Second Term Jss 2 Mathematics

NAME:…………………………………………

CLASS:…………………..

 

SECOND TERM: E– LEARNING NOTES

JS 2 (BASIC 8)

SUBJECT: MATHEMATICS

SCHEME OF WORK

WEEK TOPIC

1. Revision of First Term Work.
2. Simple Equations: Harder problems on Algebraic Fractions.
3. Word Problems on algebraic Fractions. World problem leading to Simple Algebraic Fractions.
4. Linear inequality in one variable. Graphical representations of solutions of linear inequalities in one variable.
5. Graphs: plotting points on the Cartesian plane (the axis and plotting of points). Linear Equation in two variables – plotting the graph of Linear Equations in two variables
6. More work on plotting and joining points to form plane shapes on Cartesian Planes.
7. Graphs: Linear graphs from real life situations – plotting linear graphs from real life situation. Quantitative reasoning – solving quantitative aptitude problems.
8. Plane Figures/shapes: Revision on properties of parallelogram rhombus and kite.
9. Scale drawing – Drawing to scale to represent given distances. Solving problems on quantitative aptitude related to plane shapes/ figures
10. Revision
11. Examination.

 

 

WEEK 1

Topic: Revision of last term’s work using past examinations questions papers.

NOTE: Teachers should ensure all topics that seem complex are thoroughly explained for better understanding. Evaluation should be given and marked.

 

WEEK 2

TOPIC: SIMPLE EQUATION AND WORLD PROBLEMS

DEFINITION OF EQUATIONS

Equations are open sentences which have the equal sign. Solving an equation means finding a value of the unknown which makes the equation true. Any letter a, f, c, x etc. can be used as unknown. The set of values which make an equation true is called the truth set.

Example 1

Solve the following simple equations:

1. 3a-8 = 10
2. 10x – 7 = 27
3. 4 + 5y = 19

Solution

1. 3a – 8 = 10

Method 1

3a – 8 + 8 =10 +8

3a = 18

= 6

Method 2

3a – 8 = 10

3a = 10 + 8

3a = 18

a = = 6

Note: If a number is moved from one side to an equation over the equal sign to the other side, the sign of the number changes.

1. 10x – 7 = 27

Method 1

10x – 7 + 7 = 27 + 7

10x = 34

 

x = 3.4

Method 2

10x – 7 = 27

10x = 27+ 7

x = = 3.4

EXAMPLE 2

Solve the following equations:

1. 4(3y-2) = 6
2. 2 (x-8) = 3(x+5)

Solution

1. 4 (3y-2) = 6

Removing the bracket we have

12y-8=6

Solving the equation

12y= 6 +8 =14

y= =

1. 2(x-8) = 3(x+5)

Removing the brackets

2x-16 = 3x+15

Collecting the like terms together and solving

-16-15 = 3x-2x

-31 = x

Or

x = -31

EXAMPLE 3

Solve the following equations

Solution

Multiply both sides by 3

a-2 = 12

a =12+2

a = 14

Multiply both sides by 7

5

35 + 3 = 2x

38 =2x

x =19

Collecting the like terms together

(L.C.M of 3 and x is 3x)

2 = 13x

 

WORD PROBLEM ON SIMPLE EQUATIONS

This is the same as solving simple equations.

EXAMPLE:

1. I think of a number, add 3 to it and then divide the result by 5. If the answer equals the original number, what is the number?
2. The average cost of a number of erasers is 50k. if all the erasers cost N 20, find the total number of erasers.

