1ST TERM JSS3 MATHEMATICS Scheme of Work and Note

FIRST TERM

SUBJECT: MATHEMATICS   CLASS: JSS3

SCHEME OF WORK

WEEKS TOPIC

REFERENCE BOOKS

New General Mathematics by M. F Macrae et al bk 3

Essential Maths by AJS OluwasanmiBk 3

WEEK ONE

NUMBER BASE CONVERSIONS

People count in twos, fives, twenties etc. Also the days of the week can be counted in 24 hours. Generally people count in tens. The digits 0,1,2,3,4,5,6,7,8,9 are used to represent numbers. The place value of the digits is shown in the number example: 395:- 3 Hundreds, 9 Tens and 5 Units. i.e. 3X10² + 9 X 10¹ +5 X 10⁰.

Since the above number is based on the powers of tens it is called the base ten number system i.e. 300 + 90 + 5

Also 4075 = 4 Thousand 0 Hundred 7 Tens 5 Units i.e. 4 x 10³ + 0 X 10² + 7 X 10¹ + 5 X 10⁰ Other Number systems are sometimes used.

For Example: The base 8 system is based on the power of 8. For example: Expand 647₈, 26523₇, 1011012,

(a)  647₈ = 6 x 8² + 4 x 8¹ + 7 X 8⁰ =6 x 64 + 4 x 8 + 7 x 1

(b)  26523₇ =2 x 7⁴ + 6 x7³ + 5 x 7² + 2x 7¹ + 3 x 7⁰

(c)  101101₂= 1 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 0 x2¹ + 1 x 2⁰

EVALUATION

Expand The Following

1.  434₃  

2.  101111₂

CONVERSION TO DENARY SCALE (BASE TEN)

When converting from other bases to base ten the number must be raised to the base and added.

Worked Examples:

Convert the following to base 10

(a)  17₈  

(b)  11011₂

Solutions:

(a)  17₈ = 1 X 8¹ + 7 X 8⁰ = 1 X 8 + 7 X 1 = 8 + 7 = 15

(b)  11011₂ = 1 X 2⁴ + 1 X 2³ + 0 X 2² + 1 X 2¹ + 1 X 2⁰ = 1 X 16 + 1 X 8 + 0 X 4 + 1 X 2 +

1 X 1= 16 + 8 + 0 + 2 + 1 = 27

EVALUATION

Convert The Following To Base Ten:

(a)  10100₂

(b)  2120₃

CONVERSION FROM BASE TEN TO OTHER BASES

To change a number from base ten to another base

1.  Divide the base ten number by the new base number.

2.  Continue dividing until zero is reached

3.  Write down the remainder each time

4.  Start at the last remainder and read upwards to get the answer.

Worked Examples:

1.  Convert 68₁₀ to base 6

2.  Covert 129₁₀ to base 2

Solutions:

1.   6 68

  6 11 R 2

  6 1 R 5

0 R 1

= 152₆

2.   2 129

  2  64 R 1

  2  32 R 0

  2  16 R 0

  2  8 R 0

  2  4 R 0

  2  2 R 0

  2  1 R 0

  2  0 R 1

= 10000001₂

EVALUATION

1.  Convert 569₁₀ to base 8

2.  Convert 100₁₀ to base 2

GENERAL EVALUATION

Convert the following to base seven

1. 405_ten

2. 876_ten

Evaluate the following

3. 5 – 3 + 4

READING ASSIGNMENT

New Gen Math Book 3 pg 15-17

Essential Mathematics for J.S.S.3 Pg 5 -9

WEEKEND ASSIGNMENT

1.  Express 342₆ as number in base 10 (a) 134 (b) 341 (c) 143

2.  Change the number 10010₂ to base 10 (a) 1001 (b) 40 (c) 18

3.  Express in base 2, 100₁₀ (a) 100100 (b) 1100100 (c) 11001

4.  Convert 120 base 10 to base 3 (a) 11110₃ (b) 1210₃ (c) 12110₃

5.  Convert 25 base 10 to base 2 (a) 11001₂ (b) 1001₂ (c) 1100₂

THEORY

1.  Convert 1264₈ to base 10

2.  Convert 105₁₀ to base 2

WEEK TWO

SOLVING EQUATION EXPRESSIONS

WORD PROBLEMS

Worked Examples:

1. Find 1/4 of the positive difference between 29 & 11

1. The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number
2. The sum of 35 and a certain number is divided by 4 the result is equal to double the number. Find the number.

Solutions:

1.  Positive Difference 29 – 11 = 18

 1/4 of 18 = 4 ²/5

2.  Let the number be x

 x X 5 = 20 – 2x

 5x = 20 – 2x

 5x + 2x = 20

 7x = 20

 x = 20/7 = 2

3.  Let the number be n

 sum of 35 and n = n + 35

 divided by 4 = n + 35

4

 result = 2 X n

 therefore n + 35 = 2n

4

 n + 35 = 8n

 8n – n = 35  

 7n = 35

 n = 35/7 = 5

EVALUATION

1.  From 50 subtract the sum of 3 & 5 then divide the result by 6

2.  The sum of 8 and a certain number is equal to the product of the number and 3 find

the number.

SOLVING EQUATION EXPRESSIONS WITH FRACTION

Always clear fractions before beginning to solve an equation.

To clear fractions, multiply each term in the equation by the LCM of the denominations of the fractions.

Examples:

Solve the following

1.  x = 2

 9

2.  x + 9 + 2 + x = 0

5   2

3.  2x = 5x + 1 + 3x – 5

7   2

Solutions:

1.  x = 2

 9

 Cross multiply

x = 18

2.  x + 9 + 2 + x = 0

  5   2

 Multiply by the LCM (10)

 10 X (x + 9) + 10 X (2 + x) = 0 X 10

5 2

 2 (x + 9) + 5 (2 + x) = 0

 2x + 18 + 10 + 5x = 0

 2x + 5x + 28 = 0

 7x = -28

 x = -28/7 = -4

3.  2x = 5x + 1 + 3x – 5

7   2

 Multiply by the LCM (14)

 14 X 2x = 14 (5x + 1) + 14 (3x – 5)

7 2

 28x = 2 (5x + 1) + 7 (3x – 5)

28x = 10x + 2 + 21x – 35

 28x = 31x – 33

 28x – 31x = -33

 -3x = -33

 x = 33/3 = 11

EVALUATION

Solve the following equations.

