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Ministry of Education
MINISTRY OF EDUCATION
EASTER TERM 2021
GRADE 9
MATHEMATICS
WORKSHEETS
Ministry of Education
WEEK 1
LESSON 1
Topic: Consumer Arithmetic โ Percentage Profit and Percentage Loss
๐ป๐๐ ๐ท๐๐๐๐๐ = ๐ป๐๐ ๐บ๐๐๐๐๐๐ ๐ท๐๐๐๐ โ ๐ป๐๐ ๐ช๐๐๐ ๐ท๐๐๐๐
๐ท = ๐บ. ๐ท โ ๐ช. ๐ท
๐ป๐๐ ๐ท๐๐๐๐๐ % =๐ป๐๐ ๐ท๐๐๐๐๐
๐ป๐๐ ๐ช๐๐๐ ๐ท๐๐๐๐ ร ๐๐๐%
๐ท% =๐บ.๐ทโ๐ช.๐ท
๐ช.๐ท ร ๐๐๐%
๐ป๐๐ ๐ณ๐๐๐ = ๐ป๐๐ ๐ช๐๐๐ ๐ท๐๐๐๐ โ ๐ป๐๐ ๐บ๐๐๐๐๐๐ ๐ท๐๐๐๐
L = ๐ช. ๐ท โ ๐บ. ๐ท
๐ป๐๐ ๐ณ๐๐๐ % =๐ป๐๐ ๐ณ๐๐๐
๐ป๐๐ ๐ช๐๐๐ ๐ท๐๐๐๐ ร ๐๐๐%
๐ณ% =๐ช.๐ทโ๐บ.๐ท
๐ช.๐ท ร ๐๐๐%
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 ร $4 200
= $105 000
P = S.P โ C.P
= $105 000 โ $87 500
= $ 17 500
๐% =๐.๐โ๐ถ.๐
๐ถ.๐ ร 100%
=$17 500
$87 500 ร 100%
= 20%
b) C.P of balls = $87 500
S.P of balls = 25 ร $2 700
= $67 500
L = C.P โ S.P
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= $87 500 โ $67 500
= $ 20 000
๐ฟ% =๐ถ.๐โ๐.๐
๐ถ.๐ ร 100%
=$20 000
$87 500 ร 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 ร $42 000 115% =
42 000
100 ร115
= $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 ร $42 000
= $3 360
S.P = C.P โ L
= $42 000 โ $3 360
= $38 640
Ministry of Education
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?
Ministry of Education
WEEK 1
LESSON 2
Topic: Consumer Arithmetic โ 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% โ 5%
= 95%
โด 95% = $7 600
So 1% = $7 600
95
Then C.P (100%) = $7 600
95 ร 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%
โด 125% = $85 000
So 1% = $85 000
125
Then C.P (100%) = $85 000
125 ร 100
= $68 000
Ministry of Education
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.
Ministry of Education
WEEK 1
LESSON 3
Topic: Consumer Arithmetic โ 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 ร 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.
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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 ร $60 000
= $24 000
Instalments = $4 000 ร 12
= $48 000
Total = $24 000 + $48 000
= $72 000
Amount extra paid = $72 000 โ $60 000
= $12 000
b) Interest is paid on = $60 000 โ $24 000
= $36 000
Rate of interest = $12 000
$36 000 ร 100
= 33.33%
Ministry of Education
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 yearsโ
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.
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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.
Ministry of Education
WEEK 1
LESSON 4
Topic: Consumer Arithmetic โ 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 ร $4 800 000
= $960 000
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b) mortgage = sales price โ deposit
= $4 800 000 โ $960 000
= $3 840 000
c) total amount to be repaid = monthly installments ร 12 ร 15
= $27 733 ร 12 ร 15
= $4 991 900
d) interest = total amount to be repaid โ mortgage
= $4 991 900 โ $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 ร $300 000
= $54 000
The amount to be repaid = $300 000 + $54 000
= $354 000
Amount of each instalment = $354 000
36
= $9 833
Ministry of Education
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.
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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
Ministry of Education
WEEK 2
LESSON 1
Topic: Consumer Arithmetic โ 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 โ amount of money borrowed
Rate, R โ percentage rate charged by the bank
Time, T โ period for the loan to be repaid
๐๐๐๐๐๐ ๐๐๐ก๐๐๐๐ ๐ก = ๐๐๐๐๐๐๐๐๐ ร๐ ๐๐ก๐ ร๐๐๐๐
100
๐ผ = ๐๐ ๐
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, ๐ผ = ๐๐ ๐
100
= $100 000 ร8 ร3
100
=$24 000
b) Amount accruing, ๐ด = ๐ + ๐ผ
= $100 000 + $24 000
= $124 000
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c) The amount of each instalment = ๐โ๐ ๐๐๐๐ข๐๐ก ๐๐๐๐๐ข๐๐๐
๐กโ๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐กโ๐
= $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.
Ministry of Education
WEEK 2
LESSON 2
Topic: Consumer Arithmetic โ 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) ๐ผ = ๐๐ ๐
100
$16 800 = ๐ ร0.5 ร5
100
$16 800 ร100
0.5 ร5= ๐
๐ = $1 680 000
2.5
= $672 000
b) ๐ผ = ๐๐ ๐
100
$38 675 = $850 000 ร 3.5 ร๐
100
$38 675ร100
$850 000 ร 3.5= ๐
๐ = $3 867 500
$2 975 000
= $672 000
c) ๐ผ = ๐๐ ๐
100
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$189 000 = $540 000 ร ๐ ร8.75
100
$189 000ร100
$540 000 ร 8.75= ๐
๐ = $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.
Ministry of Education
WEEK 2
LESSON 3
Topic: Consumer Arithmetic โ 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 โ 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
๐ด = ๐(1 + ๐
100)๐
= $100 000(1 + 8
100)3
= $100 000(1 + 0.08)3
= $100 000( 1.08)3
๐ด = ๐(1 + ๐
100)๐
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= $125 971.20
The amount accruing after 3 years is $125 971.20
The compound interest
C. I = A โ P
= $125 971.20 โ $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.
Ministry of Education
WEEK 2
LESSON 4
Topic: Consumer Arithmetic โ 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 โ 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 ร $2 500 000
๐ด = ๐(1 โ ๐
100)๐
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= $625 000
Value of vehicle at the beginning of 2nd year = cost price โ depreciation
= $2 500 000 โ $625 000
= $1 875 000
Depreciation during the 2nd year = 25
100 ร $1 875 000
= $468 750
Value of vehicle at the beginning of 3rd year = $1 875 000 โ $468 750
= $1 406 250
Depreciation during the 3rd year = 25
100 ร $1 406 250
= $351 562. 50
Value of vehicle at the beginning of 4th year = value of vehicle after 3 years
= $1 406 250 โ $351 562.50
= $1 054 687.50
Using the formula
The value after 3 years, ๐ด = ๐(1 โ ๐
100)๐
= $2 500 000(1 โ 25
100)3
= $2 500 000(1 โ 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%?
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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.
Ministry of Education
WEEK 3
LESSON 1
Topic: Algebra 2 - Factorization of Quadratic expressions
๐ป๐๐ ๐ฎ๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐ ๐๐๐๐ ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐ + ๐๐ + ๐
Examples of quadratic expressions:
1. ๐ฅ2 + 7๐ฅ + 6
Here: a = 1, b = 7 and c = 6
2. 3๐ฅ2 + 5๐ฅ + 2
a = 3, b = 5 and c = 2
3. ๐ฅ2 โ 9
a = 1, b = 0 and c = -9
Example 1.
Factorize completely ๐ฅ2 + 5๐ฅ + 6
Solution
We multiply the coefficient of ๐ฅ2 โaโ by the constant โcโ
I.e. ac = 1 ร 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.
๐ฅ2 + 5๐ฅ + 6
Ministry of Education
= ๐ฅ2 + 3๐ฅ + 2๐ฅ + 6
= (๐ฅ2 + 3๐ฅ) + (2๐ฅ + 6)
= ๐ฅ(๐ฅ + ๐ฅ) + 2(๐ฅ + 3)
= (๐ฅ + 3)(๐ฅ + 2)
Example 2.
