𝑐𝑚

tan 40° × 13.2𝑐𝑚 = 𝑥𝑐𝑚

x cm = 0.8816 × 13.1cm

x cm = 11.64cm (4 s.f)

Exercises


x

29.1˚

6cm 5.3cm

62˚

b

29˚ p

6cm

7cm

9˚

m


WEEK 11

LESSON 2

Topic: Trigonometry – Trigonometric Ratios (mixed problem)

Acronym for remembering ratios

𝑺𝑶𝑯 𝑪𝑨𝑯 𝑻𝑶𝑨


𝑎𝑑𝑗


ℎ𝑦𝑝


ℎ𝑦𝑝

OR

S – Some

O – old Sine

H – horses

C – Can

A – always Cosine

H – hear

T – Their

O – owner’s Tangent

A – approach

Mixed word problems involving Trig. Ratios

To solve these problems, you need to draw an illustration of the problem. In these cases, you will

get a right-angled triangle. Then you decide on which ratio is suited to solve the problem. In

some cases, more than one ratio can be used to solve a problem.


Exercises

1. Determine the angle of a slope of a road if a man after walking 30m up the road rise 3m

vertically.

2. Calculate the height of a kite if the string, 100m long is inclined at an angle of 32˚.

3. A ladder 50m long leans against a wall and is inclined at an angle of 45˚ to the ground.

What is the distance between the foot of the ladder and the wall?

4. The ladder in question 3 is adjusted and the distance between the foot of the ladder and

the wall is now 22m. What angle does the foot of the ladder now makes with the ground?

5. L and M are directly opposite each other on the bank of a river. N is on the same bank as

L at a distance of 100m from L. if LM subtends an angle of 38.5˚ at M, what is the

distance of MN?

6. A ladder rest with its foot on the horizontal ground and its top against a vertical wall. The

ladder makes an angle of 49.5˚ with the ground and its foot is 5m from a wall. How far

above the ground is the top of the ladder.

7. L R M

P Q

In the diagram above, PQ and LM represent parallel edges of an east-west river bank.

Angle PRQ = 90˚

Given that PR = RQ = 5m, calculate:

a) The size of the angle RPQ

b) The width of the river

c) The distance PQ


WEEK 11

LESSON 3

Topic: Trigonometry – Complementary Angles

Complementary angles are angles whose sum is 90˚

Example

A

b

c

B a C

From the triangle above, 𝐴 + 𝐶 = 90°

Then 𝐶 = 90° − 𝐴

Considering:

𝒔𝒊𝒏 𝑨 =𝒂

𝒃

And 𝒄𝒐𝒔 𝑪 = 𝒄𝒐𝒔(𝟗𝟎 − 𝑨) =𝒂

𝒃

Thus we can conclude that the sine of an angle is equal to the cosine of its

complementary angle and vice versa.

Example

Given that sin 25˚ = 0.423, determine what cos 65˚ is equal to. State a conclusion.

Solution

Cos 65˚ = 0.4226 = 0.423 (3 s.f)

We can conclude that 25˚ plus 65˚ gives 90˚ which make them complementary. Hence the sine of

one is equal to the cosine of the other.

Exercises

1. Given that sin 27˚ = 0.454, without the use of calculator determine the value of cos 63˚.

2. Given that sin 59˚ = 0.857, without the use of calculator determine the value of cos 31˚.

3. Compare the cosine for each below:

a) sin 34˚ b) sin 40˚ c) sin 60˚ d) sin 30 e) sin 45˚ f) sin 90˚


WEEK 11

LESSON 4

Topic: Trigonometry – Angle of elevation

The angle of elevation is a widely used concept related to height and distance, especially in

trigonometry. It is defined as an angle between the horizontal plane and oblique line from the

observer’s eye to some object above his eye. Eventually, this angle is formed above the surface.

As the name itself suggests, the angle of elevation is so formed that it is above the observer’s eye.

For example, an observer is looking at a bird sitting at the rooftop, then there is an angle

formed, which is inclined towards the bird from the observer’s eye. This elevation angle is used

in finding distances, heights of buildings or towers, etc with the help of trigonometric ratios,

such as sine, cosine and tangent.

Example 1.

The angle of elevation of the top of a vertical tree from a man standing on level ground 25m

from the base of the tree is 38.5˚.

a) Draw an illustration of the above problem

b) Calculate the height of the tree.

