Free Mathematics Lesson Note JSS 3

Free Mathematics Lesson Note JSS 3

This MATHEMATICS Lesson Note was pulled from our book (Lesson Note on MATHEMATICS for JSS3 MS-WORD); Compiled to serve as a reference material to help teachers draw out their lesson plan easier, saving you valuable time to focus on the core job of teaching.

The Lesson notes are based on the current NERDC curriculum (UBE compliant)

This MATHEMATICS Lesson Notes CoversThe Following Topics

1. WHOLE NUMBERS
2. WHOLE NUMBERS CONT’D
3. ADDITION AND SUBTRACTION IN BASE 2
4. MULTIPLICATION AND DIVISION IN BASE 2
5. RATIONAL AND NON-RATIONAL NUMBERS
6. FACTORIZATION
7. FORMULAE: SUBSTITUTION AND CHANGE OF SUBJECT
8. SIMPLE EQUATIONS INVOLVING FRACTIONS
9. SOLVING ON NUMBER BASES EXPANSION, CONVERSION AND RELATIONSHIP
10. FACTORISATION OF QUADRATIC EXPRESSIONS
11. TANGENT OF AN ANGLE
12. FACTORIZATION: COMMON FACTORS
13. SIMILARITY AND ENLARGEMEN
14. SIMULTANEOUS LINEAR EQUATION
15. TRIGONOMETRY – SINE AND COSINE OF ANGLES
16. AREA OF PLANE SHAPES
17. AREAS AND VOLUMES OF SIMILAR SHAPES
18. GEOMETRICAL CONSTRUCTION
19. TANGENT OF AN ANGLE
20. MEASURE OF CENTRAL TENDENCY
21. VARIATION
22. STATISTICS

 

Sample note

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 improper 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.

6a X (1/3a) + 6a X ½ = 6a X (1/2a)

2 + 3a =3

3a = 1

a= 1/3

Check: when a = 1/3

LHS = 1/3 X 1/3 + ½  = 1 + ½ = 1 ½

RHS = 1/3 X 1/3 = 3/2 = 1 ½ = LHS

The two examples above show that when unknowns, such as x or a, appear in the denominator, they are treated like numbers.

Clear fractions by multiplying each term of the equation by the LCM of the denominators of the fractions. Then solve the equations in the usual way.

Exercise

1. 1/x = 1/5
2. 1/9 = 1/r
3. 1/m – ¼ = 0

 

Factors with binomials in the denominator

Note: You should always check your answers to equations with fractions, since it’s possible to produce bogus solutions.

Example

Solve 2/x – 1 + 3 = 4x /x – 1.

When you have an equation with fractions, it’s often good to start by clearing denominators:

(x – 1)( 2/x – 1 + 3) = (x – 1) . 4x /x – 1.

I can cancel  x – 1 ‘s provided that x ≠ 1. Assuming that this is true,

2 + 3(x – 1) = 4x

2 + 3x – 3 = 4

3x – 1 = 4

Then

3x – 1 = 4x

–   3x          3x

———————

– 1 =   x

Check: If  x = -1,

2/x – 1 + 3 = 2/ -2 + 3 = -1 + 3 = 2,

4x/x – 1= -4/-2 = 2

 

Example

Solve 4 + 5/x – 1= 5x/x – 1.

Clear denominators:

(x – 1)(4 + 5/x – 1) = (x – 1).5x/x – 1

4(x – 1) + 5 = 5x

4x – 4 + 5 = 5x

4x + 1 = 5x

Then

4x + 1 = 5x

–  4x          4x

——————-

1 =  x

Check: If x = 1, 4 + 5/x – 1 is undefined.

Therefore, there are no solutions.

 

Word problems involving fractions

In this lesson, we will learn how to solve fraction word problems that deal with fractions i.e. parts of a whole. Remember to read the question carefully to determine the numerator and denominator of the fraction.

We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.

 

Example

A class has 20 girls and 30 boys. What part of the class are boys?

 

 

Solution:

Step 1: Numerator: boys = 30
Step 2: Denominator: class = 20 + 30 = 50
Step 3: Part or fraction

30/50 = 3/5

Answer: 3/5 of the class are boys

Example

If John earns $x in a week and spend $y, what part of his weekly salary did he save?

 

Solution

Step 1: Numerator: amount saved= x – y
Step 2: Denominator: salary = x
Step 3: Part or fraction

x – y/x

 

1. Direct and InverseProportion

Direct Proportion

If two quantities are in direct proportion, as one increases, the other increases by the same percentage.

If y is directly proportional to x, this can be written as y ∝x

A simple example of two things that are in the same proportion is the amount of apples you might buy and the amount you pay for them. If you buy twice as many apples as your friend, you pay twice as much.