Factorize the expression ๐2 โ 2๐ โ 8
๐2 โ 2๐ โ 8 ๐๐ = 1 ร โ8 = โ8
= ๐2 โ 4๐ + 2๐ โ 8
= (๐2 โ 4๐) + (2๐ โ 8)
= ๐(๐ โ 4) + 2(๐ โ 4)
= (๐ โ 4)(๐ + 2)
Example 3.
Factorize the expression 6๐ก2 โ 7๐ก โ 3
6๐ก2 + 7๐ก โ 3 ๐๐ = 6 ร โ3 = โ18
= 6๐ก2 + 9๐ก โ 2๐ก โ 3
= (6๐ก2 + 9๐ก) โ (2๐ก + 3)
Leave the negative sign outside the brackets and
change the signs inside
= 3๐ก(2๐ก + 3) โ 1(2๐ก + 3)
= (2๐ก + 3)(3๐ก โ 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
Ministry of Education
Exercises
Factorize the following quadratic expressions completely:
1. ๐ฆ2 + 6๐ฆ + 5
2. ๐2 + 14๐ + 13
3. 7๐ฅ2 + 15๐ฅ + 2
4. 12 + 8๐ฆ + ๐ฆ2
5. 4๐ฆ2 + 10๐ฆ โ 6
6. 8๐ฅ2 + 13๐ฅ โ 6
7. ๐ฅ2 + 11๐ฅ โ 60
8. ๐ฅ2 + ๐ฅ โ 6
9. 6๐ฅ2 + 2๐ฅ โ 4
10. ๐ฅ2 โ 5๐ฅ + 6
11. 11๐2 + 14๐ + 3
12. ๐ฅ2 โ 11๐ฅ + 30
13. ๐ก2 + 9๐ก โ 10
14. ๐2 โ 2๐ โ 15
15. 8๐ฅ2 โ 3๐ฅ โ 5
Ministry of Education
WEEK 3
LESSON 2
Topic: Algebra 2 - Factorization of Quadratic expressions that can be expressed as perfect
squares
๐โ๐ ๐๐ฅ๐๐๐๐ ๐ ๐๐๐ (๐ + ๐)2 ๐๐ ๐ ๐๐๐๐๐๐๐ก ๐ ๐๐ข๐๐๐
When expanded, we get ๐๐ + ๐๐๐ + ๐๐
Similarly (๐ โ ๐)2 ๐๐ ๐ ๐๐๐๐๐๐๐ก ๐ ๐๐ข๐๐๐
When expanded, we get ๐๐ โ ๐๐๐ + ๐๐
To factorize a perfect square we follow the same procedure as previous lesson, then we write our
factors of the expression like (๐ + ๐)2 ๐๐ (๐ โ ๐)2.
Example 1.
Factorize each quadratic expression below completely.
a) ๐ฅ2 + 2๐ฅ + 1
b) ๐ฅ2 โ 6๐ฅ + 9
c) 9๐ฅ2 + 30๐ฅ + 25
Solutions
๐) ๐ฅ2 + 2๐ฅ + 1 ๐๐ = 1 ร 1 = 1
= ๐ฅ2 + ๐ฅ + ๐ฅ + 1
= (๐ฅ2 + ๐ฅ) + (๐ฅ + 1)
= ๐ฅ(๐ฅ + 1) + 1(๐ฅ + 1)
= (๐ฅ + 1)(๐ฅ + 1)
= (๐ฅ + 1)2
๐) ๐ฅ2 โ 6๐ฅ + 9 ๐๐ = 1 ร 9 = 9
= ๐ฅ2 โ 3๐ฅ โ 3๐ฅ + 9
= (๐ฅ2 โ 3๐ฅ) โ (3๐ฅ โ 9)
= ๐ฅ(๐ฅ โ 3) โ 3(๐ฅ โ 3)
= (๐ฅ โ 3)(๐ฅ โ 3)
= (๐ฅ โ 3)2
Factor of ac (1) sum of factors
Must be = b (2)
1 and 1 2
Factor of ac (9) sum of factors
Must be = b (-6)
9 and 1 10
-9 and -1 -10
3 and 3 6
-3 and -3 -6
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๐) 9๐ฅ2 + 30๐ฅ + 25 ๐๐ = 9 ร 25 = 225
= 9๐ฅ2 + 15๐ฅ + 15๐ฅ + 25
= (9๐ฅ2 + 15๐ฅ) + (15๐ฅ + 25)
= 3๐ฅ(3๐ฅ + 5) + 5(3๐ฅ + 5)
= (3๐ฅ + 5)(3๐ฅ + 5)
= (3๐ฅ + 5)2
Exercises
Factorize the following quadratic expressions completely:
1. ๐ฅ2 + 10๐ฅ + 25
2. ๐2 + 8๐ + 16
3. ๐ฅ2 + 2๐ฅ + 1
4. 64 + 16๐ฆ + ๐ฆ2
5. 4๐ฆ2 + 4๐ฆ + 1
6. 4๐ฅ2 + 12๐ฅ + 9
7. ๐ฅ2 โ 8๐ฅ + 16
8. ๐ฅ2 + 2๐ฅ๐ฆ + ๐ฆ2
9. 25๐ฅ2 + 40๐ฅ + 16
10. ๐ฅ2 + 12๐ฅ + 36
11. 64๐2 + 80๐ + 25
12. 81๐ฅ2 + 126๐ฅ + 49
13. ๐ก2 + 18๐ก + 81
14. 64๐2 + 144๐ + 81
Factor of ac (225) sum of factors
Must be = b (30)
After writing all the factors we realize:
15 and 15 30
Ministry of Education
WEEK 3
LESSON 3
Topic: Algebra 2 - Factorization of Quadratic expressions that can be expressed as
difference of squares
The product of (๐ + ๐)(๐ โ ๐) = ๐2 โ ๐2
Hence ๐2 โ ๐2 = (๐ + ๐)(๐ โ ๐)
Example 1.
๐ฅ2 โ 36
Solution
๐ฅ2 โ 36 = ๐ฅ2 โ 62
= (๐ฅ + 6)(๐ฅ โ 6)
Example 2.
Factorize the expression completely:
144 โ 25๐ฆ2
Solution
144 โ 25๐ฆ2 = 122 โ (5๐ฆ)2
= (12 + 5๐ฆ)(12 โ 5๐ฆ)
Example 3.
Factorize completely:
36๐ฅ2 โ 1
Solution
36๐ฅ2 โ 1 = (6๐ฅ)2 โ 12
= (6๐ฅ + 1)(6๐ฅ โ 1)
Ministry of Education
Exercises
Factorize each completely
1. ๐ฆ2 โ 1
2. ๐ฅ2 โ 22
3. ๐ฅ2 โ 4๐ฆ2
4. ๐ฅ2 โ 25๐2
5. 36 โ 9๐2
6. 36๐2 โ 169๐ฅ2
7. 81๐ฅ2 โ 49๐ฆ2
8. 49๐2 โ 16๐ 2
9. 144๐ฅ2 โ 49
10. 64๐ฅ2 โ 25
11. 81๐2 โ 64๐2
12. 1 โ 16๐ฅ2
13. 49๐ฅ2 โ 1
14. (๐ + ๐)2 โ 22
15. (๐ โ ๐)2 โ 1
Ministry of Education
WEEK 3
LESSON 4
Topic: Algebra 2 โ Changing the subject of the formula
A formula is really an equation that tells the relationship between two quantities. For instance,
๐จ = ๐๐ , 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 ๐จ = ๐๐, make l the formula
๐จ = ๐๐
๐จ
๐=
๐๐
๐
๐จ
๐= ๐
๐ =๐จ
๐
Example 2.
Given ๐ฃ = ๐๐2โ, make:
a) h the subject of the formula
b) r the subject of the formula
Solution
a) ๐ฃ = ๐๐2โ
๐ฃ
๐๐2 =๐๐2โ
๐๐2
๐ฃ
๐๐2 = โ
h =๐ฃ
๐๐2
b) ๐ฃ = ๐๐2โ
๐ฃ
๐โ=
๐๐2โ
๐โ
๐ฃ
๐โ= ๐2
โ๐2 = โ๐ฃ
๐๐2
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๐ = โ๐ฃ
๐๐2
Example 3.