Solution

a) T top of the tree

38.5˚

Man’s eye level X 25m B base of the tree

https://byjus.com/maths/trigonometric-ratios/


b) Considering the ∆𝑇𝐵𝑋:


𝑎𝑑𝑗

Then: tan 𝑋 = 𝑇𝐵

𝐵𝑋

𝑡𝑎𝑛 38.5° = 𝑇𝐵

25𝑚

𝑡𝑎𝑛 38.5° × 25𝑚 = 𝑇𝐵

𝑇𝐵 = 0.7954 × 25𝑚

=19.88m

=20m (nearest metre)

Exercises

1. A sailor sights the top of a cliff at an angle of elevation of 12˚. He knows that the height

of the cliff is about 90m. above sea level.

a) Draw an illustration of the information above.

b) Calculate his distance from the base of the cliff to the nearest metre.

2. From a point P on the ground, which is 100m from the foot of a church tower, the angle

of elevation of the top of the tower is 50˚.

a) Sketch an illustration of the above information.

b) Calculate the height of the tower.

3. From a point P on ground level which is 50m for the foot of the front of a school, the

angle of elevation is 42.5˚. Use a scale of 1cm to represent 10m to make a scale drawing.

Use your scale drawing to calculate the height of the school.

4. The height of an electric post is 15m standing vertical from the ground. A boy standing a

30m from the base of the post spots a bird on top the post. Calculate the angle of

elevation from looking at the bird.


WEEK 12

LESSON 1

Topic: Trigonometry – Angle of depression

The angle of depression is formed when the observer is higher than the object, he/she is looking

at. When an observer looks at an object that is situated at a distance lower than the observer, an

angle is formed below the horizontal line drawn with the level of the eye of the observer and line

joining object with the observer’s eye.

Example 1.

The angle of depression of the top of a vertical tree to a stone on level ground, 25m from the

base of the tree is 32˚.

a) Draw an illustration of the above problem

b) Calculate the height of the tree.

Solution

The horizontal line WT is parallel to the level ground XB

a) W T top of the tree

32˚

32˚

Man’s eye level X 25m B base of the tree


b) Considering the ∆𝑇𝐵𝑋:


𝑎𝑑𝑗

Then: tan 𝑋 = 𝑇𝐵

𝐵𝑋

𝑡𝑎𝑛 32° = 𝑇𝐵

25𝑚

𝑡𝑎𝑛 32° × 25𝑚 = 𝑇𝐵

𝑇𝐵 = 0.6249 × 25𝑚

=15.6m

=16m (nearest metre)

Exercises

1. From the top of a cliff a man sights a boat at an angle of depression of 22˚. He knows that

the height of the cliff is about 90m above sea level.

a) Draw an illustration of the information above.

b) Calculate his distance from the base of the cliff to the nearest metre.

2. From a coastal lookout point P on the ground, which is 100m above sea level, a sailor

sights a boat at an angle of depression of 27˚.

a) Sketch an illustration of the above information.

b) Calculate the horizontal distance of the boat from the sailor.

3. An instrument in an aircraft flying at height of 400m measures the angle of depression of

the beginning of the runway as 25˚. Calculate the horizontal distance of the aircraft from

the runway.

4. A bird siting at the top of a post looks down at a cat on the ground. The cat is sitting 20m

from the foot of the post. Calculate the height of the post.

5. A marksman spots a car 200m away from the foot of the building he is lying on at an

angle of depression of 19˚. Calculate the height of the building.


WEEK 12

LESSON 3

Topic: Statistics – Measures of Central tendency

Sometimes there is need to use a single value which can represent or characterize a group (set of

data) as a whole. This single value is called a statistical average (or measure of Central

Tendency).

Three common statistical averages or measure of central tendency that we need to know are:

1. The arithmetical mean or simply called the mean for short.

2. The median

3. The mode

The mean – the average value for a set of data

𝑇ℎ𝑒 𝑚𝑒𝑎𝑛, 𝑥 ̅ = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑐𝑜𝑟𝑒𝑠

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑐𝑜𝑟𝑒𝑠

𝑥 ̅ = ∑ 𝑥

∑ 𝑓

Where: ∑ 𝑥 – sum of scores

∑ 𝑓 - number of scores (sum of frequency)

Example 1. Given the set of tests scores 2, 3, 1, 4, 6, 2, 5, 7, 2, 4, calculate the mean value.