1.  7/3c = 21/2

2.   6   = 11

 y + 3   y – 2

3.  3 – 4   = 0

 2b – 5   b – 3

Furthermore, we can consider the word equations or expressions into:

• Sum & Differences
• Products
• Expressions with fractions & equations

SUM & DIFFERENCES

The sum of a set of numbers is a result obtained when the numbers are added together. The difference between two numbers is a result of subtracting one number from the other.

Worked Examples:

1.  Find the sum of -2 & -3.4

2.  Find the positive difference between 19 & 8

3.  The difference between two numbers is 7. If the smaller number is 7 find the other.

4.  The difference between -3 and a number is 8, find the two possible values for the number.

5.  Find the three consecutive numbers whose sum is 63.

Solutions:

1.  -2 + -3.4 = -5.4

2.  19 – 8 = +11

3.  let the number be Y i.e. Y -7 = 7

 i.e Y = 7 + 7 = 14

4.  Let M represent the number

 M – (-3) = 8

 m + 3 = 8

 m = 8 – 3

 m = +5

 also -3 – m = 8

 -m = 8 + 3

 -m = 11

 m = -11

 the possible values are +5 & -11

5.  Consecutive numbers are 1,2,3,4,5,6,………….. Consecutive odd numbers are

1,3,5,7,9……….. consecutive even numbers are 2, 4, 6, 8,10……….

Representing in terms of X, we have 2X, 2X + 2, 2X + 4, 2X + 6, 2X + 8, 2X + 10…………

 for consecutive even numbers, we have X, X + 2, X + 4, X + 6…….

 for consecutive odd numbers, we have X + 1, X + 2, X + 3, X + 4…

for consecutive numbers.

 let the first number be x,

 let the second number be x + 1

 let the third number be x + 2

 Therefore x + x + 1 + x + 2 = 63

 3x + 3 = 63

 3x = 63 – 3  

3x = 60

 x = 60 /3

 = 20

The numbers are 20, 21, and 22.

EVALUATION

1.  Find the sum of all odd numbers between 10 and 20

2.  The sum of four consecutive odd numbers is 80 find the numbers
3.  The difference between 2 numbers is 9, the largest number is 32 find the numbers.

PRODUCTS

The product of two or more numbers is the result obtained when the numbers are multiplied together.

Worked Examples:

1.  Find the product of – 6, 0.7, &

2.  The product of two numbers is 8 .If one of the numbers is 1/4 find the other.

3.  Find the product of the sum of -2 & 9 and the difference between -8 & -5.

Solutions:

1.  Products -6 x 0.7 x

 -6 x 7/10 x 20/3 = -6 x 7 x 20

10 x 3

 = -2 x 7 x 2 = -28

2.  Let the number be x

 X x = 8multiply both sides by 4

 x = 8 x 4 = 33

3.  Sum = -2 + 9 = 7

 Difference = -5-(-8) = -5 + 8 = 3

 Products= 7 x 3 = 21

EVALUATION

1. The product of three numbers is 0.084 if two numbers are 0.7 & 0.2 find the third number.
2. Find the product of the difference between 2 & 7 and the sum of 2 & 7.
3. From 50 subtract the sum of 3 & 5 then divide the result by 6.
4. The sum of 8 and a certain number is equal to the product of the number and 3 find the number.

Reading Assignment

New Gen Maths for J.S.S 3 Pg 20- 24

Essential Mathematics for J.S.S 3 Pg 85-87

PROPORTION

Proportion can be solved either by unitary method or inverse method. When solving by unitary method, always

• Write in sentence the quantity to be found at the end.
• Decide whether the problem is either an example of direct or inverse method
• Find the rate for one unit before answering the problem.

Examples

1. A worker gets N 900 for 10 days of work, find the amount for (a) 3 days (b) 24 days (c) x days

Solution

For 1 day =N 900

1 day = 900/10 = N90

a. For 3 days =3 x 90 = 270

b. For 24 days = 24×90 = N 2,160

c. For x days =X x 90 = N 90 x

INVERSE PROPORTION

Example

1. Seven workers dig a piece of ground in 10 days. How long will five workers take?

Solution:

For 7 workers =10 days

For 1 worker =7×10=70 days

For 5 workers=70/5 =14 days

1. 5 people took 8 days to plant 1,200 trees, How long will it take 10 people to plant the same number of trees

Solution:

For 5 people =8 days

For 1 person =8×5=40 days

For 10 people =40/10 =4 days

CLASS WORK

1. A woman is paid N 750 for 5 days, Find her pay for (a) 1 day (b) 22 days
2. A piece of land has enough grass to feed 15 cows for x days. How long will it last (a) 1 cow (b) y cows
3. A bag of rice feeds 15 students for 7 days .How long would the same bag feed 10 students

Note on direct proportion: this is an example of direct proportion .The less time worked (3 days) the less money paid (#270) the more time worked (24 days) the more money paid (N 2,160)

COMPOUND INTEREST

Interest is a payment given for saving or borrowing money. It can either be simple interest or compound interest. It is simple interest when the interest is calculated on the principal while it is compound interest if interest is calculated on the amount at the end of each period. Amount is the sum of the principal and the interest.

Example:

Find the amount on borrowed for at simple interest.

Solution:

and so that . Substituting the values, we will obtain

Example:

Find the amount that becomes if saved for at per annum simple interest.

Solution:

1st year Principal  

interest

2nd year Principal  

interest

3rd year Principal  

interest

AMOUNT  

Alternatively, we can also solve the question with the use of the formula where represent the time or duration.