Given = ๐ก +๐ฅ
๐ฆ , make x the subject of this formula.
Solution
๐ = ๐ก +๐ฅ
๐ฆ
๐ โ ๐ก = ๐ก +๐ฅ
๐ฆโ ๐ก
(๐ โ ๐ก) ร ๐ฆ =๐ฅ
๐ฆร ๐ฆ
(๐ โ ๐ก)๐ฆ = ๐ฅ
๐ฅ = ๐ฆ(๐ โ ๐ก)
Example 4.
Make c the subject of the formula in the relation ๐ =7
2๐+5
Solution
๐ =7
2๐+5
๐(2๐ + 5) =7
2๐+5ร 2๐ + 5
๐(2๐ + 5) = 7
2๐๐ + 5๐ = 7
2๐๐ + 5๐ โ 5๐ = 7 โ 5๐
2๐๐
2๐=
7โ5๐
2๐
๐ =7โ5๐
2๐
Ministry of Education
Exercises
1. Given ๐ด = ๐๐๐, ๐๐๐๐ ๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
2. Given ๐ =๐ ๐
๐, ๐๐๐๐ ๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
3. Given ๐ผ =๐๐ ๐
100, ๐๐๐๐ ๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
4. Given ๐ =๐๐2โ
4, ๐๐๐๐ ๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
5. Given ๐ = ๐ข + ๐๐ก, ๐๐๐๐ ๐ก ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
6. Given ๐ =๐ด
๐ โ๐, ๐๐๐๐ ๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
7. Given ๐ =๐ (๐ตโ๐)
๐ต, ๐๐๐๐ ๐ต ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
8. Given ๐ โ ๐๐ =๐๐ฃ2
๐, ๐๐๐๐:
a) v the subject of the formula.
b) m the subject of the formula.
9. Given ๐2 =๐+๐
๐โ๐, ๐๐๐๐:
c) b the subject of the formula.
d) a the subject of the formula.
10. Given ๐ด =โ(๐+๐)
2, ๐๐ฅ๐๐๐๐ ๐ ๐ ๐๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ข๐๐.
11. ๐ = โ๐๐โ๐
๐+๐๐, make m the subject of the formula.
12. Given ๐ =1
๐2 ,make d the subject of the formula.
13. ๐ = โ๐2โ๐ฅ2, make x the subject of the formula.
14. Given that ๐ = ๐2โ, make h the subject of the formula.
15. If =๐๐๐ก
๐ฃโ๐ , make r the subject of the formula.
Ministry of Education
WEEK 4
LESSON 1
Topic: Algebra 2 โ Solution of Simultaneous Equations by the method of Substitution
The method of Substitution
Example 1.
Solve the equations ๐ฅ + 2๐ฆ = 4 ๐๐๐ 2๐ฅ + 3๐ฆ = 7
Solution
Step 1. Rewirte the equations
๐ฅ + 2๐ฆ = 4 โฆโฆโฆโฆeq(1)
2๐ฅ + 3๐ฆ = 7 โฆโฆ.....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
๐ฅ + 2๐ฆ = 4
๐ฅ = 4 โ 2๐ฆ โฆโฆโฆโฆeq(3)
Step 3. Substitute for x in eq(2) 4-2y and solve for y
2๐ฅ + 3๐ฆ = 7
2(4 โ 2๐ฆ) + 3๐ฆ = 7
8 โ 4๐ฆ + 3๐ฆ = 7
โ๐ฆ = 7 โ 8
โ๐ = โ๐
๐ = ๐
Step 4. Substitute for y in eq(3)
๐ฅ = 4 โ 2๐ฆ
๐ฅ = 4 โ 2(1)
๐ฅ = 2
Hence ๐ฅ = 2, ๐ฆ = 1
Note: It is better to use this method when the coefficient of one of the variable is equal to 1.
Ministry of Education
Exercises
Solve the following pairs of simultaneous equations by method of substitution.
1. ๐ฅ โ ๐ฆ = 1
๐ฅ + ๐ฆ = 3
2. ๐ฆ โ ๐ฅ = 1
๐ฅ + ๐ฆ = 5
3. ๐ฆ = 4๐ฅ + 4
๐ฅ + ๐ฆ = 14
4. ๐ = 4 โ ๐
2๐ + ๐ = 1
5. 5๐ฅ โ 2๐ = 1
๐ฅ = 5 โ 2๐
6. ๐ฅ โ ๐ฆ = 4
๐ฅ + 2๐ฆ = 10
7. ๐ โ 2๐ = 1
3๐ โ 4๐ = 1
8. 4๐ฅ โ 3๐ฆ = 1
๐ฅ โ 2๐ฆ = 4
9. ๐ + 2๐ = 4
๐ฅ โ 2๐ฆ = 6
10. 2๐ฅ + 3๐ฆ = 5
โ๐ฅ + 2๐ฆ = 8
11. ๐ฆ โ ๐ฅ = 2
2๐ฆ โ 4๐ฅ = 10
12. ๐ฅ + 6๐ฆ = 22
2๐ฅ + 10๐ฆ = 38
13. 2๐ฅ โ ๐ฆ = 5
๐ฅ + ๐ฆ = 4
14. ๐ฅ + ๐ฆ = 7
2๐ฅ + ๐ฆ = 10
15. ๐ฅ + ๐ฆ = 144
2๐ฅ โ 3๐ฆ = 63
Ministry of Education
WEEK 4
LESSON 2
Topic: Algebra 2 โ 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 5๐ฅ + 2๐ฆ = 16 ๐๐๐ โ 3๐ฅ + 4๐ฆ = 6
Solution
Step 1. Rewirte the equations
5๐ฅ + 2๐ฆ = 16 โฆโฆโฆโฆeq(1)
โ3๐ฅ + 4๐ฆ = 6 โฆโฆ.....eq(2)
Step 2. Make x the subject of eq(1)
Using 5๐ฅ + 2๐ฆ = 16 โฆโฆ..eq (1)
๐ฅ =16โ2๐ฆ
5โฆ โฆ โฆ โฆ . . ๐๐(3)
Step 3. Substitute for x in eq(2) 4-2y and solve for y
โ3๐ฅ + 4๐ฆ = 6
โ3(16โ2๐ฆ
5) + 4๐ฆ = 6
(โ48+6๐ฆ
5) + 4๐ฆ = 6
(โ48+6๐ฆ
5) ร 5 + 4๐ฆ ร 5 = 6 ร 5
โ48 + 6๐ฆ + 20๐ฆ = 30
26๐ฆ = 78
๐ฆ =78
26
๐ = ๐
Step 4. Substitute for y in eq(3)
๐ฅ =16โ2๐ฆ
5
๐ฅ =16โ2(3)
5
๐ฅ =16โ6
5
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๐ = ๐
Hence ๐ฅ = 2, ๐ฆ = 3
Exercises
Solve the following pairs of simultaneous equations by method of substitution.
1. 4๐ฅ + 3๐ฆ = 17
5๐ฅ โ 2๐ฆ = 4
2. 3๐ฅ โ 5๐ฆ = โ13
โ2๐ฅ + 3๐ฆ = 8
3. 5๐ฅ + 3๐ฆ = 27
2๐ฅ + 5๐ฆ = 26
4. 2๐ฅ + 3๐ฆ = โ8
5๐ฅ โ 2๐ฆ = 18
5. 9๐ฅ + 5๐ = 15
3๐ฅ โ 2๐ = โ6
6. 3๐ฅ + 2๐ฆ = 21
2๐ฅ + 3๐ฆ = 13
7. 7๐ฅ โ 2๐ฆ = 19
3๐ฅ + 5๐ฆ = 14
8. 5๐ฅ + 2๐ฆ = 16
โ3๐ฅ + 4๐ฆ = 6
9. 7๐ฅ + 2๐ฆ = 17
2๐ฅ โ 2๐ฆ = 1
10. 3๐ฅ โ 2๐ฆ = โ1
4๐ฅ + 7๐ฆ = 18
11. 3๐ฅ + 4๐ฆ = 27
5๐ฅ โ 2๐ฆ = 19
12. 7๐ฅ + 5๐ = 20.15
5๐ฅ + 7๐ = 18.85
Ministry of Education
WEEK 4
LESSON 3
Topic: Algebra 2 โ 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 3๐ฅ + ๐ฆ = 18 ๐๐๐ 2๐ฅ โ ๐ฆ = 7
Solution 2.