Solution

𝑥 ̅ = ∑ 𝑥

∑ 𝑓

𝑥 ̅ = 2+3+1+4+6+2+5+7+2+4

10

�̅� = 36

10

= 3.6


Mean for ungrouped data

Sometimes we have lots of data to calculate the mean from so it is best for us to organize our

scores first before calculating the mean. We do this with the help of a frequency table.

𝒙 ̅ = ∑ 𝒇𝒙

∑ 𝒇

Example 2.

The information below shows the daily collection of oranges by farmers.

200 100 300 100 400 200 300 400 200 100 200 100 300 100 200

100 300 200 200 100 400 200 100 200 300 400 100 200 400 200

100 300 400 100 200 300 100 200 100 200 300 100 100 300 400

200 500 300 400 500 200 200 100 300 200 400 500 300 100 500

Find the mean daily collection

Solution

Set up a frequency table

x Tally Frequency

(f)

fx

100 17 1700

200 18 3600

300 12 3600

400 9 3600

500 4 2000

∑ 𝑓 = 60 ∑ 𝑓𝑥 = 14500

Note: fx means f × x

𝑥 ̅ = ∑ 𝑓𝑥

∑ 𝑓

𝑥 ̅ = 14500

60

= 242 (nearest whole number)


Exercises

1. Calculate the mean for the following sets of numbers.

a) 5, 7, 1, 0, 2, 0, 6

b) 16, 7, 10, 0, 3, 4

c) 20, 15, 10, 12, 3

d) 13, 14, 15, 16, 17, 18, 19

2. Ian scored 22, 25, 35, 46, 17 and 15 runs in 6 innings during a cricket season. Find his

mean score.

3. Five suitcases have masses of 26kg, 30kg, 28kg, 35kg, and 36kg were loaded on a North

American Airplane. When another suitcase was loaded the mean became 32kg. Find the

mass of the additional suitcase.

4. The information below shows the number of minutes each child in a particular class took

to solve a particular problem.

1 2 3 4 2 2 3 3 3 4 4 4 6 6 6

6 7 7 7 8 10

Construct a frequency table and hence calculate the mean.

5. An electrical repair serviced the following number of calls from customers on 28

consecutive working days. Calculate the mean number of calls

20 21 19 17 18 23 21 22 30 21 27 17 18 19 25 20 24

25 29 26 28 25 27 19 23 21 30

6. A farmer collected a total of 600 eggs on the first three days of the week. For the

remaining days he collected 849 eggs. Calculated the eggs he collected daily.

7. Three children have an average age of 7 years 11 months. If the youngest child is not

included, the average age increases to 8 years 11 months. Find the age of the youngest

child.

8. The frequency table below shows the number of tickets sold for a calypso show.

# of tickets sold Frequency

1

2

3

4

5

6

12

35

44

18

8

3

Calculate the mean number of tickets sold


WEEK 12

LESSON 2

Topic: Statistics – Frequency table for ungrouped data

The frequency of an event (or observation or score) is defined as the number of times an event

has occurred.

Example 1.

A die is rolled 10 times and the results are as follows:

1, 2, 1, 3, 4, 2, 5, 3, 6, 1

Notice that 1 played 3 times, so we say the frequency of 1 is 3.

The frequency of 6 is 1, the frequency of 3 is 2 and so on.

Organizing raw data

The best method of organizing raw data is to arrange them in form of a frequency distribution

and a tally chart.

f – Frequency

x – scores

fx – f × x

: ∑ 𝑓 – sum of scores

∑ 𝑓𝑥 - sum of frequency

Example 2.

The information below shows the test scores for an entrance examination.

10 11 12 11 13 10 12 13 10 11 10 11 12 11 10 11 12 10 10 11 13 10

11 10 12 13 11 10 13 10 11 12 13 11 10 12 11 10 11 10 12 11 11 12

13 10 14 12 13 14 10 10 11 12 10 13 14 12 11 14

Draw a frequency table to show the data above.


