Class: Junior Secondary School 2 (JSS 2 / JS2)
Term: 3rd Term
Week: 9
Age: 13 years
Duration: 45 minutes
Subject: Mathematics
Curriculum Theme: Mathematics
Previous Lesson: Probability: Determining the Probability of Certain Events.
Topic: SIMPLE PROPORTION
Subject Matter: Direct proportion: Inverse proportion

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

Define simple proportion, direct proportion, and inverse proportion.
Identify situations that involve direct proportion and inverse proportion.
State the methods for solving problems involving direct and inverse proportion.

Affective Domain:

Appreciate the real-life applications of direct and inverse proportion.
Show interest in solving problems related to proportion.

Psychomotor Domain:

Solve practical problems involving direct proportion.
Solve practical problems involving inverse proportion.

Work collaboratively to solve problems on proportion.
Discuss and explain proportional relationships to peers.

Reference Materials

The following resources were used in planning this lesson:

9 Years Basic Education Curriculum
State Unified Scheme of Work
New General Mathematics for Junior Secondary Schools 2

Instructional Materials

The teacher will teach this lesson with the aid of:

Charts showing examples of direct and inverse proportion.
Real-life scenarios like cost of goods, speed and time.

Rationale for the Lesson

This lesson enables pupils to understand how quantities relate to each other in everyday situations. It helps pupils develop problem-solving skills necessary for various practical applications in life and further studies in mathematics.

Prerequisite/Previous Knowledge

Pupils are expected to have prior knowledge of ratios, fractions, and basic arithmetic operations.

Lesson Content/Board Summary

SIMPLE PROPORTION

Introduction to Proportion

Proportion describes a relationship between two quantities where the ratio of the quantities is constant. It helps us compare and relate different amounts.

Direct Proportion

Two quantities are in direct proportion if an increase in one quantity leads to an increase in the other quantity, and a decrease in one quantity leads to a decrease in the other, such that their ratio remains constant.

If ‘a’ is directly proportional to ‘b’, we write a ∝ b, which means a/b = k (a constant) or a₁/b₁ = a₂/b₂.

Examples of Direct Proportion:

Example 1: If 5 exercise books cost N400, what will be the cost of 8 exercise books?

Step 1: Identify the relationship. As the number of books increases, the cost increases. This is direct proportion.
Step 2: Set up the proportion. Let the cost of 8 books be Nx.
5 books / N400 = 8 books / Nx
5/400 = 8/x
Step 3: Cross-multiply and solve for x.
5 * x = 8 * 400
5x = 3200
x = 3200 / 5
x = N640
Therefore, 8 exercise books will cost N640.

Example 2: A car travels 150 km in 3 hours. How far will it travel in 5 hours at the same speed?

Step 1: Identify the relationship. As time increases, the distance traveled increases. This is direct proportion.
Step 2: Set up the proportion. Let the distance traveled in 5 hours be y km.
150 km / 3 hours = y km / 5 hours
150/3 = y/5
Step 3: Cross-multiply and solve for y.
3 * y = 150 * 5
3y = 750
y = 750 / 3
y = 250 km
Therefore, the car will travel 250 km in 5 hours.

Inverse Proportion

Two quantities are in inverse proportion if an increase in one quantity leads to a decrease in the other quantity, and a decrease in one quantity leads to an increase in the other, such that their product remains constant.

If ‘a’ is inversely proportional to ‘b’, we write a ∝ 1/b, which means a * b = k (a constant) or a₁b₁ = a₂b₂.

Examples of Inverse Proportion:

Example 1: If 4 men can dig a trench in 6 days, how long will it take 8 men to dig the same trench?

Step 1: Identify the relationship. As the number of men increases, the time taken to dig the trench decreases. This is inverse proportion.
Step 2: Set up the proportion. Let the time taken by 8 men be x days.
Men₁ * Days₁ = Men₂ * Days₂
4 * 6 = 8 * x
Step 3: Solve for x.
24 = 8x
x = 24 / 8
x = 3 days
Therefore, 8 men will dig the trench in 3 days.

Example 2: A tap fills a tank in 10 hours at a flow rate of 5 litres per minute. How long will it take to fill the same tank if the flow rate is increased to 25 litres per minute?