Step 1. Rewirte the equations
3๐ฅ + ๐ฆ = 18 โฆโฆโฆโฆeq(1)
2๐ฅ โ ๐ฆ = 7 โฆโฆ.....eq(2)
Step 2. Coefficients of y are the same; signs are different so we will add the equations
3๐ฅ + ๐ฆ = 18
2๐ฅ โ ๐ฆ = 7
5๐ฅ = 25
Then: ๐ฅ =25
5
๐ฅ = 5
Step 3. Substitute for x in eq(1) and solve for y (any equation could have worked, itโs your
choice)
3๐ฅ + ๐ฆ = 18
3(5) + ๐ฆ = 18
15 + ๐ฆ = 18
๐ฆ = 18 โ 15
๐ = ๐
Hence ๐ฅ = 5, ๐ฆ = 3
Ministry of Education
Exercises
Solve the following pairs of simultaneous equations by method of elimination.
1. ๐ฅ โ ๐ฆ = 1
๐ฅ + ๐ฆ = 3
2. ๐ฆ โ ๐ฅ = 1
๐ฅ + ๐ฆ = 5
3. ๐ฆ = 4๐ฅ + 4
๐ฅ + ๐ฆ = 14
4. ๐ = 4 โ ๐
2๐ + ๐ = 1
5. 5๐ฅ โ 2๐ = 1
๐ฅ = 5 โ 2๐
6. ๐ฅ โ ๐ฆ = 4
๐ฅ + 2๐ฆ = 10
7. โ4๐ฅ โ 6๐ฆ = 7
4๐ฅ + ๐ฆ = โ2
8. 4๐ฅ โ 2๐ฆ = 6
๐ฅ + 2๐ฆ = 4
9. ๐ฅ + 2๐ฆ = 3
2๐ฅ โ 2๐ฆ = 6
10. 2๐ฅ โ 2๐ฆ = 8
2๐ฅ + 2๐ฆ = 20
11. 4๐ฅ โ 3๐ฆ = โ4
2๐ฅ โ 3๐ฆ = 4
12. 2๐ฅ + 4๐ฆ = 5
10๐ฅ + 4๐ฆ = 52
Ministry of Education
WEEK 4
LESSON 4
Topic: Algebra 2 โ 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 5๐ฅ + 3๐ฆ = 21 ๐๐๐ 2๐ฅ + 7๐ฆ = 20
Solution 2.
Step 1. Rewirte the equations
5๐ฅ + 3๐ฆ = 21 โฆโฆโฆโฆeq(1)
2๐ฅ + 7๐ฆ = 20 โฆโฆ.....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(5๐ฅ + 3๐ฆ = 21) โฆโฆโฆโฆeq(3)
5(2๐ฅ + 7๐ฆ = 20) โฆโฆโฆโฆ.eq(4)
So we get
10๐ฅ + 6๐ฆ = 42 โฆโฆโฆโฆeq(3)
10๐ฅ + 35๐ฆ = 100โฆโฆโฆeq(4)
Step 3. Signs are the same so we will subtract the equations
10๐ฅ + 6๐ฆ = 42
- (10๐ฅ + 35๐ฆ = 100)
So we get โ29๐ฆ = โ58
๐ฆ =โ58
โ29
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๐ฆ = 2
Step 3. Substitute for y in eq(1) and solve for x (any equation could have worked, itโs your
choice)
5๐ฅ + 3๐ฆ = 21
5๐ฅ + 3(2) = 21
5๐ฅ + 6 = 21
5๐ฅ = 21 โ 6
5๐ฅ = 15
๐ฅ =15
5
๐ = ๐
Hence ๐ฅ = 3, ๐ฆ = 2
Exercises
Solve the following pairs of simultaneous equations by method of elimination.
1. 4๐ฅ + 3๐ฆ = 17
5๐ฅ โ 2๐ฆ = 4
2. 3๐ฅ โ 5๐ฆ = โ13
โ2๐ฅ + 3๐ฆ = 8
3. 5๐ฅ + 3๐ฆ = 27
2๐ฅ + 5๐ฆ = 26
4. 2๐ฅ + 3๐ฆ = โ8
5๐ฅ โ 2๐ฆ = 18
5. 9๐ฅ + 5๐ = 15
3๐ฅ โ 2๐ = โ6
6. 3๐ฅ + 2๐ฆ = 21
2๐ฅ + 3๐ฆ = 13
7. 7๐ฅ โ 2๐ฆ = 19
3๐ฅ + 5๐ฆ = 14
8. 5๐ฅ + 2๐ฆ = 16
โ3๐ฅ + 4๐ฆ = 6
9. 7๐ฅ + 2๐ฆ = 17
2๐ฅ โ 2๐ฆ = 1
Ministry of Education
Mathematics Grade 9
WEEK 5
LESSON 1
Topic: Algebra 2 โ Solution of Simultaneous Equations โ 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 ๐ + ๐ = ๐๐ โฆโฆโฆโฆโฆ.eq(1)
So ๐๐๐ + ๐ = ๐๐ โฆโฆโฆโฆโฆ.eq(2)
Using the substitution method
Finding s in eq(1) gives
๐ = ๐๐ โ ๐โฆโฆโฆโฆโฆโฆ.eq(3)
Sub. for t in eq(2)
๐๐๐ + ๐ = ๐๐๐
๐๐(๐๐ โ ๐) + ๐ = ๐๐(๐๐ โ ๐)
๐๐๐ โ ๐๐๐ + ๐ = ๐๐๐ โ ๐๐๐
โ๐๐ + ๐๐๐ = ๐๐๐ โ ๐๐๐
๐๐ = ๐๐
๐ =๐๐
๐
๐ = ๐ Sub for s in eq(3)
๐ = ๐๐ โ ๐
๐ = ๐๐ โ ๐
๐ = ๐
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 โ 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 โ one and only one arrow must leave each element of the domain.
1 3
2 5
3 7
2. A set of ordered pairs โ 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 โ1โ.
3. A graph โ 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.
๐ฆ = ๐ฅ2 โ 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 โ 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)
๐บ๐๐ฃ๐๐ ๐กโ๐ ๐๐ข๐๐๐ก๐๐๐ ๐ฆ = 2๐ฅ + 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 โ Use of functional notation
We use symbols to describe a function. For example, ๐: ๐ โ ๐๐. This states that โ๐ is a function
such that x is mapped onto 3๐ฅโ. Here we have been supplied with the rule which must be used in
order to determine the image of any given value for โ ๐ฅโ.
The function ๐: ๐ โ ๐๐ could also be expressed as ๐(๐) = ๐๐ or ๐ = ๐๐ which means the same.
Example 1.
Given ๐: ๐ โ ๐๐ + ๐ find the image of 3
Solution
๐(๐ฅ) = 3๐ฅ + 2
๐(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
๐ฅ ๐ฅ + 1 ๐(๐ฅ) 0 0+1 1
1 1+1 2
2 2+2 3
3 3+1 4 ๐: โ ๐ฅ + 1
or
๐(๐ฅ) = ๐ฅ + 1
or
๐ฆ = ๐ฅ + 1
Exercises
1. Given ๐: โ ๐ฅ + 1, find the image of 2
2. Given ๐: โ 3๐ฅ โ 1, find the images for the domain {1,2,3,4,5}
3. Given ๐(๐ฅ) = ๐ฅ2, find:
a) ๐(โ1)
b) ๐(3)
4. Copy and complete the table
๐ฅ 1 2 3 4 5 6 7 8
๐(๐ฅ) -2 0 3 4 5
From the table write down
a) The rule of ๐(๐ฅ)
b) The values ๐(7)
c) The image of 5
d) The first element whose image is 5
5. a) Given ๐(๐ฅ) = ๐ฅ โ 1, for what value of ๐ฅ will ๐(๐ฅ) = 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 โ 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 โxโ.