Solution:

1. Let the number be x.

3 added to the number becomes: x+3

Divide the result by 5:

And since the result equals original number, we have,

Solving, we have,

x +3 = 5x

3 = 5x-x

3 = 4x

= x

1. Let the number of erasers be y

Average cost of erasers = N

Average cost of erasers =

50 =

50y =

50y = 2000

y = 40 erasers

:

Solve the following equations;

READING ASSIGNMENT:

New General Mathematics for Junior Secondary Schools 2 by M. F. Macrae et al

WEEKEND ASSIGNMENT:

1. Solve
2. A sportswoman has a body of mass of xkg. Her mass is i4kg less than that of her friend. If their total mass is 76kg. Find the mass of each of them.
3. A girl has 12 sweets. She eats n out of them. Find how many sweets she has left if n=2

WEEK 3

TOPIC: WORD PROBLEMS ON ALGEBRAIC FRACTIONS

CONTENT:

• Word problems leading to simple algebraic fractions
• Problem solving

 

WORD PROBLEMS LEADING TO SIMPLE ALGEBRAIC FRACTIONS

When solving word problems, the following points are important:

1. Use a letter to represent the unknown
2. Translate problem from words to algebraic form
3. Make an equation out of it
4. Solve equation completely to get the unknown
5. Where it involve fraction, first clear the fraction using LCM
6. Give the answer in a written form
7. Check the result against the information given in the question

PROBLEM SOLVING

1. Ojo add 15 to a number and then divide the sum by three. The result is twice the first number. Find the number.

Solution:

Let the number be

Ojo add 15 to

Ojo divide the result by 3:

The result is

This implies that:

Multiply both sides 3 to clear the fraction

Collecting like terms,

Dividing both side by 5, the result is 3 ie

Therefore, the number is 3

1. I add 60 to a certain number and then divide the result by 12, the quotient is 6. Find the original number.

Solution:

Let the number be

I add 60 to the number i.e

Dividing result by 12 implies

Quotient means answer

Multiply through by 12 to clear the fraction: x12

This implies:

The original number is 12.

1. Juliet is old and her brother is half of her age. The sum of their ages is half of their father’s age who is now 60. How old is Juliet and her brother?

Solution:

Juliet age is

Her brother’s age is half of hers i.e

Their Father’s age is 60 years and half of 60 is 30.

Sum of their ages is half of their father:

Multiply through by 2 to clear the fraction:

This means that: Juliet is 10 years and her brother is 5 years.

Evaluation:

1. Find a number such that when 20 is added to ¼ of it, the result is times the original number.
2. One-fifth of an even number added tone-sixth of the next even number makes a total of 15. Find the two numbers. (Hint: let the numbers be and ).

Reading Assignment:

New General Mathematics JSS-2 UBE Edition. Page 114 to 117.

WEEKEND ASSIGNMENT:

New General Mathematics JSS-2 UBE Edition. Ex. 13h: Attempt Q1 to Q16 Page 117 to 118.

 

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WEEK 4

TOPIC: LINEAR INEQUALITIES

CONTENT:

• Inequalities
• Graphs of inequalities
• Solution of inequalities

INEQUALITIES

In mathematics, we use the equal s sign, = , to show that quantities are the same. How ever, very often, quantities are different, or unequal. The inequalities symbols are as follows:

unequal to

less than

 

Take a and b to be any number points on number line below:

If a = +3 and b = +3

a b

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

a and b have the same position on the number line.

Hence, a = b

But if a = 1

and b = -4

then a b

or b a

b a

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

A linear inequality is an open sentence with an inequality sign.

For example: x + 2 7

Example:

The radius of a circle is less than 5m. What can be said about?

1. Its circumference?
2. Its area?

Solution

r 5m

1. Circumference = 2r

Circumference 10m

Area 25m²

EVALUATION:

1. State whether each of the following is true or false.
2. 12 5 (c) -5 7
3. 0-4 -3 (d) -2 -4
4. Eraser cost 7 kobo each, x kobo is not enough to buy an eraser but a boy with y kobo is able to buy an eraser. Write down 3 different inequalities in terms of a or b or both.