We can write the connection between the cost and the amount as an equation:

Cost of apples = price per apple × number of apples bought.

This can also be written as y = kx, where k is the cost (the price per apple).

This means that, for some constant k, y = kx for all values of x and k is called the constant of proportionality.

Example

If y is directly proportional to x.

When x = 12 then y = 3

Find the constant of proportionality and the value of x when y = 8.

We know that y is proportional to x so y = kx

We also know that when x = 12 then y = 3

To find the value of k substitute the values y = 3 and x = 12 into y = kx

3 = k × 12

So k = ³/₁₂ = ¹/₄

To find the value of x , when y = 8 substitute y = 8 and k = ¹/₄ into y = kx

8 = (¹/₄) x

So x = 32 when y = 8

 

Direct Proportion to Powers

y can be directly proportional to x² , x³ and other powers of x.

They can always form an equation with k, a constant multiplier (the constant of proportionality), at the start.

eg y = kx²

 

Example

y ∝ x ³

If y = 1 when x = 2, find the value of y when x = 4

 

Solution

y ∝ x ³

So y = kx³

Substitute the value y = 1 and x = 2 into y = kx³ to find the value of k.

1 = k × 2³

So k = ¹/₈

Now use the values k = ¹/₈ and x = 4. y = ¹/₈ × 64

Gives the answer y = 8

 

Inverse Proportion

Inverse proportion is when one value increases at the other value decreases.

A simple example of inversely proportional quantities is the lengths and widths of rectangles with the same area. As the length of one side doubles, the width has to be halved for the area to stay the same.

Example

y is inversely proportional to x. When y = 3, x = 12 .

Find the constant of proportionality, and the value of x when y = 8.

y ∝ ¹/_x

y = ^k/_x

So xy = k

Substitute the values x = 12 and y = 3 into xy = k

3 × 12 = 36

So k = 36

To find the value of x when y = 8, substitute k = 36 and y = 8 into xy = k

8x = 36

So x = 4.5

Again, you can have questions involving squares, cubes or other powers of the variables.

 

Exercise

v is inversely proportional to r³. When r = 2, v = 25. Find r when v = 60.

Answer

v = ∝  so v =

Re-arrange the above to get k on its own.

k = vr³

k = 25 × 2³

So k = 200

When v = 60

60r³ = 200

r³ = ²⁰⁰/₆₀

r³ = 3.333

So r equals the cube root of 3.333

So r = 1.494

 

Graphical Representation

When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation.

In simpler terms, that means if A is always twice as much as B, then they directly vary. If a gallon of milk costs $2, and I buy 1 gallon, the total cost is $2. If I buy 10 gallons, the price is $20. In this example the total cost of milk and the number of gallons purchased are subject to direct variation — the ratio of the cost to the number of gallons is always 2.

To be more “mathematical” about it, if y varies directly as x, then the graph of all points that describe this relationship is a line going through the origin (0, 0) whose slope is called the constant of the variation. That’s because each of the variables is a constant multiple of the other, like in the graph shown below:

 

Inverse Variation
(The Opposite of Direct Variation)

In an inverse variation, the values of the two variables change in an opposite manner – as one value increases, the other decreases.

For instance, a biker traveling at 8 mph can cover 8 miles in 1 hour.  If the biker’s speed decreases to 4 mph, it will take the biker 2 hours (an increase of one hour), to cover the same distance.

Inverse variation:  when one variable increases,
the other variable decreases.

Notice the shape of the graph of inverse variation.
If the value of x is increased, then y decreases.
If x decreases, the y value increases.  We say that y varies inversely as the value of x.

An inverse variation between 2 variables, y and x, is a relationship that is expressed as:

Y = k/x

where the variable k is called the constant of proportionality.

As with the direct variation problems, the k value needs to be found using the first set of data.

 

The Reciprocal of a Number

Clearly, 3 X 1/3 = 1

1/3 is called the reciprocal of 3

3 is called the reciprocal of 1/3

One number is the reciprocal of another if their product is 1.

Example, the reciprocal of 5/6 is 6/5 since 5/6 X 6/5 = 1.

In general:

The reciprocal of a fraction is obtained by interchanging the numerator and the denominator, i.e. by inverting the fraction.

Example

Find the reciprocal of 20.

Solution:

Reciprocal of 20 is 1/20
Example

Find the reciprocal of 3/7.

Solution:

Example 9

Solution:

Note:

To find the reciprocal of a mixed number, change it into an improper fraction and then invert it.

Additive Inverse

If we add 0 (zero) to any number, the result is the same as the given number.

For example,

3 + 0 = 3,       0 + 8 = 8

We say that 0 is the identity for addition. If the sum of two numbers is 0, we say that each number is the additive inverse of the other.

For example, (+3) + (-3) = 0.

(-3) is the additive inverse of (+3).

