Download Teaching Guide for Book 7

Get Ahead Mathematics
Bilingual Teaching Guide
for Book 7
Parveen Arif Ali
Contents
Unit I: Sets..................................................................................01
Unit II: Operations on Directed Numbers...................................07
Unit III: Rational Numbers..........................................................19
Unit IV: Square Roots.................................................................37
Unit V: Percentage.....................................................................43
Unit VI: Ratio and Proportion.....................................................47
Unit VII: Algebra.........................................................................57
Unit VIII: Geometry.....................................................................79
Unit IX: Information Handling.....................................................81
iii
Introduction
Get Ahead Mathematics is a series of eight books from levels one to
eight. The accompanying Teacher’s Guides contain guidelines for the
teachers. Answer keys to Books 6, 7 and 8 are included in the textbooks
and should be referred to by the teachers.
The teacher should devise means and ways of reaching out to the
students so that they have a thorough knowledge of the subject without
getting bored.
The teachers must use their discretion in teaching a topic in such a way
as they find appropriate, depending on the intelligence level as well as
the academic standard of the class.
Encourage the students to relate examples to real things. Don’t rush.
Allow time to respond to questions and discuss particular concepts.
Come well prepared to the class. Read the introduction to the topic to be
taught in the pupils’ book. Prepare charts if necessary. Practice diagrams
to be drawn on the blackboard. Collect material relevant to the topic.
Prepare short questions, homework, tests and assignments.
Before starting the lesson make a quick survey of the previous knowledge
of the students, by asking them questions pertaining to the topic. Explain
the concepts with worked examples on the board. The students should
be encouraged to work independently, with useful suggestions from the
teacher. Exercises at the end of each lesson should be divided between
class work and homework.The lesson should conclude with a review of
the concept that has been developed or with the work that has been
discussed or accomplished.
Blackboard work is an important aspect of teaching mathematics.
However, too much time should not be spent on it as the students lose
interest. Charts can also be used to explain some concepts, as visual
material helps students make mental pictures which are learnt quickly
and can be recalled instantly.
Most of the work will be done in the exercise books. These should be
carefully and neatly set out so that the processes can easily be seen.
Careful setting out leads to better understanding.
The above guidelines for teachers will enable them to teach effectively
and develop in the students an interest in the subject.
These suggestions can only supplement and support the professional
judgement of the teacher. In no way can they serve as a substitute for
it. It is hoped that your interest in the subject together with the features
of the book will provide students with more zest to learn mathematics
and excel in the subject.
iv
v
Unit I: Sets Pages 1-5
Discuss the use of collective nouns such as a pack of cards, a
team of players, etc in our day-to-day lives. Words such as pack,
flock, team, group, etc are used to denote a collection of objects.
Explain that the word ‘set’ is used in mathematics to describe a
‘collection of objects’.
‘A set is a well-defined collection of objects’.
The phrase ‘well-defined’ means that a set must have some
specific property so that we can easily identify whether an object
belongs to the given set or not. A book, for example, does not
belong to a tea set, and a ball does not belong to a set of playing
cards. So we can say that a tea set and a set of playing cards
are well-defined sets.
Write a set of vowels of th­e English alphabet on the blackboard.
Explain that it is a well-defined set because we know that specific
letters are being referred to: i.e. a, e, i, o, u.
Now consider the example of ‘a set of interesting books in the
library’.
It is not a set because the term interesting is not clearly defined
by any fixed standards. A book may be interesting to one person
but boring to another. Thus, this is not well defined and hence it
is not a set.
Explain that in a well-defined set, the objects are not repeated.
Notation of a set
We generally use capital letters A,B,C etc. to denote sets.
1
2
Elements of a set
The objects of a set are called ‘members’ or ‘elements’ of a
set. All the elements of a set are enclosed in brackets { } and
separated by commas e.g. a set of vowels of the English alphabet;
A = {a, e, i, o, u}.
The elements of a set are said to ‘belong to’ the set. Write the
symbol for ‘belongs to’ on the board.
Write a set of even numbers on the board. B = {0, 2, 4, 6, 8}.
Ask: Does 1 belong to this set?
Explain that 1 is not an even number so it does not belong to the
set of even numbers. Write the symbol for ‘does not belong to’.
Null or empty set
Ask the pupils if they can list the elements of the following set:
‘The set of cats with two tails’.
Explain that the set does not have elements which can be listed.
Such a set is known as an ‘empty set’ or a ‘null set’.
