Name of the teacher : Aswathy. A
Subject : Mathematics
Unit : Equations
Subunit : Different problems
Name of the school : MTHSS
Standard : VIII
Strength : 40/43
Date : 10/08/2015
Time : 30 minutes
To understand about different problems in equations and its importance in mathematics through
discussion, observation, organization of chart and by analyzing prepared notes of the pupil.
NEW TERMS : equations, algebraic form.
FACTS : algebraic method of solving problems
CONCEPTS : concept of algebraic method in equations.
PROCESS : process of finding solution to different problems using algebraic equations.
The pupil will be able to,
1. Recall the term algebra.
2. Recognize algebraic form of different equations.
3. Explain the term algebra and equations.
4. Observe all aspects of equations keenly.
5. Discuss the different problems through algebraic equations.
6. Read chart quickly and accurately.
7. Ask questions to know more about algebraic equations.
8. Give illustration of algebraic equations.
9. Plan to do problems on algebraic equations.
PRE-REQUISITES: The students have knowledge on algebraic method, its addition, subtraction,
TEACHING LEARNING RESOURCES: Usual classroom aids, activity cards.
LEARNING STRATEGIES: Individual work, group discussion, observation and explanation by teacher.
CLASSROOM INTERACTION PROCEDURE EXPECTED PUPIL RESPONSE
ACTIVITY - 1
i. Yesterday we study to make algebraic form
of different equations.
iii. Today we are going to make and solve
v. Will you interest?
ACTIVITY – 2
Then teacher presents a problem
“Ticket rate for the science exhibition is 10 rupees
for a child and 26 rupees for an adult, 740 rupees
were got from 50 persons. How many children
among them? ”
Teacher writes answer on blackboard after give an
Let x be the number of children.
Number of adults =50-x.
Ticket rate for a child = 10.
Ticket rate for x children’s = 10x.
Ticket rate for an adult = 26.
Ticket rate for (50-x) adults = 26(50-x).
Total price = 740.
10x + 26(50-x) = 740.
10x + 1300 – 26x = 740
10x – 26x = 740 - 1300
16x = 560.
X = 560/16
Number of children = 35.
Number of adult = 50-35
ACTIVITY – 3
Teacher gives another problem.
“A class has the same number of boys and girls.
Only 8 boys were present on a particular day and
then the number of girls was double the number of
boys. What is the number of boys and girls?”
Teacher gave opportunity to students to explain the
question. Then teacher explain the answer and
writes on blackboard.
Let the number of boys and girls be x.
If 8boys were absent,
then the number of boys = x-8.
Number of girls = 2(number of boys).
x = 2(x-8).
x = 2x-16.
2x-x = 16.
x = 16.
Number of boys = 16.
Number of girls = 16.
Teacher divides the class into 5 groups and gives
activity cards containing questions.
”Ajayan is 10 years older than Vijayan. Next year,
Ajayan’s age would be double that of Vijayan. What
are their ages now?”
“5 times a number is equal to 3 times the sum of
the number and 4. What is the number? ”
ACTIVITY – 5
Teacher concludes the class by telling what have
learned in the class. Teacher clarifies all the doubts
ACTIVITY – 6
Teacher ask questions to students and ask to form
questions related to the topic
FOLLOW UP ACTIVITY
1. In a co-operative society the number of
men is thrice the number of women .29
women and 16 men more joined the
society and now the number of men is
double the number of women. How many
women were there in the society at first?
2. 8 times a number is equal to 4 times the
sum of the number and 4.what is the