Example 1.
Using the domain x={-2, 0, 2} plot the graph of the function ๐ฆ = 2๐ฅ โ 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)
๐ฆ = 2๐ฅ โ 1
Exercises
Draw the graph for each function, using a scale of 1cm to represent 1 unit on both axis.
a) ๐ฆ = ๐ฅ โ 5
b) ๐(๐ฅ) = 2๐ฅ โ 1
c) ๐: ๐ฅ โ 2๐ฅ โ 1
d) ๐: ๐ฅ โ 5๐ฅ + 2
e) ๐ฆ = โ2๐ฅ + 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 โ 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. ๐บ๐๐๐๐๐๐๐ก ๐๐ ๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐, ๐ = ๐๐๐ ๐
๐๐ข๐
Example 1.
Run=8 units (2, 3)
Rise= 8 units
(-2, -5)
๐ = 8
8= 1
Using the formula:
๐ป๐๐ ๐๐๐๐ ๐๐๐๐, ๐ = ๐๐โ๐๐
๐๐โ๐๐
Considering (-2, -5) and (2, 3)
๐ = ๐ฆ2โ๐ฆ1
๐ฅ2โ๐ฅ1
๐ = 3โโ5
2โโ2
๐ = 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 โ Equation of a straight line
๐โ๐ ๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ข๐๐ก๐๐๐ ๐๐ ๐ ๐ ๐ก๐๐๐๐โ๐ก ๐๐๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ: ๐ = ๐๐ + ๐
Where: โmโ is the gradient
And โcโ 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
๐ = ๐๐ + ๐ ; 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
๐ฆ = ๐๐ฅ + ๐
m = 3 and c = 4
Then the equation of this line is
๐ฆ = 3๐ฅ +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 ๐ = ๐๐ โ ๐, write down the gradient of this line and the y-intercept.
Solution
๐ = ๐๐ + ๐
Since the equation is given as: ๐ = ๐๐ โ ๐
Then:
m =2
c = -1
Exercises
1. State the gradient and the y-intercept for each of the following equations:
a) ๐ = ๐๐ + ๐
b) ๐ = โ๐๐ โ ๐
c) ๐ = ๐ +๐
๐
d) ๐ = โ๐ โ ๐
e) ๐ = ๐๐
f) ๐ = ๐
g) ๐ = ๐
h) ๐ = โ๐
๐๐ +
๐
๐
2. Write the specific equation for each straight line given:
a) ๐ = 4 ๐๐๐ ๐ = 2
b) ๐ = โ2 ๐๐๐ ๐ = โ1
c) ๐ = 1 ๐๐๐ ๐ =2
3
d) ๐ = โ9 ๐๐๐ ๐ = 0
e) ๐ = 0 ๐๐๐ ๐ = 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 โ 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
๐ = ๐๐ + ๐ ; 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, ๐ = ๐๐โ๐๐
๐๐โ๐๐
๐ = ๐โ๐
๐โ๐
๐ = ๐
๐
m = 2
now finding c, we apply the second procedure given above
using (4, 3)
we get; 3 = 2(4) + ๐
3 = 8 + ๐
3 โ 8 = ๐
๐ = โ5
Finally, we rewrite the equation
The general form of the equation is ๐ = ๐๐ + ๐
Then the specific equation is given by ๐ = ๐๐ โ ๐
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 ๐ = ๐๐ + ๐ 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 ๐ = ๐๐ + ๐ 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 ๐ = ๐๐ + ๐
WEEK 7
LESSON 1
Topic: Relation and Function โ Using graphs to solve linear simultaneous equations
Example 1.
Solve graphically, the pair of simultaneous equations:
2๐ฅ + ๐ฆ = 8 and 3๐ฅ โ ๐ฆ = 7
Solution
Step 1: Rewrite the equations
2๐ฅ + ๐ฆ = 8 โฆโฆeq(1)
3๐ฅ โ ๐ฆ = 7 โฆโฆeq(2)
Step 2: make y the subject of the formula in both equations (i.e. rewrite the equations in the form
๐ฆ = ๐๐ฅ + ๐)
We get: ๐ฆ = โ2๐ฅ + 8
๐ฆ = 3๐ฅ โ 7
Step 3: construct a table of values for each equation (Assume values for x).
๐ฆ = โ2๐ฅ + 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)
๐ฆ = 3๐ฅ โ 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. ๐ฅ โ ๐ฆ = 1
๐ฅ + ๐ฆ = 3
2. ๐ฆ โ ๐ฅ = 1
๐ฅ + ๐ฆ = 5
3. ๐ฆ = 4๐ฅ + 4
๐ฅ + ๐ฆ = 14
4. ๐ฆ = 4๐ฅ ๐๐๐ ๐ฅ + ๐ฆ = 5
5. 2๐ฅ + ๐ฆ โ 5 = 0 ๐๐๐ ๐ฆ = ๐ฅ โ 4
6. 3๐ฅ + ๐ฆ = 0 ๐๐๐ 4๐ฅ + ๐ฆ = 2
7. 4๐ฅ = ๐ฆ โ 2 ๐๐๐ 3๐ฅ โ ๐ฆ = 8
8. ๐ฅ + 2๐ฆ = 7 ๐๐๐ ๐ฆ โ 2๐ฅ โ 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 โ Graphs of linear inequalities
Example 1.
Draw the graph of ๐ฆ < ๐ฅ + 2
Solution
Write equation as ๐ฆ = ๐ฅ + 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 ๐ โค ๐๐ โฅ.
Then we shade the region (the arrows indicate the shaded part)
Notice also the arrows are not touching the broken lines that represents ๐ฆ < ๐ฅ + 2
Example 2.
Draw the graph of 2๐ฅ + ๐ฆ โค 7
0
1
2
3
4
5
6
7
-3 -2 -1 0 1 2 3 4 5
Solution
Write equation as ๐ฆ = โ2๐ฅ + 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 ๐ฆ โค โ2 + 7
Exercises
Draw graphs to represent each of the following inequations.
a) ๐ฅ + ๐ฆ > 5
b) ๐ฅ + ๐ฆ < 5
c) ๐ฆ > ๐ฅ + 4
d) ๐ฆ < ๐ฅ + 4
e) ๐ฅ + 2๐ฆ โฅ 8
f) 2๐ฅ + ๐ฆ โค 5
g) ๐ฆ โค 5
-2
0
2
4
6
8
10
12
-3 -2 -1 0 1 2 3 4 5
WEEK 7
LESSON 3
Topic: Geometry โ Pythagorasโ Theorem
Pythagorasโ 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 Pythagorasโ 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. ๐2 = ๐2 + ๐2
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:
๐ฉ๐ช๐ = ๐จ๐ฉ๐ + ๐จ๐ช๐
Which is the same as ๐๐ = ๐๐ + ๐๐
Solution
๐2 = ๐2 + ๐2
๐2 = (6๐๐)2 + (8๐๐)2
๐2 = 36๐๐2 + 64๐๐2
๐2 = 100๐๐2
๐ = โ100๐๐2
๐ = 10๐๐
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 โ Pythagorasโ 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:
๐ฉ๐ช๐ = ๐จ๐ฉ๐ + ๐จ๐ช๐
Which is the same as ๐๐ = ๐๐ + ๐๐
Solution
๐2 = ๐2 + ๐2
(10๐๐)2 = ๐2 + (8๐๐)2
100๐๐2 = ๐2 + 64๐๐2
100๐๐2 โ 64๐๐2 = ๐2
๐2 = 36๐๐2
๐ = โ36๐๐2
๐ = 6๐๐
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 = 45ห 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 โ 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) ๐๐
๐๐ =
๐๐
๐๐ =
๐๐
๐๐
5๐๐
7๐๐=
๐๐
6๐๐
๐๐ = 30
7
๐๐ = 4.3๐๐
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 โ 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 โ๐ด๐ต๐ถ โก โ๐๐๐
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, โ๐ด๐ต๐ถ is such that BD = CD. Prove that AB = AC.