(+3) is the additive inverse of (-3).

(-8) + (+8) = 0.

(+8) is the additive inverse of (-8).

(-8) is the additive inverse of (8+).

Example 1

State the additive inverse of:

1. -19
2. 0.32
3. -7/8
4. 6 x 10⁷
5. – 3.1 x 10^-5

 

Given number                  Additive inverse

1. -19                                       + 19
2. 0.32 -0.32
3. -7/8 +7/8
4. 6 x 10⁷-6 x 10⁷
5. – 3.1 x 10^-5         +3.1 x 10^-5           

In parts d and e of example 1, remember that the power of 10 in a number in standard form places the decimal the decimal point. It is not significant in deciding whether the number is positive or negative.

Solve the following equations:

1. x + 7 = 2

 

Solution

x + 7 = 2 is the same as x + (+7) = 2

Add (-7) to both sides.

X + (+7) + (-7) = 2 + (-7)

X + 0 = 2 – 7

X = – 5

 

Multiplicative Inverse

If we multiply any number by 1 the result is the same as the given number. For example,

1 x 9 = 9, -5 x 1 = 15, 1 x /34 = ¾

We say that 1 is the identity for multiplication.  If the product of two numbers is the  multiplicative inverse  of the other. For example,

9 x 1/9 = 1

1/9 is the multiplicative inverse of 9.

9 is the multiplicative inverse of 1/9.

(-5) x(-1/5) = 1

-1.5 is the multiplicative inverse of 3/4.

-5 is the multiplicative inverse of -1/5.

¾ x 4/3 = 1

4/3 is the multiplicative inverse of ¾.

¾ is the multiplicative inverse of 4/3.

You have already used multiplicative inverses. In Book 1 you used reciprocals. The reciprocal of a fraction is that fraction turned upside down. The reciprocal of 2/3 is 3/2. Thus the multiplicative inverse of a number is the same as its reciprocal. 1/8 is the reciprocal of 8/1 or 8. 1/8 is the multiplicative inverse of 8.

 

Example

Find the multiplicative inverses of the following:

1. -32/  b. 0.3  c. 2 ½  d. n

Reciprocal of -3/2 = -2/3

-2/3 is the multiplicative inverse of -3/2.

Check: (-3/2) x (-2/3) = + (3/2 x 2/3) = 1.

1. 0.3 = 3/10

Reciprocal of 3/10 = 10/3.

10/3 (or 3 1/3) is the multiplicative inverse of 0.3.

Check: 0.3 x 3 1/3 = 3/10 x 10/3 = 1.

1. 2 ½ = 5/2

2/5 is the multiplicative inverse of 2 ½.

The check is left as an exercise.

1. i/n is the multiplicative inverse of n.

n x 1/n = 1.

Example

Solve -5x = 20.

 

Method I:

Notice that -5 is the multiplying x. Multiply both sides by the multiplicative inverse of -5.

Multiply both sides by -1/5.

(-1/5) x (-5) X x = (-1/5) x (+20)

1 X x = -(1/5 x 20)

x = -4

 

Method II:

Notice that multiplying by -1/5 is equivalent to dividing by -5. The example can be solved as follows.

-5x = 20

Divide both sides by -5.

(-5) X x/(-5) = +20/-5

1 X x = -(20/5)

x = -4

The second method is usually quicker.

 

Inverse Operation

Do the following:

1. Stand up. Sit down.
2. Add 3 to 15. Subtract 3 from the result.
3. Multiply 7 by 2. Divide the result by 2.

In each case you should end where you start. When this happens, we say that the two actions are inverse operations.

Sitting down is the inverse operation of standing up. Adding a number is the inverse operation of subtracting the same number. Multiplying a number and dividing by the same number are inverse operations.

operation	Inverse operation
Shut the door	Open the door
Add 20	Subtract 20
Subtract -3	Add -3
Multiply by 4	Divide by 4
Divide by 0.3	Multiply by 0.3

 

1. Compound Interest

Simple Interest

Interest is the payment given for saving money. It can also be the price paid for borrowing money. When interest is calculated on the basic sum of money saved (or borrowed) it called simple interest.

To find the simple interest, use this formula: Interest = Principal × Rate of interest × Time

The principal is the amount of money you borrow or invest.

The rate of interest is the percent charged for the use of money, the percentage charged will be divided by hundred to get the actual value for application in solving a problem.

 

Exercises

Compute the interest if the principal is 2000 dollars at a rate of interest of 5% for 4 years.

Using a calculator,

Interest = 2000 × 5% × 4 = 2000 × 0.05 × 4

Interest = 100 × 4 = 400

 

Exercises

Compute the interest if the principal is 2,000,000 dollars at a rate of interest of 4% for a year

Using a calculator,

Interest = 2,000,000 × 4% × 1

Interest = 2,000,000 × 0.04 × 1

Interest = 80,000 × 1 = 80,000

If you have 2 million dollars and your bank pay you 4% interest every year, you will earn 80,000 dollars every year.