An empty set is denoted by { } or the Greek letter ‘phi ’.
Write the set, A = {0}, on the board.
Ask: Is this an empty set?
Explain that it is not an empty set because it contains the
element ‘0’.
Methods of writing sets
Explain that sets are described or written in two ways. These are:
1. Tabular form
In this form, all elements of the set are listed,
e.g. A = {a, e, i, o, u} (a set of vowels).
2.
Descriptive form
In this form, the elements of the set are not listed, but a
specific property satisfied by the elements of the set is given,
3
4
e.g. A = {the vowels of the English alphabet}.
Write some standard sets on the board and explain their
properties.
Types of sets
Sets can be differentiated into three types depending on the
number of elements they have.
1.
Finite sets
A set that has a limited number of elements is called a ‘finite set ’,
e.g. A = {1, 2, 3, 4}.
Set A has 4 elements.
2.
Infinite sets
A set that has an unlimited number of elements is called an
‘infinite set’, e.g. A = {0, 2, 4, 6, 8,…}.
This set is an infinite set because it contains endless even
numbers.
3.
Empty sets
It is a set that has no members, as explained earlier.
Union of two sets
A union of two sets is a set which contains the elements of both
sets. A and B are written as one set in which none of the elements
are repeated e.g. A = {1, 2, 3}, B = {2, 3, 4}.
The union set of A and B is {1, 2, 3, 4}.
Intersection of sets
The intersecton of two sets A and B is a set which contains the
common elements of both the sets e.g. A = {1, 2, 3}, B = {2, 3, 4}.
The intersection of A and B is {2, 3}.
5
6
Commutative property of union and intersection of sets
For any two sets A and B, the union of A and B is equal to the
union of B and A e.g. A = {1, 2}, B = {1, 3}.
A union B = {1, 2, 3} and B union A = {1, 2, 3}. Both the sets are
equal.
For any two sets A and B, the intersection of A and B is equal to
the intersection of B and A e.g. A = {1, 2, 3}, B = {2, 3}.
A intersection B = {2, 3} and B intersection A = {2, 3}. Both the
sets are equal.
Difference of two sets
Difference of sets is like subtraction. The elements of set B are
subtracted from A and the elements left in A is the solution set
e.g. A = {1, 2, 3, 4} and B = {2, 3, 4, 5}.
The difference between A and B is A – B = {1}. The difference
between B and A is B – A = {5}.
Unit II: Operations on Directed Numbers Pages 6-12
The number line
When we count or measure, we use real numbers. These numbers
can be pictured as points on a line, called a ‘number line’.
To construct a number line
Choose a starting point on a line and label it ‘0’ (zero). This
point is called the ‘origin’. The origin separates the line into two
horizontal sides, the ‘positive side’ and the ‘negative side’. If
the line is horizontal, the side to the right of the origin is taken to
be the positive side and the side to the left as the negative side.
Mark off equal units of distance on both sides of the origin. On
the positive side, pair the end points with positive integers +1, +2,
+3, +4,… and so on.
7
8
On the negative side, pair the end points with negative integers
–1, –2, –3, –4,… and so on.
The positive integers, the negative integers and zero make up the
set of ‘integers’.
The positive integers and zero are often called ‘whole numbers’.
Any number that is either a positive number, a negative number
or zero is called a ‘real number ’.
When we see the pairing of points on a number line, the paired
points are at the same distance from the origin but on the opposite
sides of it.
Each number in a pair such as +7 and –7 is called the ‘opposite
of ’ the other number.
Order of integers
On a horizontal number line, the numbers increase from left to
right and decrease from right to left.
The ‘symbols of inequality’ are used to show the order of pairs
of real numbers.
< means ‘less than’
> means ‘greater than’
By studying a number line we can see that –5 is less than –2 and
0 is greater than –5.
–5 < –2 and 0 > –5
+5 > –5 and 3 > 0
Absolute value of numbers
In any pair of non-zero opposites, such as –5 and +5, one number
is negative and the other is positive. The positive number of any
pair of non-zero real numbers is called the ‘absolute value’ of each
number in the pair.
9
–5 < –2 0 > –5
+5 > –5 3 > 0
10
The absolute value of a number ‘a’ is denoted by |a| e.g. |–5| = 5
and |+5| = 5.
Notice that the absolute value of a real number ‘a’ is ‘a’ if ‘a’ is
non-negative and also if ‘a’ is negative.
The absolute value of 0 is defined as 0 itself.
|0| = 0.