Graphs of Inequalities in one Variable

Examples

1. The inequality x 2 means that x can have any value less than 2. We can show these values on the number line below:

-2 -1 0 1 2 3

The heavy arrowed line on the number line illustrated above shows the range of values that x can have. The empty circle shows that the value 2 is not included; x can have any other value to the left of 2.

 

1. The inequality means that x can have the value -1 or any value greater than -1. From the illustration below, the shaded circle shows that the value -1 is included.

-3 -2 -1 0 1 2

Evaluation:

Sketch the graphs of the following inequalities:

1. X 4 c. X -3
2. X d. X 5

SOLUTION OF INEQUALITIES

Example1:

Solve the inequality x + 4 6

Solution

x + 4 6

Subtract 4 from both sides.

X + 4-4 6 – 4

X 2

Example 2:

Solve 19 4 – 5x

Solution:

19 4 – 5x

Subtract 4 from both sides

19-4 4-4 -5x

15 -5x

Divide both sides by -5 and reverse the inequality sign.

 

-3 x

If -3 x, then x -3.

READING ASSIGNMENT:

NEW GENERAL MATHEMATICS FOR JSS 2 (UBE EDITION)

PG. 209 – 213.

WEEKEND ASSIGNMENT:

Ex. 22f 1-5; 22g 1-5; 22h 1-5

JSS 2 SECOND TERM MATHEMATICS LESSON NOTE

WEEK 5

TOPIC: GRAPH

Content:

• Plotting points on Cartesian plane
• Linear equation in two variables
• Graph of linear equation in two variables

PLOTTING POINTS ON CARTESIAN PLANE

Graph is a picture of numerical data. A familiar example is the representation of numbers on the number line. The positions of the number on the line are called points.

CARTESIAN PLANE

This is plane surface with axes drawn on it. The Cartesian is derived from a French philosopher and mathematician who work out the possibility of presenting geometry in a numerical form. His full name is Rene Descartes, hence the name Cartesian (+)y-axis

 

Vertical line Point of intersection (origin)

 

(-) (0, 0) (+) x-axis

 

Horizontal line

(-)

We describe a point on a Cartesian plane by two numbers say () called coordinates. is the distance of the point from vertical line called -axis and is the distance of the point from horizontal line called -axis. The figure illustrates more.

+3

+2 .A

B+1

 

-4 -3 -2 -1 +1 +2 +3 +4 +5

-1

.C -2 .D

-3

The position of each point is given by an ordered pair of number. These are called co-ordinates of the point. The first number is called the – coordinate and second number is called -coordinate. The coordinates are separated by a comma.

A(3,2) B(-2,1) C(-3,-2) D(3,-2)

 

Evaluation:

Plot the following points on a graph:

1. (2,4), (5,1), (3,4),(1,1)
2. (6,6), (1,6), (4,4), (1,4)

LINEAR GRAPH

This is also called straight line graph. A linear graph is obtained by plotting the points whose coordinates () satisfying a linear equation of the form where and are given constants and

gradient and

PLOTTING OF GRAPH

This is the process by which those recorded pairs of points in the in the table of values are now transferred onto the graph paper appropriately.

To do this accurately the following points are considered:

1. Prepare a table of value for function or equation given
2. Draw the axes (x-axis and y-axis)
3. Choose the appropriate scale if not given
4. Choose any convenient range of your choice for x if not given.
5. Plot the points according to the calculated coordinate pairs
6. Join the plotted points using rule for a straight line graph and curve for quadratic graph

Example:

Plot the graph of the function using values of from -3 to 2

Solution:

Note: Teacher should demonstrate the plotting of this graph in the class and extend examples to two distinct line graphs on this plane using single table of value.

EVALUATION:

Draw the graphs of the following functions:

1. b. c.

READING ASSIGNMENT:

New General Mathematics JSS-2 UBE Edition page102-104.