Great, you can quit your day time job!

 

Exercises

Compute the interest if the principal is 100 dollars at a rate of interest of 2% for 10 year

Using a calculator,

Interest = 100 × 2% × 10

Interest = 100 × 0.02 × 10

Interest = 2 × 10 = 20

With little money invested and low interest, 10 years investment gives you a mere 20 dollars

This might be a waste of time!

 

Compound Interest

When money is saved with simple interest, the interest is paid at regular intervals and the principal remains the same.

With compound interest the interest is added to the principal at the end of each interval.

Thus, the principal increases and so the interest becomes greater for each interval. Most savings schemes give compound interest, not simple interest.

 

Example

Find the compound interest on N60 000 for 2 years at 8% per annum.

Note: ‘per annum’ means ‘each year’.

The interest is added at 1 year intervals.

1^st year: I₁ = N 60 000 X 8 X 1/100 = N 4 800

Amount at end of 1^st year = N60 000 + N4 800 = N64 800

2^nd year: The principal is now N64 800

= N648 X 8 = N5 148

Amount at end of 2^nd year = N64 800 + N5 184

= N69 984

Compound interest = N69 984 – N60 000 = N9 984

The working is easier if arranged in a table. The annual interest can be calculated by inspection. For example, 6% of N21 000 is found by multiplying N21 000 by 6, and moving the digits two places to the right (to divide by 100:; i.e.

6% of N21 000 = N210 X 6 = N1 260

The example below shows how to arrange the working

 

Example

Find the amount that N5 000 becomes if saved for 3 years at 6% per annum compound interest.

1^st year: Principal      N5 000

6% Interest                  +  300        (6/100 X 5 000)

—————-

2^nd year: Principal   N5 300

6% Interest               +     318          (6/100 X 5 300)

—————-

3^rd year: Principal       5 618

6% Interest             +       337.08    (6/100 X 5 618)

——————

Amount                   N  5 955.08

 

Assessment

Find the a. amount b. the compound interest, for each of the following

1. N40 000 for 2 years at 8% per annum
2. N60 00 for 2 years at 7% per annum
3. N50 000 for 2 years at 6% per annum

When calculating compound interest, the arithmetic often gives final answers to many decimal places. Final answers should be rounded to the nearest naira. Such rounding should be left to the last line of the working. If possible, use a calculator to calculate interest.

When money is borrowed, interest must be paid back as well as the principal. When a large sum of money is paid back over a number of years, the principal gradually reduces.

 

Depreciation

Many items, such as cars, clothes, electrical goods, lose values and time passes. This loss in value is called depreciation. Depreciation is usually given as a percentage of the item’s value at the beginning of the year. For example, if a radio costing N10 000 depreciates by 20% per annum, then its value will be N8 000 at the end of the first year. At the end of the second year, its value will be N8 000 less 20% of N8 000, i.e. N8 000 – N1 6000 = N 6 400.

Example

A car costing N680 000 depreciates by 25% in its first year and 20% in its second year, Find its avalue after 2 years.

1^st year:

Value of car              N680 000

25% depreciation    – 170 000        (1/4 of 680 000)

——————–

2^nd year:

Value of car                510 000

20% depreciation    – 102 000        (1/5 of 510 000)

Value after 2 yr =    N408 000

 

Inflation

Due to rising prices, money loses its value as time passes. Loss in value of money is called inflation. Inflation is usually given as the percentage increase in the cost of buying things from one year to the next. For example, if the rate of inflation 15% per annum, then a CD player which cost N10 000 a year ago will now cost N11 500. Money has lost it’s a value since it now costs more to buy the same thing.

Example

How long will it take for prices to double if the rate of inflation is 20% per annum?

Start with an initial cost of 100 units.

Initially, cost = 100

rise =   20

—————-

after 1 year, cost = 120

rise =   24               (i.e. 20% of 120)

—————

after 2 years, cost = 144

rise =     28.8        (20% of 144)

—————-

after 3 years, cost = 172.8

rise =    34.56      (20% of 172.8)

The cost after 4 years is a little more than double the initial cost. Hence prices will double in just under 4 years.

 

Assessment

1. 6 times a number is 48. What is the number?
2. Find the number which, when multiplied by 10, gives 70.
3. A number divide by 5 gives 9, what is the number?
4. 12/ x – 1 = 3
5. 4/ 1 + 4 = 1
6. 2 = 7/ y + 2

Solve the following equations:

7. 4x = 28
8. 3x = 18
9. -4d = 20
10. 7x = 4 2/3

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