Solution

x Tally Frequency

(f)

fx

10 17 170

11 18 198

12 12 144

13 9 117

14 4 56

∑ 𝑓 = 60 ∑ 𝑓𝑥 = 685

Exercises

1. The heights of 50 students correct to the nearest centimetre are given below:

150 151 152 153 153

151 153 154 152 155

153 154 151 153 152

154 155 153 154 154

152 153 155 151 153

153 152 153 152 155

154 153 154 155 152

155 152 156 153 151

153 154 153 156 154

152 153 152 154 153

Construct a frequency distribution table for the data above.

2. There are 25 participants in a shooting competition. The score of each participant is listed

below.

1 3 5 0 2

2 1 6 5 6

0 3 5 1 1

5 2 1 0 6

1 4 0 3 5

Draw a frequency table to represent the information above.


WEEK 12

LESSON 4

Topic: Statistics – Measures of Central Tendency (continued)

The Median – the middle or the central value of a set of data arranged in ascending or

descending order. It is represented by 𝑄2.

Finding the median from a set of data

Example 1.

Find the median of the following heights which are stated in centimetres:

a) 163, 158, 154, 161, 156, 159, 155

b) 158, 163, 154, 161, 157, 156, 159, 155

Solution

a) The heights in ascending order

154, 155, 156, 158, 159, 161,163

𝑄2 = 158

∴ 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑎𝑛 ℎ𝑒𝑖𝑔ℎ𝑡, 𝑄2 = 158𝑐𝑚

b) The heights in ascending order

154, 155, 156, 157,158, 159, 161,163

𝑄2 =157+158

2

∴ 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑎𝑛 ℎ𝑒𝑖𝑔ℎ𝑡, 𝑄2 = 157.5𝑐𝑚

Exercises

1. Find the median of the following numbers

a) 1, 1, 2, 2, ,3 ,4, 5, 6, 7, 6, 8

b) 4, 1, 7, 3, 4, 8, 9, 11, 2, 9

c) 27, 23, 25, 23, 24, 26

d) 7, 9, 8, 9, 10, 12, 10, 10


2. Find the median of the following numbers: 2, 1, 9, 8, 3, 4, 11, 2, 5, 3, 5.

3. Find the median of the following numbers: 7, 10, 2, 15, 3, 4

If 15 is taken out from the list of numbers, find the new median.

4. The masses of six men in kg are: 80, 77, 82, 77, 83, 84. Calculate the median mass.

5. A number of children were asked to estimate the height of a tree in metres. The following

results were obtained from their responses.

9 11 12 13 8 9 7 11

12 7 12 10 10 12 11 14

6 10 11 13 8 9 12 10

Find the median of the estimated heights.

6. In a shooting contest in which 50 persons participated, the following frequency table was

obtained.

Score Frequency

1 3

2 5

3 10

4 6

5 5

6 2

7 3

8 1

9 1

a) Find the median score.

b) Determine the probability that if a participant is chosen at random he scored less than

6


WEEK 13

LESSON 1


Determining the median from an ungrouped frequency table

Consider the question 6 in the last exercise, you may have found some difficulties in arriving at

the median.

The process is very simple; we draw up a frequency table with another column for cumulative

frequency and generate the cumulative frequency.

The cumulative frequency is obtained by adding in ascending order. It is the position of each

number when arranged in ascending order.

Using question 6 as an example

In a shooting contest in which 50 persons participated, the following frequency table was

obtained.

Score Frequency

1 3

2 5

3 10

4 6

5 5

6 2

7 3

8 1

9 1

Find the median score.


Solution

We first draw up a cumulative frequency table

Score Frequency Cumulative frequency

1 3 3

2 5 3 + 5 = 8

3 10 8 + 10 = 18

4 6 18 + 6 = 22

5 5 22 + 5 = 27

6 3 27 + 3 = 30

7 6 30 + 6 = 36

8 4 36 + 4 = 40

9 10 40 + 10 = 50

Notice that the last number generated is 50, which is the number of persons participating in the

event (sum of frequency).