WEEKEND ASSIGNMENT:

New General Mathematics JSS-2 UBE Edition page102, Ex. 12a Q1 to Q4

WEEK 6

TOPIC: PLANE SHAPES ON CARTESIAN PLANE

Content:

• Plotting and joining of points to form plane shapes

PLOTTING AND JOINING OF POINTS TO FORM PLANE SHAPES

 

Examples:

1. Plot the points: P (-5, 3) Q (3, 5) R (-4,-4)
2. Join the plotted points
3. Name the shape obtained

Solution:

Q(3,5)

P(-4,3) 5

3

 

 

-4

R(4,-4) -4

Note: Teacher should give more examples in the class while course teaching this topic.

EVALUATION:

Using a graph paper, draw axes OX and OY and label the x-axis -7 to 7 and y-axis from -5 to 5. Using the same scale for each axis, plot the following points and join them together in the given order.

1. A(2,5) B(-5,-2) C(4,4)
2. P(-5,2) Q(0,5) R(7,2) S(0,0)
3. Name each in questions (1) and (2) above

READING ASSIGNMENT:

New General Mathematics JSS-2 UBE Edition page104-109

WEEKEND ASSIGNMENT:

New General Mathematics JSS-2 UBE Edition page108, Ex. 12d Q1 to Q6

WEEK 7

TOPIC: GRAPHS

CONTENT:

• linear graphs from real life situation
• Quantitative reasoning on linear graph

LINEAR GRAPHS FROM REAL LIFE SITUATION

Example

A student walks along a road at a speed of 120m per minute.

1. Make a table of values showing how far the students have walked after 0, 1, 2, 3, 4, 5 minutes.
2. Using a scale of 1cm to 1min on the horizontal axis and 1cm to 100cm on the vertical axis, draw a graph of this information.
3. Use the graph to find the following:
4. How far the student has walked after 2.6min
5. How long it takes the student to walk 500m.

Solution:

1. Table of values is illustrated below:

Time (s)	0	1	2	3	4	5	6
Speed (km/h)	0	15	20	45	60	75	90

b.

600 x

500 x

Distance (m)

400

x

300

x

200

x

100

0 1 2 3 4 5

Time (min)

(c) i. 2.6 min corresponds to 310m (approximately). The student has walked about 310m after 2.6min.

ii. 500m corresponding to 4.2min (approximate). The student takes about 4.2min to walk 500m.

EVALUATION:

A girl walks along a road at a speed of 100m per minute.

1. Copy and complete the table below:

 

Time (min)	0	1	2	3	4	5	6
Distance (m)	0	100	200

1. Using a scale of 2cm to 1min on the horizontal axis and 2cm to 100m on the vertical axis, draw a graph of the information.
2. Use your graph to find i. how far the girl has walked after 5.7 min, ii. how long it takes her to walk 335m

READING ASSIGNMENT:

New General Mathematics JSS 2 (UBE Edition), pg 119- 136

WEEKEND ASSIGNMENT:

New General Mathematics JSS 2 (UBE Edition), pg 119- 136

Ex. 14a. Q5; 14b.Q 2; 14c. Q1 page 121-125

 

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WEEK 8

TOPIC: PLANE FIGURES/SHAPES

CONTENT:

Properties of Quadrilateral (i.e. parallelogram, rhombus, and kite)

^{QUADRILATERAL}

Any four sided plane figure is called a quadrilateral.

Examples of quadrilaterals are:

1. PARALLELOGRAM

A parallelogram is a quadrilateral whose pairs are equal and parallel.

PROPERTIES OF PARALLELOGRAM

1. Its opposite sides are parallel and equal
2. Its opposite angles are equal
3. Its diagonals bisect each other
4. Each diagonal bisects the parallelogram into two congruent triangles
5. The (4) angles together add up to 360⁰

Consider the parallelogram below:

A 10cm B

E

4cm

4cm

4cm

D 10cm C

Line AB =Line DC = 10cm

Line AB is parallel to line DC

Line AD = Line BC = 4cm

Line AD is parallel to line BC

Angle BAD = Angle BCD

Angle ADC = Angle ABC

Diagonal AC divides the parallelogram ABCD into two congruent triangles ABC and ADC

Diagonal DB divides the parallelogram ABCD into two congruent triangles DAB and DCB

The diagonals bisect each other at point E i.e.