Then, substituting into the formula, we can have

()

EVALUATION

1. What is simple interest?
2. Define compound interest.
3. Calculate of

WEEKEND ASSIGNMENT

1. Esther is 3 times as old as her sister Tolu, if the sum of their ages is 20 years. Find the difference between their ages.(a) 20 years  (b) 8 years  (C)  10 years

   2.  9 was subtracted from a certain number and the result was divided by 4 if the finalanswer is 5 what was the original number? (a) 29  (b) 18  (c) 20

   3.  A woman is 4 times as old as her son. In five years time she will be 3 times as old as her son. How old is the woman (a) 50 yrs  (b) 40 yrs  (c) 45 yrs

   4.  Bayo is 4 times as old as his sister Tolu. If the sum of their ages is 20 years, find the difference between their ages. (a) 12 yrs  (b) 15 yrs  (c) 18 yrs

   5.  Subtract the square root of 4 from the square of 4 and divide the result by 2

   (a) 2  (b) 4  (c) 7

THEORY

1.  Divide 36 by the difference between the product of 3 & 6 and the square root of 36.

2.  When I add 45 to a certain number, and divide the sum by 2, the result is the same as five times the number, what is the number?

WEEK THREE

BINARY NUMBERS (BASE 2 NUMBERS)

• Addition in base 2
  
• Subtraction in base 2
  
• Multiplication & Division in base 2
  

ADDITION IN BASE TWO

We can add binary numbers in the same way as we separate with ordinary base 10 numbers.

The identities to remember are:-

0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 10, 1 + 1 + 1 = 11, 1 + 1 + 1 + 1 = 100

Worked Examples

Simplify the following

1.  1110 + 1001  2.  1111 + 1101 + 101

Solutions:

1.  1110

+  1001

10111

2.  1111

+ 1101

101

100001

Note: 11 take 1 carry 1

10 take 0 carry1

100 take 0 carry 10

EVALUATION

1.  Simplify the following 101 + 101 +111

2.  10101 + 111

ADDITION IN BICIMALS

In bicimals, the binary point are placed underneath each other exactly the same way like ordinary decimals.

Example:

1. 1.1011_two + 10.1001_two + 10.01

2. 10.001_two+ 101.111

Solution:

1. 1 . 1011

10 . 1001

10 . 0100

 110. 1000_two

2. 101.111

10.001

1000.000

SUBTRACTION IN BASE TWO

The identities to remember on subtraction are: 0 – 0 = 0, 1 – 0 = 1, 10 – 1 = 1, 11 – 1 = 10, 100 – 1 = 11

Worked Examples

Simplify the following:-

(a)  1110 – 1001  (b) 101010 – 111

Solutions:

(a)  1110

– 1001

101

SUBTRACTION IN BICIMAL

Example

101.101_two – 11.011_two

101.101

11.011

10.010_two

EVALUATION

1.  10111÷110  

2.  10001 x 11

READING ASSIGNMENT

New Gen Maths Book 3, chapter 1 Exercise 1e pg 18 Nos 1-12

Essential Mathematics for J.S.S.3 Pg 8-10

WEEKEND ASSIGNMENT

1.  Express 342₆ as a number in base 10.  (a) 342   (b) 3420 (c) 134

2.  Change the number 10010 to base 10  (a) 18 (b) 34 (c) 40

3.  Express in base two the square of 11  (a) 1001 (b) 1010 (c) 1011

4.  Find the value of (101)² in base two (a) 1010 (b) 1111 (c) 1001

5.  Multiply 100001₂ by 11 (a) 1001 (b) 1100011 (c) 10111

THEORY

1.  Calculate 110₂ x (1011₂ + 1001₂ – 101₂)

2.  Convert 110111 to base five

WEEK FOUR

MULTIPLICATION AND DIVISION IN BASE TWO

In multiplication, 0 x 0 = 0, 1 x 0 = 0, 1 x 1 = 1.

When there is long multiplication of binary numbers, the principle of addition can be used to derive the answer. Under division, the principle of subtraction can be used.

Worked Examples:

1.  1110 x 111  

Solution:

1.  1110

X 110

0000

1110

1110

1010100

2. Multiply 11.01 by 1.1 in base two

Solution:

1101

x 11

1101

1101

100.111

DIVISION IN BASE TWO

Example:

Divide 110010101 by 1111

Solution:

110010101₂ ÷ 1111₂ = 11011₂

Example:

Find the square root of 1010001₂

Solution:

First convert it to base 10

10100012 = 1 x 2 ⁶+ 0 + 1 x 2 ⁴+ 0 + 0 + 0 + 1 x 2⁰

= 64 + 16 + 1 = 81_ten

The square root of 81 = 9_ten

2 9  1

2 4  0

2 2  0

2 1  1

0 = 1001₂

EVALUATION

1.  10111÷110  

2.  10001 x 11

READING ASSIGNMENT

New Gen Maths Book 3, chapter 1 Exercise 1e pg 18 Nos 1-12

Essential Mathematics for J.S.S.3 Pg 8-10

WEEKEND ASSIGNMENT

1.  Express 342₆ as a number in base 10.  (a) 342   (b) 3420 (c) 134

2.  Change the number 10010 to base 10  (a) 18 (b) 34 (c) 40

3.  Express in base two the square of 11  (a) 1001 (b) 1010 (c) 1011

4.  Find the value of (101)² in base two (a) 1010 (b) 1111 (c) 1001

5.  Multiply 100001₂ by 11 (a) 1001 (b) 1100011 (c) 10111

THEORY

1.  Calculate 110₂ x (1011₂ + 1001₂ – 101₂)

2.  Convert 110111 to base

WEEK FIVE

RATIONAL AND NON-RATIONAL NUMBERS AND COMPOUND INTEREST

RATIONAL AND NON-RATIONAL NUMBERS

Numbers which can be written as exact fractions or ratios in the form are called rational numbers. For example, we can write these numbers as

In addition, rational numbers are also numbers that can be written as recurring decimals, for instance: is equivalent respectively to the following:

WeNumbers which cannot be written as exact fractions or recurring decimals are called non-rational numbers. Examples of non-rational numbers are

SQUARE ROOTS

Since rational numbers are not perfect squares, so their square roots cannot be obtained easily except by trial and error method or by the use of Table of Square Roots in the four-figure table.

Example 1:

Find to three significant figures by the use of tables.