Working with real numbers
The rules used in adding and multiplying real numbers are based
on several properties. These are:
1. Every pair of real numbers has one and only one sum that
is a real number.
2. Every pair of real numbers has one and only one product that
is a real number.
3. When you add two real numbers, you get the same sum no
matter what order you use in adding them.
4. When you multiply two real numbers, you get the same
product no matter what order you use in multiplying them.
Addition of integers
We already know how to add two positive numbers. We can use
a horizontal number line to help us find the the sum of any two
real numbers.
e.g. To find the sum of –2 and –5, draw a number line. Starting at
the origin move the pencil along the number line 2 units to the left.
Then from that position move the pencil 5 units to the left. Moves
to the left represent negative numbers. Together, the two moves
amount to a move of 7 units to the left from the origin.
Thus: –2 + (–5 ) = –7.
The expression –(2 + 5) represents the opposite of the sum of 2
and 5.
11
12
Since
2+5=7
–(2 + 5) = –7
The expression –2 + (–5) represents the sum of the opposite of
2 and the opposite of 5.
Using the number line we know that –2 + (–5) = –7.
It follows that –(2 + 5) = –2 + (–5).
Using the property of the opposite of a sum and the familiar
addition facts for positive numbers, we can compute sums of any
real numbers without thinking of a number line.
e.g. To simplify Solution
–8 + (–3)
–8 + (–3) = –(8 + 3)
= –11.
Subtraction of integers
Write the following examples of the subtraction of 2:
3–2=1
4–2=2
5 – 2 = 3
Now write examples of the addition of –2:
3 + (–2) = 1
4 + (–2) = 2
5 + (–2) = 3
Comparing the entries in the two columns shows that subtracting
2 gives the same result as adding the opposite of 2.
This suggests the following definition of subtraction for all real
numbers.
The difference a – b is equal to the sum of a + (–b).
That is to subtract b, add the opposite of b.
13
3 + (–2) = 1
4 + (–2) = 2
5 + (–2) = 3
3 + (–2) = 1
4 + (–2) = 2
5 + (–2) = 3
a
b
a
b
14
e.g. To simplify 12 – 3
Solution 12 + (–3)
12 – 3 = 9.
Multiplication of integers
When we multiply any given real number by 1, the product is
identical to the given number.
e.g. 3 x 1 = 3
1x3=3
When we multiply any given real number by 0, the product is 0:
e.g. 3 x 0 = 0
0x3=0
When we multiply any given real number by –1, the product is the
opposite of that number.
e.g. 3 x –1 = –3
Rules for multiplication of positive and negative numbers
1.
The product of a positive and a negative number is a negative
number.
2. The product of two positive or two negative numbers is a
positive number.
Distributive property of multiplication over addition
For all real numbers a, b and c:
ab + ac = a (b + c) and ba + ca = (b + c) a.
e.g. If 4, 5
then
15
and 7 are any three numbers,
4 x 5 + 4 x 7
or
= 4 (5 + 7)
= 20 + 28
= 48
5
=
=
=
x4+7x4
(5 + 7) 4
20 + 28
48
16
Verifying the distributive property of multiplication over
addition
e.g. To simplify
Solution
–5 x (8 + 6)
–5 x (8 + 6) = (–5 x 8) + (–5 x 6)
–5 x 14
= –40 + (–30)
–70
= –70
Division of integers
Before going on to division of integers, the terms ‘reciprocal ’ or
‘multiplicative inverse’ must be explained.
Two numbers whose product is 1 are called reciprocals or
multiplicative inverses of each other.
e.g. 3 and 1/3 are reciprocals because 3 x 1/3 = 1.
0 has no reciprocal because the product of 0 and any real number
is 0, not 1.
The symbol for the reciprocal or multiplicative inverse of a nonzero real number a is ‘1/a’.
The reciprocal of a product of non-zero real numbers is the
product of the reciprocals of the numbers.
That is, for all non-zero real numbers a and b:
1/ab = 1/a x 1/b
To divide a by b we multiply a by the reciprocal of b
The quotient is often represented as a fraction:
e.g. a divided by b = a/b
We can use the definition of division to replace any quotient by
a product.
17
b
a
b
a
18
e.g. 21/7 = 21 x 1/7 = 3
21/–7 = 21 (–1/7) = –3
–21/7 = –21 x 1/7 = –3
–21/–7 = (–21) (–1/7) = 3
Rules for dividing positive and negative integers
1. The quotient of two positive numbers or two negative
numbers is positive.
