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2022 Mathematics Lesson Note for First Term JSS 3

 

Mathematics Lesson Notes JSS3 First Term

SCHEME OF WORK

 

WEEK 1&2 WHOLE NUMBERS

 

WEEK 3 WHOLE NUMBERS CONT’D

 

WEEK 4 ADDITION AND SUBTRACTION IN BASE 2

 

WEEK 5 MULTIPLICATION AND DIVISION IN BASE 2

 

WEEK 6 RATIONAL AND NON-RATIONAL NUMBERS

 

WEEK 7 FACTORIZATION

 

WEEK 8 FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT

 

WEEK 9 SIMPLE EQUATIONS INVOLVING FRACTIONS

WEEK 10 CONVERTING NUMBERS TO BASES

 

Week 1&2

 Topic: Whole Numbers

Contents

  • Binary Number System
  • Using Computer
  • Word Problems

A. Number bases

Most people count in tens. For instance, the place value of the digits in number 4956 in this system

Thousands     hundreds    tens      units

↓                       ↓                   ↓             ↓

4                      9                   5             6

The place values are in powers of ten. It is called the base ten system.

4956 = 4 X 1000 + 9 X 100 + 5 X 10 + X 1

4 X103 + 9 X 102 + X 101 + X 1

To learn more, click here 

Week 3

Topic: Whole Numbers Cont’d

Contents

  • Expression Involving Fractions
  • Direct and Inverse Proportion
  • Compound Interest

By an equation with fractions, I’ll mean an equation to solve in which the variable appears in the denominator of one or more fractions. As you’ve seen with equations involving number fractions, the natural approach is to multiply to clear denominators. To do this, you should:

Factor any denominator that can be factored.

Multiply both sides of the equation by the least common multiple of the denominators to clear the fractions.

Fractions with unknowns in the denominator

Example

2 ¾ + 33/2x = 0

2 ¾ + 33/2x = 0

Express 2 ¾ as an imporper fraction.

11/4 + 33/2x = 0

The denominators are 4 and 2x. Their LCM is 4x. Multiply each term in the equation by 4x.

4x(11/4) + 4x(33/2x) = 4x X 0

11x + 66 = 0

11x = -66

X = -6

Check: when x = -6,

LHS = 2 ¾ + 33/-12 = 2 ¾ – 11/4 = 0 = RHS

Example

Solve 1/3a + ½ = 1/2a

1/3a + ½ = 1/2a

The denominators are 3a, 2 and 2a. Their LCM is 6a. Multiply each term in the equation by 6a. To learn more, click here

Week 4

Topic: Addition and Subtraction in Base 2

Let’s first take a look at decimal addition.

As an example we have 26 plus 36,
26
+36

To add these two numbers, we first consider the “ones” column and calculate 6 plus 6, which results in 12. Since 12 is greater than 9 (remembering that base 10 operates with digits 0-9), we “carry” the 1 from the “ones” column to the “tens column” and leave the 2 in the “ones” column.

Considering the “tens” column, we calculate 1 + (2 + 3), which results in 6. Since 6 is less than 9, there is nothing to “carry” and we leave 6 in the “tens” column.

    26
+36
62

Binary addition

Works in the same way, except that only 0’s and 1’s can be used, instead of the whole spectrum of 0-9. This actually makes binary addition much simpler than decimal addition, as we only need to remember the following:

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10

To learn more, click here

Week 5

Topic: Multiplication and Division in Base 2

Binary multiplication

Is actually much simpler than decimal multiplication. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. In binary multiplication, we only need to remember the following,

0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1

Note that since binary operates in base 2, the multiplication rules we need to remember are those that involve 0 and 1 only. As an example of binary multiplication we have 101 times 11,

  101
x11

First we multiply 101 by 1, which produces 101. Then we put a 0 as a placeholder as we would in decimal multiplication, and multiply 101 by 1, which produces 101.

  101
x11
101
1010  <– the 0 here is the placeholder

To learn more, click here

Week 6

Topic: Rational and Non-rational Numbers

Rational Numbers

We can write numbers such as 8, 4½, 1/5, 0.211, √49/16, 0.3 as exact fractions or ratios:

8/1, 9/2, 1/5, 211/1 000, 7/4, 1/3.