Solution:

gives from the table. Hence, answer is to .

Example2:

Find to the nearest tenth by the use of tables.

Solution:

is equivalent to . This is equal to. We can now look up from the table to give. So that.

Hence, answer is to the nearest tenth.

EVALUATION

1. Which of the following is an irrational number?
2. Which of the following is a rational number?
3. Find the square root of, leaving your answer in one decimal place.

DIRECT AND INVERSE VARIATION

DIRECT VARIATION

This is used to describe quantities which vary in proportions to each other, such that as one increases the other increases, and as one decreases the other decreases. Thus, if P varies directly as R, then the expression symbolically becomes. The expression can now be written in equation form as

Where has been replaced by is a constant of variation. It can also be expressed as

The equation is the equation of variation.

Example 1:

If varies directly as the square of, find the law of variation between given that when Find the value of when and the value of

Solution:

 and  the law of variation becomes

For substitution gives  .

Then .

For substitution gives  

such that

then

GRAPHICAL REPRESENTATION OF DIRECT VARIATION

Data collected from quantities that vary directly can be represented graphically. This will give a straight line graph through the origin as shown below.

Example 2:

Given that distance varies directly with time, consider the table below and plot a graph for such relationship.

Solution:

EVALUATION

1. varies directly as and when Find when
2. If increases by from question find the percentage change in .

INVERSE VARIATION

This variation means that related quantities vary inversely or as reciprocal to each other. Hence as one increases the other decreases; and as one decreases, the other increases. Thus if varies inversely as , symbolically this is written as .The expression can now be written in equation form as.

Where has been replaced by is a constant of variation. It can also be expressed as

The equation is the equation of variation.

Example 3:

Given that is inversely proportional to, and that, find the (a) relationship between and (b) value of when

Solution:

such that and

(a) is the required relationship between and (b)

EVALUATION

1. The current in in an electric circuit varies inversely with the resistance If a current of is produced by a resistance of what current will be produced by a resistance of ?
2. Find the percentage change in the current from question if the resistance is decreased by

GRAPHICAL REPRESENTATION OF INVERSE VARIATION

The graph here will not be a straight line from the origin instead it will give us a curve.

Example 4: Given that speed varies inversely to time use the below table to plot a graph of an inverse relationship between

Solution:

GENERAL EVALUATION

1. Factorize the expression
2. Factorize
3. What is the value of the digit in the?
4. What is the highest common factor of and ?
5. Simplify

READING ASSIGNMENT

Essential Mathematics for J.S.S. 3 by Oluwasanmi A.J.S. 2014 edition; Pages

Essential Mathematics Workbook for J.S.S. 3 by Oluwasanmi A.J.S.; Exercise 7.1, numbers

WEEKEND ASSIGNMENT

1. If and when find the value ofwhen. A. B. C.
2. and when find the value of when . A. B. C.
3. If varies directly as and what is value of when ?

   A. B. C.

4. and when Find the relationship between m and n.

   A. B. C. D. m =

5. Find the value of m when. A. 6B. C.

THEORY

1. and when find (a) when (b) the percentage change in if increases by .
2. When repaying a loan, the number of monthly payments, , varies inversely with the amount of each payment, . The loan can be repaid by 10 monthly payment of Find the formula which connects. Hence find how long it takes to repay the loan with monthly payments of

JOINT AND PARTIAL VARIATION

JOINT VARIATION

Joint variation is obtained when a quantity varies with more than one other quantity either directly and/or inversely. For instance, is jointly proportional to both and as in. Also, is directly proportional to and inversely proportional to as in.

Example 1:

If .When

1. Find the relation between
2. Find when

Solution:

1.  and  

 After substituting, we have

The relation between them is given by

Example 2:

The universal gas law states that the volume of a given mass of an ideal gas varies directly with its absolute temperature and inversely with its pressure A certain mass of gas at an absolute temperature and pressure has a volume. Find:

1. the formula that connects .
2. the pressure of the gas when its absolute temperature is and its volume is 0.018m³
   

Solution:

1.  and   , such that

Substituting the values, becomes

and the relationship is

EVALUATION

1. Suppose . When Find
2. Find the percentage change in when increases by and decreases by

PARTIAL VARIATION

Partial variation problems occur everywhere around us. Some examples are described below:

• When a hairdresser makes hair, the money he/she charges M, is dependent on both the cost of the wool (thread or weavon in some cases) C which is constant, and on the time T, taken to make the hair. The less the weaves, the less the time it will take to complete and the less the charges. We can write a partial equation for this as: , where are constants.
• Domestic electricity prepaid meter bills are prepared on two components which are rental charge (independent of the amount of power consumed) and consumption charges (dependent on the quantity of power consumed). We can also write the total bill T in partial equation as: , where are constants depending on the customer.

Thus, partial variation statements can come in these formats described below:

• is partly constant and partly varies as is interpreted as
• varies partly as and partly inversely as can also be interpreted as

In these cases, and are constants that can be obtained simultaneously.

Example 3:

is partly constant and partly varies as the square of Write an equation connecting and Given that when and when Write down the law of variation. Find when

Solution:

The equation connecting and is , where and are constants.

When, becomes

When , we have   becomes

Combining the two equations and solving simultaneously,

Subtracting: and  

Substitute for into , so that

And .  The law of variation becomes

When becomes

Example 4:

varies as partly as and partly as the cube of . When and when . Write the law connecting and . Find when

Solution:

The equation connecting and is , where and are constants.

when , becomes

when , becomes

Combining the two equations and solving simultaneously to eliminate

Subtracting:  and  

Alternatively, dividing through by , gives and dividing through by , gives .

Then,

Subtracting:

And as obtained above.

Substitute for into , so that and

So that . The law of variation becomes

When becomes

Example 5:

The cost in naira of making a dress is partly constant and partly varies with the amount of time in hours it takes to make the dress. If the dress takes 3 hours to make, it costs N2700, and if it takes 5 hours to make the dress, it costs N3100. Find the cost if it takes hours to make the dress.

Solution:

Using and to represent the cost and time respectively, we can proceed by writing

From first statement:

From second statement:  

Solving the simultaneously,

Subtracting: and  

Substitute for into , so that

and.