𝑻𝒉𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒆𝒅𝒊𝒂𝒏 = 𝟏

𝟐(𝒏 + 𝟏)𝒕𝒉 𝒓𝒂𝒏𝒌

So, the position of the median, 𝑄2 = 1

2(50 + 1)𝑡ℎ 𝑟𝑎𝑛𝑘

= 25.5𝑡ℎ 𝑟𝑎𝑛𝑘

This implies that the median is the average of the 25𝑡ℎ 𝑎𝑛𝑑 𝑡ℎ𝑒 26𝑡ℎ 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠

From the table it can be seen that the 25𝑡ℎ 𝑎𝑛𝑑 𝑡ℎ𝑒 26𝑡ℎ 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠 both carries scores of 5

Then 𝑄2 = 5

NOTE: If there were two different observations at 𝟐𝟓𝒕𝒉 𝒂𝒏𝒅 𝒕𝒉𝒆 𝟐𝟔𝒕𝒉 then we would have

taken their sum and divided it by 2.

Exercises

1. The scores were obtained from a mock exam form Math.

20 15 16 19 21 22 14 17 13 17 19 20 22 15 17

16 19 20 21 22 15 20 17 16 19 20 14 15 22 13

24 15 23 19 18 17 19 18 18 16 21 20 14 19 15

a) Draw up a cumulative frequency table for the above data.

b) Determine the median from the table


2. The marks obtained by 40 students in a test are shown in the table below:

Frequency table

Marks Frequency

1 3

2 5

3 6

4 9

5 5

6 2

7 6

8 4

a) Find the median of marks shown in the frequency distribution given above.

b) Determine the probability that if a student is chosen at random, his marks are 5 or less.

3. The following table shows the number of children per family in the families of the pupils

in a class.

Frequency table

# of children per family 1 2 3 4 5 6 7

Frequency 2 3 9 5 6 4 1

a) Calculate the median

b) Determine the probability that if a family is chosen at random it has more than 5 children.

4. The shoe sizes of pupils in a class are:

4 7 4 6 5 5 5 4 8 7 8 8 7 5 7

6 8 5 8 9 9 6 5 4 5 7 7 5 9 5

a) Draw a frequency table to represent the information given

b) What is the median shoe size?


WEEK 13

LESSON 2


The Mode – the value with the greatest frequency (i.e. the score that occurs the most)

Example 1.

What is the modal value of the set of numbers?

2, 3, 4, 3, 5, 6, 3, 2, 5, 3, 6

Solution

Modal value is 3 (because it occurs 4 times; the most times)

There can be more than one modal value

Example 2.

From the set of values, identify the modal value.

3, 2, 5, 4, 1, 3, 4, 2, 5, 4, 2

Solution

The modal values are 2 and 4 (they occur the same number of times)

Sometimes there is no modal value

Example 3.

State the modal value from the set of scores

2, 4, 6, 8, 10

There is no mode.

Exercise

1. Find the mode of each of the following:

a) 1, 2, 2, 4, 6, 5, 8

b) 5, 0, 2, 10, 2, 6

c) 8, 7, 6, 5, 7, 7, 9, 8

d) 25, 15, 16, 25, 17, 15, 15, 25, 15

e) 1, 1, 6, 6, 5, 4, 3, 2

f) 6, 10, 7, 10, 6, 9, 8, 4

g) 0, 1, 2, 3, 4, 5, 6

h) 8, 10, 12, 18, 38, 67


WEEK 13

LESSON 3

Topic: Statistics – Frequency tables for grouped data

The grouped frequency table is a statistic method to organize and simplify a large set of data in

to smaller "groups." When a data consists of hundreds of values, it is preferable to group them

in a smaller chunk to make it more understandable.

Example 1.

The marks obtained by 30 students on a mathematics test are as follows:

20 80 88 25 0 15 2 60 3 90

55 60 59 57 54 51 62 63 70 94

77 43 55 44 49 81 82 35 36 98

Solution

Step 1. Determine the number of groups or class intervals; if not given.

Let us use 10 marks interval; this is called the class size or the width of the class

These can be as follows: 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, … 90-99

We use these class intervals as our score and draw our frequency table.

Marks (class interval) Frequency

0-9 3

10-19 1

20-29 2

30-39 2

40-49 3

50-59 6

60-69 4

70-79 2

80-89 4

90-99 3


Exercises

1. The masses of 40 students, chosen at random are given below to the nearest kg.

40 56 34 65 71 53 49 51

61 35 67 70 59 45 43 52

60 64 55 58 72 31 37 53

35 32 67 69 72 34 75 55

46 48 59 62 43 71 30 47

Construct a frequency table with class intervals: 30-39, 40-49, 50-59, 60-69, 70-79.