Such numbers are called rational numbers.

Numbers which cannot be written as exact fractions are called non-fractional numbers, or irrational numbers. √7 is an example of a non-rational number. √7 = 2.645 751 …., the decimals extending without end and without recurring.

π is another example of a non-rational number. Π = 3.141 592 …., again extending forever without repetition. The fraction 22/7 is often used for the value of π. However, 22/7 is a rational number and is only an approximate value of π.

All recurring decimals are rational numbers. Read the following example carefully.

Example

Write 3.17 as a rational number

Let n = 3/17

i.e. n =3.17 17 17 ………… (1)

Subtract (1) from (2),

99n = (317.17 17 . . .) – (3.17 17 . . .)

99n = 314

Thus, 3.17 = 314/99, a rational number.

A non-rational number extends forever and is non-recurring.

To learn more, click here

Week 7

Topic: Factorization

Expression of Algebraic Expressions

The expression (x + 2)(x – 5) means (x + 2) X (x – 5).  The product of the two binomials (x + 2) and (x – 5) is found by multiplying each term in the first binomial by each term in the second binomial. Read the following examples carefully.

Example 1

Find the product of (x + 2) and (x – 5).

= (x + 2)(x – 5) = x(x + 2) – 5(x + 2)

= x2 + 2x – 5x – 10

= x2  – 3x – 10

Example 2 

Expand (2c – 3m)(c – 4m)

(2c – 3m)(c – 4m)

= c(2c – 3m) – 4m(2c -3m)

= 2c2 – 3cm – 8cm + 12m2

=  2c– 11cm + 12m2

To learn more, click here

Week 8

Topic: Simple Equations Involving Fractions

Do you go blank when you see x, y and z in mathematics? Well, this is your abc to solving equations.

Solving simple equations

In an equation, letters stand for a missing number. To solve an equation, find the values of missing numbers. A typical exam question is:

Solve the equation 2a + 3 = 7

This means we need to find the value of a. The answer is a = 2

There are two methods you can use when solving this type of problem:

Trial and improvement

Using inverses

Trial and improvement

This method involves trying different values until we find one that works.

Look at the equation 2a + 3 = 7

To solve it:

Write down the equation: 2a +3 = 7

Then, choose a value for ‘a’ that looks about right and work out the equation. Try ‘3’.

a = 3, so 2 ×3 + 3 = 9.

Using ‘3’ to represent ‘a’ makes the calculation more than 7, so choose a smaller number for ‘a’.

Try a = 2

× 2 + 3 = 7

Which gives the right answer. So a = 2

To learn more, click here

Week 9

Topic: FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT

Formulae and Substitution

formula is an equation with letters which stands for quantities. For example

C = 2πr

Is the formula which gives the circumference, c, of a circle of radius r.

In science,

I = V/R

Is the formula which shows the relationship between the current I amps, voltage, V volts, and resistance, R ohms, in an electrical circuit. In arithmetic,

I = PRT/100

Is the formula which gives interest, I, gained on a principal, P, invested at R% per annum for T years. Sometimes the same letter can stand for different quantities in different formulae. For example, I stands for current in the science formula and I stands for interest in the arithmetic formula. Formulae  is the plural of formula. To learn more, click here

Week 10

Topic: PROBLEM-SOLVING ON NUMBER BASES EXPANSION, CONVERSION AND RELATIONSHIP

Converting from base b to base 10

The next natural question is: how do we convert a number from another base into base 10? For example, what does 42015 mean? Just like base 10, the first digit to the left of the decimal place tells us how many 50’s we have, the second tells us how many 51’s we have, and so forth. Therefore:

4201= (4.5+ 2.5+ 0.5+ 1.50)10

= 4.125 + 2.25 + 1

=55110

From here, we can generalize. Let x =(anan-1 … a1a0)be an n + 1 -digit number in base b. In our example (274610) a= 2, a= 7, a= 4 and a= 6. We convert this to base 10 as follows:

x = (anan-1 … a1a0)b

= (bn.a+ bn-1 . an-1 + … + b.a+ a0)10

To learn more, click here

 

 

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