So that . The law of variation becomes

If it takes to make the dress, the cost becomes

EVALUATION

1. varies partly directly with and partly varies inversely with y. When and when. Find when
2. An examination fee is partly constant and partly varies with the number of subjects entered. When the examination fee is three subjects are entered. When the fee is , five subjects are entered. Find the number of subjects entered if the fee is

    

   GENERAL EVALUATION
   

3. Express in tonnes.
4. Express in fraction in its lowest term.
5. What is the sum of and kobo expressed in kobo?
6. Factorize
7. A trader gives discount on an article in his kiosk marked . How much would a customer pay on such article?
   

READING ASSIGNMENT

Essential Mathematics for J.S.S. 3 by Oluwasanmi A.J.S. 2014 edition; Pages

Essential Mathematics Workbook for J.S.S. 3 by Oluwasanmi A.J.S.; Exercise, numbers

WEEKEND ASSIGNMENT

1. and Find when . A. B. C. D. 5
2. In the question above. A. increases by B. decreases by C. increases by D. X decreases by 25%
3. is partly constant and partly varies with y. This is statement can be represented as

   A. B. C. D. P = a + y

4. If and then A. B. C.
5. How many constants do we have in partial variation? A. B. C.

THEORY

1. If and when calculate

   (a)when (b) when

2. The charge for a pair of shoe is partly constant and partly varies as the number of pair of shoes. If pairs cost and the cost of pairs is Find the charges for pairs.

    

   GENERAL EVALUATION
   

1. Make L the subject of the formula
2. Change 84 in base ten to a number in base two.
3. Represent on the number line.
   
4. Convert in a decimal number to an octal number.
   
5. A car travelled in minutes. What is the speed of the car in ?
   

    

   READING ASSIGNMENT
   

Essential Mathematics for J.S.S. 3 by Oluwasanmi A.J.S. 2014 edition; Pages

Essential Mathematics Workbook for J.S.S. 3 by Oluwasanmi A.J.S.; Exer., numbersExercise, numbers.

WEEKEND ASSIGNMENT

1. The simple interest on for years at is _________.

   A. B. C. D. #3,000

2. The amount on the above question is A. B. C.
3. can also be written as A. B. C. D. 0.405
4. The square root of is A. B. C. D. 2695
5. Which of the following is a rational number?

    

   THEORY

6. What is the compound interest on borrowed for years at rate?
7. Find the square root of , leaving your answer in one decimal place.

WEEK SIX

FACTORISATION

CONTENT

• Factorization of simple expression
• Difference of two squares
• Factorization of quadratic expression

FACTORISATION OF SIMPLE EXPRESSION

To factorize an expression completely, take the HCF outside the bracket and then divide each term with the HCF.

Example:

Factorize the following completely.

1. 8xy + 4x²y

2. 6ab – 8a²b + 12ab

Solution:

1. 8xy + 4x²y

8xy = 2 X 2 X 2 X x X y

4x2y = 2 X 2 X xXxX y

HCF = 4xy

8xy + 4x2y = 4xy ( + )

= 4xy ( 2 + x)

2. 9a²bc³ – 12ab²c²

9a²bc³ = 3 X 3 X a X a X c X c X c

12ab²c² = 2 X 2 X 3 X b X b X c X c

HCF = 3abc²

= 3abc² (3ac – 4b)

EVALUATION

factorize the following expression

1. 9x²yz² – 12x³z³

2. 14cd + 35cd²f

3. 20m²n – 15mn²

FACTORISATION BY GROUPING

To factorize an expression containing four terms, you need to group the terms into pairs. Then factorize each pair of terns.

Example:

factorize ab – 2cb + 2cf – af

Solution:

Group ab and af together and 2cb and 2cf together

i.e. ab – 2cb + 2cf –af = ab – af – 2cb + 2cf

= a( b – f ) -2c( b – f )

= (a – 2c)( b – f)

EVALUATION

factorize these expressions;

• 16uv – 12vt + 20mu – 15mt
• ap +aq +bq + bp
• mn – pq-pn +mq

FACTORISATION OF QUADRATIC EXPRESSIONS

A quadratic expression has two (2) as its highest power hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax ² + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

NOTE

• if ax ² +bx + c= 0, this is known as quadratic equation
• a is coefficient of x², b is coefficient of x and c is a constant term.
• When an expression contains three terms, it is known as trinomial.
• To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: factorization of trinomial of the form x² +bx + c.

1. Factorise x² +7x +6

Steps:

1. Multiply the 1^st and the last term (3rd term) of the expression.
2. Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
3. Replace the middle term with these two numbers and factorize by grouping.

Solution to example:

X² x 6 = 6x²

Factors: 6 and 1

X² + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

EVALUATION

• z² – 2z + 1
• x² +10x – 24

FACTORISATION OF QUADRATIC EQUATIONS OF THE FORM ax ² +bx +c

Example: 5x ² -9x +4

Solution:

Product: 5x ² x 4 = 20x ²

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x ² – 9x + 4

5x² -5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

EVALUATION

1. 2x ² +13x +6
2. 13d ² – 11d – 2

FACTORISATION OF TWO SQUARES

To factorise two squares with difference, we need to remember the law guiding difference of two squares i.e. x ² – y² = (x + y) (x- y).

Examples:

1. P ² – Q² = (P+Q) (P-Q)
2. 36y ² – 1= 6 ² y ² – 1 ²

   = (6y)² – 1 ² = ( 6y+1) (6y-1).