2. All the students in an athletic class were all timed in a 200m race. Here are the times

recorded in seconds to the nearest tenths of a second.

22.4 23.5 30.1 26.3 31.6

32.4 35.5 25.3 33.7 27.8

29.3 22.9 23.5 31.8 32.3

23.1 26.8 27.6 28.4 21.9

Choose a suitable set of class intervals with width 4 for the above data then draw a frequency

distribution table to represent the data.

3. The data below represents scores in a test out of 100.

43 71 47 65 59 53 57 45 82 71 87 84 79 56 77

66 80 57 84 91 92 60 50 49 50 76 79 52 93 57

Draw a frequency table using the class interval 40-49, 50-59, 60-69, etc, to represent the

information given.

4. The following is a list of marks gained by 30 students in a Social Studies test in which the

highest mark possible is 20.

12 3 14 12 15 17 18 13 4 15 11 10 11 14 13

15 17 8 14 1 10 16 9 11 3 16 7 12 12 18

Using class interval: 1-4, 5-8, 9-12, 13-16, 17-20, construct a frequency table.


WEEK 13

LESSON 4

Topic: Statistics

Mixed problems involving measures of central tendency.

Exercises

1. Find the mean, mode and median values for each below.

a) 6, 8, 8, 9, 5, 4, 7, 5, 3, 2, 9

b) 14, 17, 15, 12, 14, 16, 13, 15, 18, 12

c) 24, 25, 43, 59, 11, 2, 19, 53


below:

11 21 16 19 16 20 15 14 13 17 18 16 21 22 13

12 13 20 12 22 16 15 23 14 13 23 12 15 17 16

23 14 19 11 18 17 16 22 15 21 12 18 19 17 20

a) Set up a frequency distribution table for the above data

b) From the table, determine the mean, mode and median

c) Find the probability that a competitor chosen at random has a score greater than 15.

3. The marks obtained from 50 students in a test were:

23 24 45 34 36 27 14 28 47 49

25 46 16 19 29 18 37 39 33 21

43 42 42 42 29 41 34 38 39 46

19 23 25 24 32 37 42 31 47 49

45 46 36 28 29 14 17 45 21 19

a) Construct a frequency table with class interval 10-14, 15-19, etc.

b) What is the size of the class?

c) Determine the modal class distribution.

d) What is the probability that a student scored between 25 and 40 marks at the test.


WEEK 14

LESSON 1

Topic: Statistics – Histogram for ungrouped data

A histogram is a diagram that is used to display the data contained in a frequency distribution. A

histogram is different from a bar graph. In a histogram there is no space between the bars. The

width of the bars is the same.

Example 1.

Example 2.

The following table shows the number of children per family in the families of the pupils in a

class.

Frequency table

# of children in a family 1 2 3 4 5 6 7

Frequency 2 3 9 5 6 4 1

Draw a histogram to represent the above data.


Solution

Histogram

Frequency

Number of children

Exercises

1. 50 students joined the library one week. Their ages were recorded as follows:

10 16 13 8 14 13 13 12 9 10

8 14 10 9 10 14 11 14 16 12

13 15 15 10 8 12 12 15 13 13

11 13 11 8 9 13 16 16 13 8

10 12 9 13 15 14 15 11 12 9

a) Construct a frequency table frequency table for the above

b) Draw a suitable histogram to represent the data.

2. The following are test scores for a driving class taken by 25 participants out of a possible

20 marks

13 20 17 14 20

20 14 20 20 19

14 19 15 17 15

15 18 16 18 14

13 14 16 15 17

Construct a suitable histogram for the above data.

9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7


WEEK 14

LESSON 2

Topic: Statistics – Histogram for grouped data

Example 1.

50 students joined the library one week. Their ages were recorded as follows:

10 16 13 8 14 13 13 12 9 10

8 14 10 9 10 14 11 14 16 12

13 15 15 10 8 12 12 15 13 13

11 13 11 8 9 13 16 16 13 8

10 12 9 13 15 14 15 11 12 9

a) Construct a frequency table frequency table for the above data using 5-9, 10-14, 15-19

b) Draw a suitable histogram to represent the data.