2. The quotient of a positive number and a negative number is
negative.
Dividing by zero
Dividing by zero would mean multiplying by the reciprocal of zero.
But zero has no reciprocal. Therefore, division by zero has no
meaning in the set of real numbers.
Unit III: Rational Numbers Pages 13-23
Positive integers are called ‘natural numbers’ or ‘counting
numbers’.
We have learnt earlier that the sum of two natural numbers
is always a natural number. Also, when a natural number is
subtracted from another natural number, the result is not always
a natural number. It could be a whole number or a negative
integer.
Now let us see what happens when a natural number is divided
by another natural number.
e.g. when 4 is divided by 2, the result is 2 but when 3 is divided
by 4 we get 3/4 which does not belong to any of the number
systems that we have already discussed. The number 3/4 is called
a ‘rational number ’.
2/3, –5/6, 0, 4, –2, etc. are all rational numbers.
So we can define a rational number as ‘any number that can be
expressed in the form of a/b, where a and b are integers and
b is not equal to 0 ’.
19
20
Positive integers, negative integers, zero and common fractions
all belong to the system of rational numbers.
Rational numbers can be expressed as a quotient of integers in
a number of ways.
e.g. 2 can be written as 2, 4/2, 8/4, –12/–6, etc.
To determine which of any two rational numbers is greater, we can
find a common denominator for the numbers and then compare
them.
The fraction with the greater numerator will be the greater number.
e.g. To find out which is greater, 9/2 or 13/5?
Solution: Finding the LCM of 2 and 5.
It is 10.
Changing the fractions with the new denominator
45/10 and 26/10.
Therefore,45/10 is the greater rational number.
Order of rational numbers
There is a difference between rational numbers and integers in
the sense that there is a higher and a lower integer for any given
integer. However, a rational number does not have a next higher
or previous lower number.
a + 1/2 (b – a) is the formula for calculating the number in the
middle of two said rational numbers such as a and b.
Follow the example given in book.
Representing rational numbers on the number line
We divide the number line into as many parts as indicated by the
denominator of the fraction.
Each part then represents one part of the fraction.
21
22
e.g. To represent 2/5 on the number line we must divide each
segment into 5 equal parts. Each part will then represent 1/5. 2/5
will be the second mark lying to the right of the zero.
The density property of rational numbers
Between every pair of different rational numbers there is another
rational number.
If a and b are rational numbers and a < b then:
the number halfway from a to b is c = 1/2 (b – a) and
the number one-third of the way from a to b is c = 1/3 (b – a).
Operations on rational numbers
Addition
Adding rational numbers with a common denominator is easy.
If a, b, and c are integers and c is not equal to 0 then
a/c + b/c = a + b .
c
We can extend this rule to rational numbers with unequal
denominators.
e.g. 3/4 + 2/5
The LCM is 20.
Renaming the fractions:
15/20 + 8/20
= 23/20.
Note that the sum of a rational number is also a rational
number.
Adding three or more rational numbers:
When we add three or more rational numbers, the order in which
we add does not affect the result.
23
a/c + b/c = a + b
c
24
e.g. Add 1/2, 2/3, 3/4.
(1/2 + 2/3) + 3/4
The LCM is 12.
Renaming the fractions:
(6/12 + 8/12) + 9/12
= (14/12) + 9/12
= 23/12.
By changing the order of the fractions:
1/2 + (2/3 + 3/4)
= 6/12 + (8/12 + 9/12)
= 6/12 + 17/12
= 23/12.
We see that the result is the same in both cases.
Additive inverse
For any rational number a/b, there is a number –a/b such that
a/b + (–a/b) = 0.
We say that –a/b is the additive inverse of a/b.
Similarly a/b is the additive inverse of –a/b.
For integers we have seen that 5 – 2 = 5 + (–2).
Here –2 is the additive inverse of 2.
Subtraction
Subtracting 2 from 5 is the same as adding the additive inverse
of 2 with 5. As each rational number has an additive inverse, the
concept of subtraction of integers can also be extended to rational
numbers.
The difference of rational numbers is also a rational number.
25
26
e.g. Subtract 1/2 from 3/4.
Solution: 3/4 – 1/2
The LCM is 4.
Renaming the fractions:
3/4 – 2/4
= 1/4.
To subtract three rational numbers:
e.g. 3/4, 1/3, 1/2
Solution: (3/4 – 1/3) – 1/2
The LCM is 12.