WEEK 8
LESSON 3
Topic: Geometry โ 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 180ยฐ so the angle at the centre is half of that.
Example 1.
O
C B
A
If angle AOB = 70ยฐ, 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 = 70ยฐ
Then angle ACB = 1
2(70ยฐ)
= 35ยฐ
Exercises
1.
O
C 20ห B
A
Find the size of the angle AOB.
2.
M
O
105ห 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 โ Angles in a circle (continued)
Theorem 2
Angles in the same segment are equal
Example 1.
C
D
O
A
B
If angle ACB = 59ห, determine the size of the angle ADB. State a reason for your answer.
Solution
Given that angle ACB =59ห
Then angle ADB = angle ACB = 59ห
(angles in the same segment)
Exercises
1. Calculate the size of the angles x and y, giving reasons for your answer.
58ห x
y
37ห
2. If angle AQB = 21ห , determine the magnitude of angle APB, stating a reason for your
answer.
A
B
Q
P
3.
44ห
20ห
Given the two angles in the figure above, determine the remaining four angles stating a
reason for each.
WEEK 9
LESSON 1
Topic: Trigonometry โ 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 ฮฑ
b c
Opposite to angle B Hypotenuse
ฮฒ
C a B
Opposite to side A
Adjacent to side B
Sine of an angle
๐ป๐๐ ๐๐๐๐ ๐๐ ๐๐ ๐๐๐๐๐ =๐๐๐๐๐๐๐๐ ๐๐๐ ๐
๐๐๐๐๐๐๐๐๐๐
๐๐๐ ๐ฝ = ๐๐๐
๐๐๐
i.e. From triangle above
๐๐๐ ๐จ =๐
๐ Or ๐๐๐ ๐ฉ =
๐
๐
Example 1.
Using a scientific calculator, find:
a) sin 30ห b) sin 45ห c) sin 60ห d) sin 75ห
Solution
a) sin 30ห = 0.5
b) sin 45ห = 0.7071 067811865 = .07071 (4 s.f)
c) sin 60ห = 0.8660 (4 s.f)
d) sin 75ห = 0.9659 (4 s.f)
The inverse sine of an angle
๐๐๐โ๐๐ ๐๐ ๐๐๐ ๐๐๐๐
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
๐ด = ๐ ๐๐โ10.429
A = 25.4ห
b) B = arc sin 0.905
B = 64.82ห
Exercises
1. Using a scientific calculator, find the sine for each of the following:
a) sin 8ห c) sin 24ห
b) sin 16ห d) sin 48ห
2. Using a scientific calculator, find the sine for each of the following:
a) sin 53ห c) sin 63ห
b) sin 74ห d) sin 87ห
3. Using a scientific calculator, find the sine for each of the following:
a) sin 23.2ห c) sin 67.6ห
b) sin 7.4ห d) sin 81.9ห
4. Using a scientific calculator, find the sine for each of the following:
a) sinโ1 0.235 c) sinโ1 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 ฮฒ = 0.883
b) sin A = 0.786 d) sin C = 0.7387
WEEK 9
LESSON 2
Topic: Trigonometry โ Trigonometric Ratios (Sine Ratio)
Calculating an unknown angle using Sine Ratio
Example
b 15cm
8cm
Find the magnitude of the angle marked b.
Solution
๐ ๐๐ ๐ = ๐๐๐
โ๐ฆ๐
๐ ๐๐ ๐ = 8๐๐
15๐๐
๐ ๐๐ ๐ = 0.5333
angle b = sinโ1 0.5333
angle b = 32.23ห
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 โ Trigonometric Ratios (Sine Ratio)
Finding a missing side using the Sine Ratio
Example
40ห 15cm
x cm
Find the length of the side marked x.
Solution
๐ ๐๐ ๐ = ๐๐๐
โ๐ฆ๐
๐ ๐๐ 40ยฐ = ๐ฅ ๐๐
15๐๐
sin 40ยฐ ร 15๐๐ = ๐ฅ๐๐
x cm = 0.64278 ร 15cm
x cm = 9.63cm (3 s.f)
Exercises
Determine the size of the marked angle
29.2ห
b 6.2cm
8cm y
52ห
m 6cm 12cm
7cm
50ห 9ห
WEEK 9
LESSON 4
Topic: Trigonometry โ Trigonometric Ratios (Cosine Ratio)
A
Adjacent to angle A ฮฑ
b c
Opposite to angle B Hypotenuse
ฮฒ
C a B
Opposite to side A
Adjacent to side B
Cosine of an angle
๐ป๐๐ ๐๐๐๐๐๐ ๐๐ ๐๐ ๐๐๐๐๐ =๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐
๐๐๐๐๐๐๐๐๐๐
๐๐๐ ๐ฝ = ๐๐ ๐
๐๐๐
i.e. From triangle above
๐๐๐ ๐จ =๐
๐ Or ๐๐๐ ๐ฉ =
๐
๐
Example 1.
Using a scientific calculator, find:
a) cos 30ห b) cos 45ห c) cos 60ห d) cos 75ห
Solution
a) cos 30ห = 0.8660 (4s.f)
b) cos 45ห = 0.7071 (4 s.f)
c) cos 60ห = 0.5
d) cos 75ห = 0.9659 (4 s.f)
The inverse cosine of an angle
๐๐๐โ๐๐ ๐๐ ๐๐๐ ๐๐๐๐
Example 2.
Using a scientific calculator, evaluate the following to determine the angle:
a) cos A = 0.429 b) B = arc cos 0.905
Solution
a) cos A = 0.429
๐ด = ๐๐๐ โ10.429
A = 64.6ห
b) B = arc cos 0.905
B = 25.18ห
Exercises
1. Using a scientific calculator, find the sine for each of the following:
a) cos 8ห c) cos 24ห
b) cos 16ห d) cos 48ห
2. Using a scientific calculator, find the sine for each of the following:
a) cos 53ห c) cos 63ห
b) cos 74ห d) cos 87ห
3. Using a scientific calculator, find the sine for each of the following:
a) cos 23.2ห c) cos 67.6ห
b) cos 7.4ห d) cos 81.9ห
4. Using a scientific calculator, find the sine for each of the following:
a) cosโ1 0.235 c) cosโ1 0.799
b) Arc cos 0.435 d) arc cos 0.9134
5. Using a scientific calculator, find the sine for each of the following unknown angles,
given that:
a) cos a = 0.602 c) cos ฮฒ = 0.883
b) cos A = 0.786 d) cos C = 0.7387
WEEK 10
LESSON 1
Topic: Trigonometry โ Trigonometric Ratios (Cosine Ratio)
Calculating an unknown angle using Cosine Ratio
Example
b 15cm
6cm
Find the magnitude of the angle marked b.
Solution
๐๐๐ ๐ = ๐๐๐
โ๐ฆ๐
๐๐๐ ๐ = 6๐๐
15๐๐
๐๐๐ ๐ = 0.4
angle b = cosโ1 0.4
angle b = 66.42ห
Exercises
Determine the size of the marked angle
6.2cm
8cm y 15.5cm
x
3cm
a b 4.1cm
13cm 6cm 12cm
WEEK 10
LESSON 2
Topic: Trigonometry โ Trigonometric Ratios (Cosine Ratio)
Finding a missing side using the Cosine Ratio
Example
40ห 15cm
x cm
Find the length of the side marked x.
Solution
๐๐๐ ๐ = ๐๐๐
โ๐ฆ๐
๐๐๐ 40ยฐ = ๐ฅ ๐๐
15๐๐
cos 40ยฐ ร 15๐๐ = ๐ฅ๐๐
x cm = 0.7660 ร 15cm
x cm = 11.49cm (4 s.f)
Exercises
Determine the size of the marked angle
32.1ห
8cm 6.2cm
52ห y
b
29ห
m 6cm c
7cm
9ห
WEEK 10
LESSON 3
Topic: Trigonometry โ Trigonometric Ratios (Tangent Ratio)
A
Adjacent to angle A ฮฑ
b c
Opposite to angle B Hypotenuse
ฮฒ
C a B
Opposite to side A
Adjacent to side B
Tangent of an angle
๐ป๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐ ๐๐๐๐๐ =๐๐๐๐๐๐๐๐ ๐๐๐ ๐
๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐
๐๐๐ ๐ฝ = ๐๐๐
๐๐ ๐
i.e. From triangle above
๐๐๐ ๐จ =๐
๐ Or ๐๐๐ ๐ฉ =
๐
๐
Example 1.