EVALUATION

1. 121- y ²
2. x²y² – 4²

READING ASSIGNMENT

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105-

WEEKEND ASSIGNMENT

1. The coefficient of x ² in x ² + 3x -5 is A. 3 B. 1 C. -5 D. 2
2. Simplify e ² – f ² A. (e+f)(e-f) B. (e+f)(f+e) C. (e-f)(f-e) D. e+f
3. Factorize x ² +x -6 A. (x+3)(x+2) B. (x-2)(x+3) C. (x+1)(x+5) D. x + 2
4. Solve by grouping 5h ² -20h + h – 4 A. (h-4)(5h+1) B. (h+4)(5h-1) C. (h+2)(h-5) D. h – 4
5. 49m ² – 64n ² when factorized will be A. (7m+8n)(8m+7n) B. (8m-7n)(8m+7n)

   C. (7m-8n)(7m+8n) D. 7m – 8n

THEORY

Factorise the following

1. 4p² – 12p +9q²
2. f ² – 2f + 1

FACTORISATION OF QUADRATIC EXPRESSIONS

A quadratic expression has two (2) as its highest power hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax ² + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

NOTE

1. if ax ² +bx + c= 0, this is known as quadratic equation
2. a is coefficient of x², b is coefficient of x and c is a constant term.
3. When an expression contains three terms, it is known as trinomial.
4. To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: factorization of trinomial of the form x² +bx + c.

1. Factorise x² +7x +6

Steps:

1. Multiply the 1^st and the last term (3rd term) of the expression.
2. Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
3. Replace the middle term with these two numbers and factorize by grouping.

    

Solution to example:  

X² x 6 = 6x²

Factors: 6 and 1

X² + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

EVALUATION

1. z² – 2z + 1
2. x² +10x – 24

Factorization of quadratic equations of the form ax ² +bx +c

Example: 5x ² -9x +4

Solution:

Product: 5x ² x 4 = 20x ²

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x ² – 9x + 4

5x² – 5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

EVALUATION

1. 2x ² +13x +6
2. 13d ² – 11d – 2

FACTORISATION OF TWO SQUARES

To factorize two squares with difference, we need to remember the law guiding difference of two squares i.e. x ² – y² = (x + y) (x- y).

Examples:

1. P ² – Q² = (P+Q) (P-Q)
2. 36y² – 1= 6 ² y ² – 1 ²= (6y) ² – 1 ² = ( 6y+1) (6y-1).

EVALUATION

1. 121- y ²
2. x²y² – 4²

READING ASSIGNMENT

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105-

WEEKEND ASSIGNMENT

1. The coefficient of x ² in x ² + 3x -5 is (a) 3 (b) 1 (c) -5
2. Simplify e ² – f ² (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
3. Factorize x ² +x -6 (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
4. Solve by grouping 5h ² -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
5. 49m ² – 64n ² when factorised will be (a) (7m+8n)(8m+7n) (b) (8m-7n)(8m+7n)

   (c) (7m-8n)(7m+8n)

THEORY

Factorise the following

1. 4p² – 12p +9q²
   
2. f ² – 2f + 1

FACTORISATION OF QUADRATIC EXPRESSIONS

A quadratic expression has two (2) as its highest power hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax ² + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

NB:

1. if ax ² +bx + c= 0, this is known as quadratic equation
2. a is coefficient of x², b is coefficient of x and c is a constant term.
3. When an expression contains three terms, it is known as trinomial.
4. To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: factorization of trinomial of the form x² +bx + c.

1. Factorise x² +7x +6

Steps:

1. Multiply the 1^st and the last term (3rd term) of the expression.
2. Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
3. Replace the middle term with these two numbers and factorize by grouping.

Solution to example:

X² x 6 = 6x²

Factors: 6 and 1

X² + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

Evaluation: 1. z² – 2z + 1

2. x² +10x – 24

Factorization of quadratic equations of the form ax ² +bx +c

Example: 5x ² -9x +4

Solution:

Product: 5x ² x 4 = 20x ²

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x ² – 9x + 4

5x² -5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

Evaluation:

1. 2x ² +13x +6
2. 13d ² – 11d – 2

FACTORISATION OF TWO SQUARES

To factorize two squares with difference, we need to remember the law guiding difference of two squares i.e. x ² – y² = (x + y) (x- y).

Examples:

1. P ² – Q² = (P+Q) (P-Q)
2. 36y ² – 1= 6 ² y ² – 1 ²

   = (6y)² – 1 ² = ( 6y+1) (6y-1).

Evaluation:

1. 121- y ²
2. x²y² – 4²

READING ASSIGNMENT

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105

WEEKEND ASSIGNMENT

1. The coefficient of x ² in x ² + 3x -5 is (a) 3 (b) 1 (c) -5
2. Simplify e ² – f ² (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
3. Factorize x ² +x -6 (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
4. Solve by grouping 5h ² -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
5. 49m ² – 64n ² when factorized will be (a) (7m+8n)(8m+7n) (b) (8m-7n)(8m+7n)

   (c) (7m-8n)(7m+8n)

THEORY

Factorise the following

1. 4p² – 12p +9q²
   
2. f ² – 2f + 1

WEEK SEVEN

SOLVING EQUATION EXPRESSIONS

WORD PROBLEMS

Worked Examples:

1. Find 1/4 of the positive difference between 29 & 11
2. The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number
3. The sum of 35 and a certain number is divided by 4 the result is equal to double the number. Find the number.

Solutions:

1. Positive Difference 29 – 11 = 18

 1/4 of 18 = 4 ²/5

1. Let the number be x

   x X 5 = 20 – 2x

 5x = 20 – 2x

 5x + 2x = 20

 7x = 20

 x = 20/7 = 2

1. Let the number be n

 sum of 35 and n = n + 35

 divided by 4 = n + 35

4

 result = 2 X n

 therefore n + 35 = 2n

4

 n + 35 = 8n

 8n – n = 35  

 7n = 35

 n = 35/7 = 5

EVALUATION

1. From 50 subtract the sum of 3 & 5 then divide the result by 6
2. The sum of 8 and a certain number is equal to the product of the number and 3 find the number

SOLVING EQUATION EXPRESSIONS WITH FRACTION

Always clear fractions before beginning to solve an equations: –

To clear fractions, multiply each term in the equation by the LCM of the denominations of the fractions.