Solution

a)

Class interval Frequency

5-9 10

10-14 31

15-19 9


b)

Histogram

40

35

30

25

20

15

10

5

5-9 10-14 15-19

Exercises


23 24 45 34 36 27 14 28 47 49

25 46 16 19 29 18 37 39 33 21

43 42 42 42 29 41 34 38 39 46

19 23 25 24 32 37 42 31 47 49

45 46 36 28 29 14 17 45 21 19

a) Construct a frequency table with class interval 10-14, 15-19, etc.

b) Draw a histogram to represent the data

c) Determine the modal class distribution.

d) What is the probability that a student scored between 25 and 40 marks at the test.


WEEK 14

LESSON 3

Topic: Statistics – Frequency polygon

A frequency polygon is constructed by lines joining the mid-points of each of the top lines of the

rectangles on a histogram.

Example 1.

Taking example 2 from lesson 1 week 14

The following table shows the number of children per family in the families of the pupils in a

class.

Frequency table

# of children in a family 1 2 3 4 5 6 7

Frequency 2 3 9 5 6 4 1

Draw a frequency polygon to represent the above data.

Solution

Frequency polygon

Frequency

Number of children

9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7


Example 2. Is taken from week 2 worksheet 14

Frequency Polygon

40

35

30

25

20

15

10

5

5-9 10-14 15-19

Exercises

1. Andera kept a record of the number of cherries she collected each morning. The numbers

were as follows:

10 12 15 8 0 3 5

8 13 13 18 25 27 30

35 40 50 55 58 58 50

48 42 35 30 29 29 20

a) Choose a class size of 5 and construct a frequency table.

b) Use the table to determine the mean and median values.

c) Draw a histogram and frequency polygon to represent the number of cherries

collected.

d) From the histogram determine the modal value.


WEEK 14

LESSON 4

Topic: Statistics – Cumulative Frequency Polygon

Using example from lesson one worksheet thirteen.

cumulative frequency table

Score Frequency Cumulative frequency

1 3 3

2 5 8

3 10 18

4 6 22

5 5 27

6 3 30

7 6 36

8 4 40

9 10 50

Construct a cumulative frequency polygon to represent the data.

Solution

Step 1 – draw a table to determine the values for the cumulative frequency.

Step 2 – draw the polygon

0

10

20

30

40

50

60

0 1 2 3 4 5 6 7 8 9 10

Cumulative frequency


Exercises


below:

11 21 16 19 16 20 15 14 13 17 18 16 21 22 13

12 13 20 12 22 16 15 23 14 13 23 12 15 17 16

23 14 19 11 18 17 16 22 15 21 12 18 19 17 20

a) Set up a cumulative frequency distribution table for the above data

b) Construct a cumulative frequency polygon to represent the data


23 24 45 34 36 27 14 28 47 49

25 46 16 19 29 18 37 39 33 21

43 42 42 42 29 41 34 38 39 46

19 23 25 24 32 37 42 31 47 49

45 46 36 28 29 14 17 45 21 19

a) Construct a cumulative frequency table with class interval 10-14, 15-19, etc.

b) What is the size of the class?

3. A butcher kept the following record for the number of pigs he slaughtered during the 9

weeks before Christmas.

Grouped frequency table

Days # of pigs

slaughtered

Cumulative

frequency

1-7 15 15

8-14 12

15-21 36

22-28 75

29-35 98

36-42 69

43-49 58

50-56 32

57-63 5

a) Determine the class size used in the class intervals

b) Copy and complete the table above

c) Construct a cumulative frequency table for the data


WEEK 15

LESSON 1

Topic: Statistics – Introduction to Matrices

A matrix is a rectangular array of numbers or letters consisting of m rows and n columns

enclosed in a pair of curved or squared brackets and usually denoted by a capital letter.

Example 1.

𝐴 = 2 −34 0

The numbers or letters in the matrix are called the elements

Naming a matrix

A matrix is named by the number of rows by the number of columns.

2 −5 1 𝑟𝑜𝑤 × 2 𝑐𝑜𝑙𝑢𝑚𝑛𝑠

2 × 1 𝑚𝑎𝑡𝑟𝑖𝑥

−10

2 𝑟𝑜𝑤𝑠 𝑏𝑦 1 𝑐𝑜𝑙𝑢𝑚𝑛


𝑎 𝑏 𝑐𝑑 𝑒 𝑓

2 𝑟𝑜𝑤𝑠 𝑏𝑦 3 𝑐𝑜𝑙𝑢𝑚𝑛𝑠


Exercises

1. Given that A = 2 −34 0

, state:

a) The order of the matrix

b) The first element in the second row

c) The second element in the first row

2. If 𝑃 = 5 −1

−3 07 6

, state the order of the matrix, the second element in the second

column and the third element in the first column.