Using a scientific calculator, find:
a) tan 30ห b) tan 45ห c) tan 60ห d) tan 75ห
Solution
a) tan 30ห = 0.5774 (4s.f)
b) tan 45ห = 1
c) tan 60ห = 1.732(4 s.f)
d) tan 75ห = 3.732 (4 s.f)
The inverse tangent of an angle
๐๐๐โ๐๐ ๐๐ ๐๐๐ ๐๐๐ ๐
Example 2.
Using a scientific calculator, evaluate the following to determine the angle:
a) tan A = 0.429 b) B = arc tan 2.905
Solution
a) tan A = 0.429
๐ด = ๐ก๐๐โ10.429
A = 23.22ห
b) B = arc cos 2.905
B = 71ห
Exercises
1. Using a scientific calculator, find the sine for each of the following:
a) tan 8ห c) tan 24ห
b) tan 16ห d) tan 48ห
2. Using a scientific calculator, find the sine for each of the following:
a) tan 53ห c) tan 63ห
b) tan 74ห d) tan 87ห
3. Using a scientific calculator, find the sine for each of the following:
a) tan 23.2ห c) tan 67.6ห
b) tan 7.4ห d) tan 81.9ห
4. Using a scientific calculator, find the sine for each of the following:
a) tanโ1 0.235 c) tanโ1 1.799
b) arc tan 0.435 d) arc tan 2.434
5. Using a scientific calculator, find the sine for each of the following unknown angles,
given that:
a) tan a = 0.602 c) tan ฮฒ = 3.803
b) tan A = 0.786 d) tan C = 1.2367
WEEK 10
LESSON 4
Topic: Trigonometry โ Trigonometric Ratios (Tangent Ratio)
Calculating an unknown angle using Tangent Ratio
Example
b 15cm
6cm
13.75cm
Find the magnitude of the angle marked b.
Solution
๐ก๐๐ ๐ = ๐๐๐
๐๐๐
๐ก๐๐ ๐ = 13.75๐๐
6๐๐
๐ก๐๐ ๐ = 2.292
angle b = tanโ1 2.292
angle b = 66.43ห
Exercises
Determine the size of the marked angle
6.2cm y 4.3cm
8cm
x
3cm 6.3cm
a c
6cm 3.1cm
9cm
Ministry of Education
MINISTRY OF EDUCATION
EASTER TERM 2021
GRADE 9
MATHEMATICS
WORKSHEETS
Ministry of Education
WEEK 11
LESSON 1
Topic: Trigonometry โ Trigonometric Ratios (Tangent Ratio)
Finding a missing side using the Tangent Ratio
Example
41.4ห
x cm
13.2cm
Find the length of the side marked x.
Solution
๐ก๐๐ ๐ = ๐๐๐
๐๐๐
๐ก๐๐ 40ยฐ = ๐ฅ ๐๐
13.2๐๐
tan 40ยฐ ร 13.2๐๐ = ๐ฅ๐๐
x cm = 0.8816 ร 13.1cm
x cm = 11.64cm (4 s.f)
Exercises
Determine the size of the marked angle
x
29.1ห
6cm 5.3cm
62ห
b
29ห p
6cm
7cm
9ห
m
Ministry of Education
WEEK 11
LESSON 2
Topic: Trigonometry โ Trigonometric Ratios (mixed problem)
Acronym for remembering ratios
๐บ๐ถ๐ฏ ๐ช๐จ๐ฏ ๐ป๐ถ๐จ
๐ก๐๐ ๐ = ๐๐๐
๐๐๐
๐๐๐ ๐ = ๐๐๐
โ๐ฆ๐
๐ ๐๐ ๐ = ๐๐๐
โ๐ฆ๐
OR
S โ Some
O โ old Sine
H โ horses
C โ Can
A โ always Cosine
H โ hear
T โ Their
O โ ownerโs Tangent
A โ 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.
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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 32ห.
3. A ladder 50m long leans against a wall and is inclined at an angle of 45ห 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 38.5ห 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 49.5ห 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 = 90ห
Given that PR = RQ = 5m, calculate:
a) The size of the angle RPQ
b) The width of the river
c) The distance PQ
Ministry of Education
WEEK 11
LESSON 3
Topic: Trigonometry โ Complementary Angles
Complementary angles are angles whose sum is 90ห
Example
A
b
c
B a C
From the triangle above, ๐ด + ๐ถ = 90ยฐ
Then ๐ถ = 90ยฐ โ ๐ด
Considering:
๐๐๐ ๐จ =๐
๐
And ๐๐๐ ๐ช = ๐๐๐(๐๐ โ ๐จ) =๐
๐
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 25ห = 0.423, determine what cos 65ห is equal to. State a conclusion.
Solution
Cos 65ห = 0.4226 = 0.423 (3 s.f)
We can conclude that 25ห plus 65ห gives 90ห which make them complementary. Hence the sine of
one is equal to the cosine of the other.
Exercises
1. Given that sin 27ห = 0.454, without the use of calculator determine the value of cos 63ห.
2. Given that sin 59ห = 0.857, without the use of calculator determine the value of cos 31ห.
3. Compare the cosine for each below:
a) sin 34ห b) sin 40ห c) sin 60ห d) sin 30 e) sin 45ห f) sin 90ห
Ministry of Education
WEEK 11
LESSON 4
Topic: Trigonometry โ 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
observerโs 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 observerโs 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 observerโs 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 38.5ห.
a) Draw an illustration of the above problem
b) Calculate the height of the tree.
Solution
a) T top of the tree
38.5ห
Manโs eye level X 25m B base of the tree
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b) Considering the โ๐๐ต๐:
๐ก๐๐ ๐ = ๐๐๐
๐๐๐
Then: tan ๐ = ๐๐ต
๐ต๐
๐ก๐๐ 38.5ยฐ = ๐๐ต
25๐
๐ก๐๐ 38.5ยฐ ร 25๐ = ๐๐ต
๐๐ต = 0.7954 ร 25๐
=19.88m
=20m (nearest metre)
Exercises
1. A sailor sights the top of a cliff at an angle of elevation of 12ห. 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 50ห.
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 42.5ห. 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.
Ministry of Education
WEEK 12
LESSON 1
Topic: Trigonometry โ 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 observerโs 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 32ห.
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
32ห
32ห
Manโs eye level X 25m B base of the tree
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b) Considering the โ๐๐ต๐:
๐ก๐๐ ๐ = ๐๐๐
๐๐๐
Then: tan ๐ = ๐๐ต
๐ต๐
๐ก๐๐ 32ยฐ = ๐๐ต
25๐
๐ก๐๐ 32ยฐ ร 25๐ = ๐๐ต
๐๐ต = 0.6249 ร 25๐
=15.6m
=16m (nearest metre)
Exercises
1. From the top of a cliff a man sights a boat at an angle of depression of 22ห. 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 27ห.
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 25ห. 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 19ห. Calculate the height of the building.
Ministry of Education
WEEK 12
LESSON 3
Topic: Statistics โ 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 โ the average value for a set of data
๐โ๐ ๐๐๐๐, ๐ฅ ฬ = ๐ ๐ข๐ ๐๐ ๐ ๐๐๐๐๐
๐๐ข๐๐๐๐ ๐๐ ๐ ๐๐๐๐๐
๐ฅ ฬ = โ ๐ฅ
โ ๐
Where: โ ๐ฅ โ sum of scores
โ ๐ - 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
๐ฅ ฬ = โ ๐ฅ
โ ๐
๐ฅ ฬ = 2+3+1+4+6+2+5+7+2+4
10
๏ฟฝฬ ๏ฟฝ = 36
10
= 3.6
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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.