Examples:

Solve the following

1.  x = 2

 9

2.  x + 9 + 2 + x = 0

  5   2

3.  2x = 5x + 1 + 3x – 5

7   2

Solutions:

1.  x = 2

 9

 Cross multiply

x = 18

2.  x + 9 + 2 + x = 0

  5   2

 Multiply by the LCM (10)

 10 X (x + 9) + 10 X ( 2 + x) = 0 X 10

5 2

 2 (x + 9) + 5 (2 + x) = 0

 2x + 18 + 10 + 5x = 0

 2x + 5x + 28 = 0

 7x = -28

 x = -28/7 = -4

3.  2x = 5x + 1 + 3x – 5

7   2

 Multiply by the LCM (14)

 14 X 2x = 14 (5x + 1) + 14 ( 3x – 5)

7 2

 28x = 2 (5x + 1) + 7 (3x – 5)

28x = 10x + 2 + 21x – 35

 28x = 31x – 33

 28x – 31x = -33

 -3x = -33

 x = 33/3 = 11

EVALUATION

Solve the following equations.

1.  7/3c = 21/2

2.   6   = 11

 y + 3   y – 2

3.  3 – 4   = 0

 2b – 5   b – 3

Furthermore, we can consider the word equations or expressions into:

• Sum & Differences
• Products
• Expressions with fractions & equations

SUM & DIFFERENCES

The sum of a set of numbers is a result obtained when the numbers are added together. The difference between two numbers is a result of subtracting one number from the other.

Worked Examples:

1.  Find the sum of -2 & -3.4

2.  Find the positive difference between 19 & 8

3.  The difference between two numbers is 7. If the smaller number is 7 find the other.

4.  The difference between -3 and a number is 8, find the two possible values for the number.

5.  Find the three consecutive numbers whose sum is 63.

Solutions:

1.  -2 + -3.4 = -5.4

2.  19 – 8 = +11

3.  let the number be Y i.e. Y -7 = 7

 i.e. Y = 7 + 7 = 14

4.  Let M represent the number

 M – (-3) = 8

 m + 3 = 8

 m = 8 – 3

 m = +5

 also -3 – m = 8

 -m = 8 + 3

 -m = 11

 m = -11

 the possible values are +5 & -11

5.  Consecutive numbers are 1, 2, 3,4,5,6, Consecutive odd numbers are

1, 3,5,7,9……….. consecutive even numbers are 2, 4, 6, 8,10……….

 Representing in terms of X, we have 2X, 2X + 2, 2X + 4, 2X + 6, 2X + 8, 2X + 10…………

 for consecutive even numbers, we have X, X + 2, X + 4, X + 6…….

 for consecutive odd numbers, we have X + 1, X + 2, X + 3, X + 4…

for consecutive numbers.

 let the first number be x,

 let the second number be x + 1

 let the third number be x + 2

 Therefore x + x + 1 + x + 2 = 63

 3x + 3 = 63

 3x = 63 – 3  

3x = 60

 x = 60 /3

 = 20

 The numbers are 20, 21, and 22.

EVALUATION

1.  Find the sum of all odd numbers between 10 and 20

2.  The sum of four consecutive odd numbers is 80 find the numbers
3.  The difference between 2 numbers is 9, the largest number is 32 find the numbers.

PRODUCTS

The products of two or more numbers is the result obtained when the numbers are multiplied together.

Worked Examples:

1.  Find the product of – 6, 0.7, &

2.  The product of two numbers is 8 .If one of the numbers is 1/4 find the other.

3. Find the product of the sum of -2 & 9 and the difference between -8 & -5.

Solutions

1.  Products -6 x 0.7 x

 -6 x 7/10 x 20/3 = -6 x 7 x 20

10 x 3

 = -2 x 7 x 2 = -28

2.  Let the number be x

 X x = 8multiply both sides by 4

 x = 8 x 4 = 33

3.  Sum = -2 + 9 = 7

 Difference = -5-(-8) = -5 + 8 = 3

 Products= 7 x 3 = 21

EVALUATION

1. The product of three numbers is 0.084 if two numbers are 0.7 & 0.2 find the third number
2. Find the product of the difference between 2 & 7 and the sum of 2 & 7
3. From 50 subtract the sum of 3 & 5 then divide the result by 6
4. The sum of 8 and a certain number is equal to the product of the number and 3 find the number

Reading Assignment

New Gen Maths for J.S.S 3 Pg 20- 24

Essential Mathematics for J.S.S 3 Pg 85-87

WEEKEND ASSIGNMENT

1. Esther is 3 times as old as her sister Tolu, if the sum of their ages is 20 years. Find the difference between their ages.

 (a) 20 years  (b) 8 years  (C)  10 years

2.  9 was subtracted from a certain number and the result was divided by 4 if the final

answer is 5 what was the original number?

 (a) 29  (b) 18  (c) 20

3.  A woman is 4 times as old as her son. In five years time she will be 3 times as old as her son. How old is the woman(a) 50 yrs  (b) 40 yrs  (c) 45 yrs

4.  Bayo is 4 times as old as his sister Tolu. If the sum of their ages is 20 years, find the difference between their ages.

 (a) 12 yrs  (b) 15 yrs  (c) 18 yrs

5.  Subtract the square root of 4 from the square of 4 and divide the result by 2

 (a) 2  (b) 4  (c) 7

THEORY

1.Divide 36 by the difference between the product of 3 & 6 and the square root of 36.

2.  When I add 45 to a certain number, and divide the sum by 2, the result is the same as five times the number, what is the number?

WEEK EIGHT

CHANGE OF SUBJECT OF FORMULA

A formula is a general equation involving two or more unknowns. An example is the formula a = ¶r² which gives the area of the circle in terms of its radius r. In this formula, a is called the subject of the formula.

Simplifying a formula by substitution

Example:

Given mx+c=y, express x in terms of m, c, and y. Find the value of x if y=10, c=2 and m is 4

Solution: mx+c= y, mx=y-c

X= y-c

M

To find the value of X, X= 10-2

4

X= 8/4 =2

Evaluation: If I= PRT

1. Make p the subject and find the value of p if I = 10, R= 4, and T= 5.

CHANGING THE SUBJECT OF A FORMULA

When a variable which forms a part of the formula is made subject, we say we have changed the subject of the formula.