Step 1: Identify the relationship. As the flow rate increases, the time taken to fill the tank decreases. This is inverse proportion.
Step 2: Set up the proportion. Let the time taken at 25 litres/minute be y hours.
Rate₁ * Time₁ = Rate₂ * Time₂
5 litres/min * 10 hours = 25 litres/min * y hours
Step 3: Solve for y.
50 = 25y
y = 50 / 25
y = 2 hours
Therefore, it will take 2 hours to fill the tank at a flow rate of 25 litres per minute.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes
Teaching Skill: Set Induction
Teacher’s Activity: The teacher introduces the lesson by asking pupils about situations where changing one quantity affects another, such as the cost of buying more items or the time taken to complete a task with more workers.
Pupils’ Activity: Pupils share their thoughts and experiences related to the teacher’s questions.
Learning Point: Pupils are prepared for the day’s topic on proportion.

Step 2: Explanation of Direct Proportion

Time: 10 minutes
Teaching Skill: Explanation/Definition
Teacher’s Activity: The teacher defines direct proportion, explains its characteristics, and writes the definition and a simple example on the board.
Pupils’ Activity: Pupils listen attentively, copy notes, and ask questions for clarification.
Learning Point: Pupils understand the concept and definition of direct proportion.

Step 3: Solving Problems on Direct Proportion

Time: 10 minutes
Teaching Skill: Demonstration/Problem Solving
Teacher’s Activity: The teacher demonstrates how to solve problems involving direct proportion using the method of ratio equality, working through the examples from the board summary.
Pupils’ Activity: Pupils observe the steps, participate in solving by suggesting steps, and copy the worked examples.
Learning Point: Pupils learn the practical application and method for solving direct proportion problems.

Step 4: Explanation of Inverse Proportion

Time: 5 minutes
Teaching Skill: Explanation/Definition
Teacher’s Activity: The teacher defines inverse proportion, explains its characteristics, and writes the definition and a simple example on the board.
Pupils’ Activity: Pupils listen, take notes, and compare it with direct proportion.
Learning Point: Pupils understand the concept and definition of inverse proportion.

Step 5: Solving Problems on Inverse Proportion

Time: 5 minutes
Teaching Skill: Demonstration/Problem Solving
Teacher’s Activity: The teacher demonstrates how to solve problems involving inverse proportion using the method of constant product, working through the examples from the board summary.
Pupils’ Activity: Pupils pay attention, contribute to solving, and copy the worked examples.
Learning Point: Pupils learn the practical application and method for solving inverse proportion problems.

Step 6: Discussion on Differences and Applications

Time: 5 minutes
Teaching Skill: Discussion/Comparison
Teacher’s Activity: The teacher leads a brief discussion comparing direct and inverse proportion, highlighting their differences and real-life scenarios for each.
Pupils’ Activity: Pupils participate in the discussion, distinguishing between the two types of proportion.
Learning Point: Pupils can differentiate between direct and inverse proportion and identify their applications.

Step 7: Evaluation/Review

Time: 5 minutes

Teaching Skill: Questioning/Assessment

Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

Define direct proportion.
Give two examples of situations involving inverse proportion.
If 6 labourers can complete a task in 10 days, how many days will it take 15 labourers to complete the same task?
If 3 pencils cost N120, what will be the cost of 7 pencils?

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Pupils demonstrate understanding of the lesson.

Step 8: Conclusion

Time: 0 minutes
Teaching Skill: Summarization
Teacher’s Activity: The teacher summarizes the key points of the lesson and assigns homework.
Pupils’ Activity: Pupils take note of the homework assignment.
Learning Point: The lesson is reinforced and concluded.

Lesson Keywords

Proportion – A statement of equality between two ratios.
Direct Proportion – A relationship where two quantities increase or decrease together at a constant ratio.
Inverse Proportion – A relationship where an increase in one quantity leads to a proportional decrease in another, and vice-versa, such that their product is constant.
Ratio – A comparison of two quantities by division.

Differentiation

For struggling learners, the teacher will provide simpler examples and more guided practice. Advanced learners will be given more complex problems and encouraged to explore additional real-world applications of proportion.

Note for teachers using this lesson plan

Ensure pupils grasp the fundamental difference between direct and inverse proportion before moving to problem-solving. Encourage pupils to identify the type of proportion in each problem before attempting to solve it. Use visual aids to illustrate proportional relationships.