3. Given that A = −3 4 62 1 0

, state:


b) The first element in the second row

c) The second element in the first row

4. If 𝐵 = 2 1 −90 5 13 6 12

, state:


b) The third element in the second row

c) The second element in the fourth row

5. Given 𝐶 = 123

11 ,


b) The element in the second row

6. If 𝐷 = −2 3 −56 7 45 0 −4

−103


b) The third element in the second row

c) The second element in the fourth column

7. Given that the matrix 𝑅 = 12 9 −123 10 4

−5 11 2

0 6−7 514 11

, state:


b) The third element in the third row

c) The second element in the fourth column

d) The first element in the second column


WEEK 15

LESSON 2

Topic: Statistics – Matrices (continued) – Addition and Subtraction

Matrices of the same order can be added or subtracted by adding or subtracting the

corresponding elements.

Example 1.

If 𝐴 = 2 4

−1 6 𝑎𝑛𝑑 𝐵 =

−4 57 6

, determine:

a) A + B

b) A – B

Solution

a) A + B = 2 4

−1 6 +

−4 57 6

= 2 + −4 4 + 5−1 + 7 6 + 6

= −2 96 12

b) A – B = 2 4

−1 6 −

−4 57 6

= 2 − −4 4 − 5−1 − 7 6 − 6

= 6 −1

−8 0

Exercises

1. If 𝐴 = 2 4

−1 6 , 𝐵 =

4 71 4

𝑎𝑛𝑑 𝐶 = −4 57 6

, determine:

a) A + B d) B – C

b) A + B + C e) C – A

c) C + B f) A – B – C


WEEK 15

LESSON 3

Topic: Statistics – Matrices (continued) – Scalar multiplication

Given that 𝐴 = 𝑎 𝑏𝑐 𝑑

,

then 𝑘𝐴 = 𝑘 𝑎 𝑏𝑐 𝑑

= 𝑎 𝑏𝑐 𝑑

𝑘 = 𝑘𝑎 𝑘𝑏𝑘𝑐 𝑘𝑑

where k is a constant

Example 1.

Given that 𝐴 = −2 15 0

𝑎𝑛𝑑 𝑘 = 4, determine kA.

Solution

𝑘𝐴 = 4𝐴 = 4 −2 15 0

= 4 × −2 4 × 14 × 5 4 × 0

= −8 420 0

Exercises

Given 𝑃 = 3 1

−2 5 𝑎𝑛𝑑 𝑄 =

4 68 −9

, evaluate:

a) 2P

b) 3P – Q

c) 5Q

d) 4Q + 2P

e) P + 2Q – Q

f) Q + 3P + 3Q

g) 2P – Q – P + 2P


WEEK 15

LESSON 4

Topic: Statistics – Matrices (continued) – Equal matrix

Two matrices are equal if they are in the same order and the corresponding elements are equal.

i.e. Given that 𝐴 = 𝑎 𝑏𝑐 𝑑

𝑎𝑛𝑑 𝐵 = 𝑒 𝑓𝑔 ℎ

,

then 𝐴 = 𝐵 𝑖𝑓 𝑎 𝑏𝑐 𝑑

= 𝑒 𝑓𝑔 ℎ

i.e. 𝑎 = 𝑒, 𝑏 = 𝑓, 𝑐 = 𝑔, 𝑑 = ℎ

Example 1.

Given that 𝑥 −34 𝑦

= −2 −34 6

, determine the values of x and y

Solution

𝑥 −34 𝑦

= −2 −34 6

Equating the corresponding elements, we get

x = -2 and y = 6

Exercises

1. Given that x yy 6 =

−2 −3−3 6

, determine the values of x and y

2. If a bc d

= −1 −35 0

, determine the values of a , b, c, and d.

3. Given that 𝐴 = 4 31 2

𝑎𝑛𝑑 𝐵 = −2 −13 5

, determine X in the following equation

𝑿 − 𝑩 = 𝑨

easter term 2021 grade 9 mathematics ... - education.gov.gy

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