Examples:

1. In the formula S=2¶r(r+h), make h the subject.
2. Make m the subject if F= mv-mu

   T

Solution: S=2¶r(r+h)

S=2¶r ² +2¶rh

S- 2¶r ² =2¶rh

S-2¶r² =h

2¶r

2. F =mv-mu

T

Cross multiply: FT =mv-mu

FT =m(v-u)

FT

v-u = m

EVALUATION

1. Make r the subject if V =¶r ²h
2. Make u the subject if 1/f =1/v + 1/u

READING ASSIGNMENT

1. Exam focus for J.S.C.E Pg. 212 -216
2. Essential Maths for J.S.S.3 Pg. 47-54

WEEKEND ASSIGNMENT

1. Express a in terms of u, v, and t in v= u+at
   1. a = vu-t (b) a= v-u (c) a = v+u

   tt

2. If Z=2p +3, find the value of Z when p=1. (a) Z=2, (b) Z= 5, (c) Z =7

3. In the relation X = m – 6y , how would you write m in terms of x and y?

6

(a) m=x-6 (b) m = 6x -6y (c) m =6x +6y

4. Express n in terms of s, a, and L if S=n/2(a+L)

(a) n=2(a+L) (b) n= 2S (c) 2S-(a-L)

Sa+L a+L

5. Make U the subject if V ² = U² +2as.

(a) U= V ² -2as (b) U = (V ² -2as) ²  (c) V² – 2a

THEORY

1. Given L = ar^n-1, make a the subject of the formula.

2. If S + 2t , make t the subject of the formula. Find the value of t if S=1, d= 2.

d

WEEK NINE AND TEN

TOPIC: MEASURE OF CENTRAL TENDENCY

CALCULATION OF RANGE, MEAN, MEDIAN AND MODE OF UNGROUPED DATA

RANGE

The range of a set of numbers is the difference between the largest and the smallest numbers.

Example: Find the range of the following set of scores: 79, 60, 52, 34, 58, 60.

Solution

Arrange the set in rank order: 79, 60, 60, 58, 52, 34

The range is 79 – 34 = 45

THE MEAN

There are many kinds of average. The mean or arithmetic mean, is the most common kind. If there are n numbers in a set, then

Mean = sum of the numbers in the set/ n

Examples

1) Calculate the mean of the following set of numbers.

176 174 178 181 174

175 179 180 177 182

Solution

Mean = 176 + 174 + 178 +…. + 182/10

= 1776/10

= 177.6

2) Five children have an average age of 7 years 11 months. If the youngest child is not included, the average increases to 8 years 4 months. Find the age of the youngest child.

Solution

Total age of all five children

= 5 x 7 yr 11 mo

= 35 yr 55 mo

= 35 yr + 4 yr 7 mo

= 39 yr 7 mo

Total age of the four older children

= 4 x 8 yr 4 mo

= 32 yr 16 mo

= 32yr + 1 yr 4 mo

= 33 yr 4 mo

Age of youngest child

= 39yr 7 mo – 33 yr 4 mo

= 6 yr 3 mo

EVALUATION

1) Find x if the mean of the numbers 13, 2x, 0, 5x and 11 is 9. Also find the range of the set of numbers.

2) A mother has seven children. The mean age of the children is 13 years 2 months. If the mother’s age is included, the mean age rises to 17 years 7 months. Calculate the age of mother.

MEDIAN AND MODE

MEDIAN: If a set of numbers is arranged in order of size, the middle term is called the median. If there is an even number of terms, the median is the arithmetic mean of the two middle terms.

Examples

Find the median of a) 15, 11, 8, 21, 17, 3, 8 b) 3.8, 2.1, 4.4, 8.3, 9.2, 5.0.

Solution

a) Arrange the numbers in rank order (i.e. from highest to lowest).

21, 17, 15, 11, 8, 8, 3

There are seven numbers. The median is the 4^th number, 11.

b) Arrange the numbers from the lowest to highest.

2.1, 3.8, 4.4, 5.0, 8.3, 9.2

There are six numbers. The median is the mean of the 3^rd and 4^th terms.

Median = (4.4 + 5.0) /2

= 4.7

MODE: The mode of a set of numbers is the number which appears most often, i.e. the number with the greatest frequency.

Example: Twenty-one students did an experiment to find the melting point of naphthalene. The table below shows their results. What was a) the modal temperature b) the median temperature?

temperature (^oC) 78 79 80 81 82 83 90

frequency 1 2 7 5 3 2 1

a) Seven students recorded a temperature of 80^oC. This was the most frequent result.

Mode = 80^oC

b) There were 21 students. The median is the 11^th temperature. If the temperatures were written down in order, there would be one of 78^oC, two of 79^oC, seven of 80^oC, and so on. Since 1 +2 + 7 = 10, the 11^th temperature is one of the five 81^oCs.

Median = 81^o C.

Evaluation

For the following set of numbers:

13, 14, 14, 15, 18, 18, 19, 19, 19, 21

a) state the median, b) state the mode, c) calcilate the mean.

WEEKEND ASSIGNMENT

1) The number of goals scored by a team in nine handball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 15

Which of the following statements are true of these scores?

a) The mean is greater than the mode.

b) The mode and the median are equal.

c) The mean, median, and mode are all equal.

Use the table below to question 2-5

The table below shows the number of pupils (f) scoring a given mark (x) in attest.

X 2 3 4 5 6 7 8 9 10 11 12

f 3 8 7 10 13 16 15 15 6 2 5

2) Find the mode.

a) 7 b) 8 c) 9 d) 10

3) Find the median.

a) 6 b) 7 c) 8 d) 9

4) Calculate the mean.

a) 6.7 b) 6.8 c) 6.9 d) 6.95

5) Find the range.

a) 10 b) 11 c) 9 d) 12

THEORY

1)x, x, x, y represent four numbers. The mean of the numbers is 9, their median is 11. Find y

2) Students at a teacher training college are grouped by age as given in table below.

Age (years) 20 21 22 23 24 25

Frequency 4 5 10 16 12 3

a) Find the modal age.

b) Find the median age.

c) Calculate the mean age of the students.

READING ASSIGNMENT

ESSENTIAL MATHS BK 3 PG 201 – 205 Ex 22.5nos 17 – 20