AE = EC = and DE = EB

 

1. RHOMBUS

A rhombus is a quadrilateral which has four sides equal length.

 

PROPERTIES OF RHOMBUS

1. All four sides are equal in length.
2. The opposite angles are equal.
3. The diagonals are the lines of symmetry.
4. The diagonals bisect each other at right angles.

A B

P

 

D C

Line AB = line BC = line DC = line AD, i.e.

AB = BC= DC = AD

AD is parallel to BC

AB is parallel to DC

<DAB = <BCD, and <DPC = <APD

<APB = <BPC = <DPC = <APD = 90⁰

PA = PC and PB = PD

1. KITE

A kite is a quadrilateral in which one diagonal is a line of symmetry.

Lines of symmetry

NOTE: Teachers should display these shapes using various class activities

EVALUATION / ACTIVITIES:

Draw a parallelogram on a cardboard paper such that its diagonals meet C. then cut out the parallelogram along the diagonals. What do you notice?

READING ASSIGNMENT:

New General Mathematics for JSS 2 (UBE Edition) , pg 32-38

WEEKEND ASSIGNMENT

New General Mathematics for JSS 2 (UBE Edition) , pg 37

Ex. 3d Q1-Q10

WEEK 9

TOPIC: SCALE DRAWING

Content:

• Scale
• Scale drawing
• Reading scale drawing Scale

SCALE

A scale simply means the dimension or proportion of objects in comparison to its original or actual size after the drawing. It is widely use in sciences especially in Geography, Biology, Mathematics, Physics, etc.

The scale of a drawing is determined by comparing the length of the drawing with the actual length of the object.

Scale

A is original

B is drawing

Diameter of A = 25cm Diameter of B = 10cm

Scale = the scale is 2 to 5, i.e 2 : 5 meaning 2cm represents 5cm

EVALUATION:

Use measurement to find the scale of the following shapes/figures:

Original

Drawing

Original

b.

 

Drawing

 

SCALE DRAWING

In scale drawing the materials needed are pencil, ruler and set square. The drawings should be on plane sheets or white papers. The dimensions of the actual object should be written on the drawing and a title has to accompany it. However, a rough sketch must be drawn first.

Example:

A measurement of 8000m is drawn to a scale where 1cm represents 500m. Find its length on the drawing.

Solution:

500m is represented by 1cm.

1m is represented by

8000m is represented by

Length on drawing = 16cm.

READING SCALE DRAWING

Reading scale drawing is common to map reading and interpretation. For instance, scales on maps are often given in the ratio form as 1 : .

A map of scale 1 : 50 000 means 1cm represents 50 000cm.

This means 1cm on map represents 50 000cm on land.

Map scale

Rearranging the above formula we have:

Actual distance = map scale x distance on map

OR Distance on map

 

Example:

Change the scale 1 : 800 000 to the form 1cm represents ….km.

Solution:

1:800 000 means 1cm represents 800 000cm.

But 1km = 100 000cm

1cm represents 800 000 100 000 = 8km

The scale is 1cm represents 8km.

EVALUATION

1. Two cities are 70km apart. The distance between them is 20cm on the map. What is the scale of the map?
2. Rewrite the following scales in the form 1:n
3. 1cm represent 20 000cm
4. 1cm represents 500 000cm

READING ASSIGNMENT:

New General Mathematics JSS 2, UBE Edition, pg 149 – 157

WEEKEND ASSIGNMENT:

New General Mathematics JSS 2, UBE Edition, pg 153

Ex. 1-3

 

WEEK 10: Revision

 

WEEK 11: Examination

 

END OF SECOND TERM WORK