๐ ฬ = โ ๐๐
โ ๐
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
โ ๐ = 60 โ ๐๐ฅ = 14500
Note: fx means f ร x
๐ฅ ฬ = โ ๐๐ฅ
โ ๐
๐ฅ ฬ = 14500
60
= 242 (nearest whole number)
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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
Ministry of Education
WEEK 12
LESSON 2
Topic: Statistics โ 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 โ Frequency
x โ scores
fx โ f ร x
: โ ๐ โ sum of scores
โ ๐๐ฅ - 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.
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Solution
x Tally Frequency
(f)
fx
10 17 170
11 18 198
12 12 144
13 9 117
14 4 56
โ ๐ = 60 โ ๐๐ฅ = 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.
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WEEK 12
LESSON 4
Topic: Statistics โ Measures of Central Tendency (continued)
The Median โ the middle or the central value of a set of data arranged in ascending or
descending order. It is represented by ๐2.
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
๐2 = 158
โด ๐กโ๐ ๐๐๐๐๐๐ โ๐๐๐โ๐ก, ๐2 = 158๐๐
b) The heights in ascending order
154, 155, 156, 157,158, 159, 161,163
๐2 =157+158
2
โด ๐กโ๐ ๐๐๐๐๐๐ โ๐๐๐โ๐ก, ๐2 = 157.5๐๐
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
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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
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WEEK 13
LESSON 1
Topic: Statistics โ Measures of Central Tendency (continued)
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.
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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).
๐ป๐๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐ ๐๐๐ = ๐
๐(๐ + ๐)๐๐ ๐๐๐๐
So, the position of the median, ๐2 = 1
2(50 + 1)๐กโ ๐๐๐๐
= 25.5๐กโ ๐๐๐๐
This implies that the median is the average of the 25๐กโ ๐๐๐ ๐กโ๐ 26๐กโ ๐๐๐ ๐๐๐ฃ๐๐ก๐๐๐๐
From the table it can be seen that the 25๐กโ ๐๐๐ ๐กโ๐ 26๐กโ ๐๐๐ ๐๐๐ฃ๐๐ก๐๐๐๐ both carries scores of 5
Then ๐2 = 5
NOTE: If there were two different observations at ๐๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐๐ 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
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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?
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WEEK 13
LESSON 2
Topic: Statistics โ Measures of Central Tendency (continued)
The Mode โ 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
Ministry of Education
WEEK 13
LESSON 3
Topic: Statistics โ 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, โฆ 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
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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.
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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
2. There are 25 participants in a shooting competition. The score of each participant is listed
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.
Ministry of Education
WEEK 14
LESSON 1
Topic: Statistics โ 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.
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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
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WEEK 14
LESSON 2
Topic: Statistics โ 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
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b)
Histogram
40
35
30
25
20
15
10
5
5-9 10-14 15-19
Exercises
1. 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) 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.
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WEEK 14
LESSON 3
Topic: Statistics โ 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
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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.
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WEEK 14
LESSON 4
Topic: Statistics โ 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 โ draw a table to determine the values for the cumulative frequency.
Step 2 โ draw the polygon
0
10
20
30
40
50
60
0 1 2 3 4 5 6 7 8 9 10
Cumulative frequency
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Exercises
1. There are 25 participants in a shooting competition. The score of each participant is listed
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
2. 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 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
Ministry of Education
WEEK 15
LESSON 1
Topic: Statistics โ 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.
๐ด = 2 โ34 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 โ5 1 ๐๐๐ค ร 2 ๐๐๐๐ข๐๐๐
2 ร 1 ๐๐๐ก๐๐๐ฅ
โ10
2 ๐๐๐ค๐ ๐๐ฆ 1 ๐๐๐๐ข๐๐
2 ร 1 ๐๐๐ก๐๐๐ฅ
๐ ๐ ๐๐ ๐ ๐
2 ๐๐๐ค๐ ๐๐ฆ 3 ๐๐๐๐ข๐๐๐
2 ร 3 ๐๐๐ก๐๐๐ฅ
Exercises
1. Given that A = 2 โ34 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 ๐ = 5 โ1
โ3 07 6
, state the order of the matrix, the second element in the second
column and the third element in the first column.
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3. Given that A = โ3 4 62 1 0
, state:
a) The order of the matrix
b) The first element in the second row
c) The second element in the first row
4. If ๐ต = 2 1 โ90 5 13 6 12
, state:
a) The order of the matrix
b) The third element in the second row
c) The second element in the fourth row
5. Given ๐ถ = 123
11 ,
a) The order of the matrix
b) The element in the second row
6. If ๐ท = โ2 3 โ56 7 45 0 โ4
โ103
a) The order of the matrix
b) The third element in the second row
c) The second element in the fourth column
7. Given that the matrix ๐ = 12 9 โ123 10 4
โ5 11 2
0 6โ7 514 11
, state:
a) The order of the matrix
b) The third element in the third row
c) The second element in the fourth column
d) The first element in the second column
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WEEK 15
LESSON 2
Topic: Statistics โ Matrices (continued) โ Addition and Subtraction
Matrices of the same order can be added or subtracted by adding or subtracting the
corresponding elements.
Example 1.
If ๐ด = 2 4
โ1 6 ๐๐๐ ๐ต =
โ4 57 6
, determine:
a) A + B
b) A โ B
Solution
a) A + B = 2 4
โ1 6 +
โ4 57 6
= 2 + โ4 4 + 5โ1 + 7 6 + 6
= โ2 96 12
b) A โ B = 2 4
โ1 6 โ
โ4 57 6
= 2 โ โ4 4 โ 5โ1 โ 7 6 โ 6
= 6 โ1
โ8 0
Exercises
1. If ๐ด = 2 4
โ1 6 , ๐ต =
4 71 4
๐๐๐ ๐ถ = โ4 57 6
, determine:
a) A + B d) B โ C
b) A + B + C e) C โ A
c) C + B f) A โ B โ C
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WEEK 15
LESSON 3
Topic: Statistics โ Matrices (continued) โ Scalar multiplication
Given that ๐ด = ๐ ๐๐ ๐
,
then ๐๐ด = ๐ ๐ ๐๐ ๐
= ๐ ๐๐ ๐
๐ = ๐๐ ๐๐๐๐ ๐๐
where k is a constant
Example 1.
Given that ๐ด = โ2 15 0
๐๐๐ ๐ = 4, determine kA.
Solution
๐๐ด = 4๐ด = 4 โ2 15 0
= 4 ร โ2 4 ร 14 ร 5 4 ร 0
= โ8 420 0
Exercises
Given ๐ = 3 1
โ2 5 ๐๐๐ ๐ =
4 68 โ9
, evaluate:
a) 2P
b) 3P โ Q
c) 5Q
d) 4Q + 2P
e) P + 2Q โ Q
f) Q + 3P + 3Q
g) 2P โ Q โ P + 2P
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WEEK 15
LESSON 4
Topic: Statistics โ Matrices (continued) โ Equal matrix
Two matrices are equal if they are in the same order and the corresponding elements are equal.
i.e. Given that ๐ด = ๐ ๐๐ ๐
๐๐๐ ๐ต = ๐ ๐๐ โ
,
then ๐ด = ๐ต ๐๐ ๐ ๐๐ ๐
= ๐ ๐๐ โ
i.e. ๐ = ๐, ๐ = ๐, ๐ = ๐, ๐ = โ
Example 1.
Given that ๐ฅ โ34 ๐ฆ
= โ2 โ34 6
, determine the values of x and y
Solution
๐ฅ โ34 ๐ฆ
= โ2 โ34 6
Equating the corresponding elements, we get
x = -2 and y = 6
Exercises
1. Given that x yy 6 =
โ2 โ3โ3 6
, determine the values of x and y
2. If a bc d
= โ1 โ35 0
, determine the values of a , b, c, and d.
3. Given that ๐ด = 4 31 2
๐๐๐ ๐ต = โ2 โ13 5
, determine X in the following equation
๐ฟ โ ๐ฉ = ๐จ
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