Renaming the fractions:
(9/12 – 4/12 ) – 6/12
= 5/12 – 6/12
= –1/12.
Changing the order of the fractions:
3/4 – (1/3 – 1/2)
Solution: 3/4 – (2/6 – 3/6)
3/4 – (–1/6)
3/4 + 1/6
LCM is 24
18/24 + 4/24
22/24 = 11/12.
Note that by changing the order of subtraction of the numbers we
do not get the same result.
Multiplication
The product of two rational numbers is also a rational number.
For any two rational numbers a/b and c/d:
a x c = a x c = ac
bxd
b d
bd
ac/bd is a rational number.
27
a x c = a x c = ac
bxd
b d
bd
28
e.g. 2/5 x 3/4
Solution: 2/5 x 3/4
= 3/10.
If we change the order of the rational numbers, the product is the
same.
e.g. 3/4 x 2/5
Solution: 3/4 x 2/5
= 3/10.
The same is the case when we multiply three or more rational
numbers.
e.g. Multiply –5/8 x 4/15 x –3/4.
We can multiply them in different orders
(–5/8 x 4/15) x –3/4
or –5/8 x (4/15 x –3/4)
Solution: –1/6 x –3/4
–5/8 x –1/5
= 1/8.
= 1/8.
We can see that the order of the rational numbers does not
affect the product.
Multiplicative identity
The product of 1 and any rational number is equal to the same
rational number.
1 is called the ‘multiplicative identity’ for rational numbers.
Division
The dividend is multiplied by the multiplicative inverse of the
divisor for dividing one rational number by another.
e.g. Divide 2/7
Solution:
by 4/3
2/7 ÷ 4/3
= 2/7 x 3/4
= 6/28.
Verifying commutative and associative properties of addition
of rational numbers
The commutative property states that the order of adding rational
numbers does not alter the result.
29
30
e.g. 2 + 1 = 1 + 2
3
4
4
3
8+3 = 3+8
12
12
11 = 11
12 12
Whereas, the associative property implies that the order of
grouping of rational numbers does not alter the result.
(
( (
(
e.g. 1 + 1 + 2 = 1 + 1 + 2
3
4
3
4
3
3
1+ 4+6 = 1+1 + 2
3
12
3
4
1 + 10 = 2 + 2
3
12
3
4
4 + 10 = 8 + 6
12
12
14 = 14
12
12
Verifying commutative and associative properties of
multiplication of rational numbers
The commutative property of multiplication states that if two
rational numbers are multiplied, irrespective of the order in which
they are multiplied; the products will be the same.
e.g. 1/2 x 1/4 = 1/4 x 1/2
Solution: 1/8 = 1/8
The associative property says that when multiplying, the order of
rational numbers does not affect the product.
( (
e.g. 1 x
3
1
2 x
9
31
2 x 5 = 1 x
6 3
3
5 = 1 x 5
63 3 9
5
5
=
27
27
( (
2+5
3 6
2 + 1 = 1 + 2
3
4
4
3
8+3 = 3+8
12
12
11 = 11
12 12
1+
3
(
( (
(
1 + 2 = 1 + 1 + 2
4
3
4
3
3
1+ 4+6 = 1+1 + 2
3
12
3
4
1 + 10 = 2 + 2
3
12
3
4
4 + 10 = 8 + 6
12
12
14 = 14
12
12
( (
1x 2 x 5 = 1 x
6 3
3 3
1
2 x 5 = 1 x 5
9 63 3 9
5
5
=
27
27
( (
2+5
3 6
32
Verifying distributive property of multiplication over addition
and subtraction
Distributive property of multiplication over addition:
For any three rational numbers we can see that multiplication is
distributive over addition.
e.g. Simplify the expression 1/2 x (2/3 + 2/5).
Solution: (1/2 x 2/3) + (1/2 x 2/5)
= 1/3 + 1/5
= 8/15.
Distributive property of multiplication over subtraction:
For any three rational numbers we can see that multiplication is
distributive over subtraction.
e.g. Simplify the expression1/2 x (1/4 – 1/3).
Solution: (1/2 x 1/4) – (1/2 x 1/3)
= 1/8 – 1/6
= –1/24.
Properties of rational numbers
The explanation and examples pertaining to rational numbers
discussed above can be summarised as the properties of rational
numbers. These are:
1.
2.
Commutative property of addition of rational numbers.
The order of adding rational numbers does not affect the result.
Associative property of addition of rational numbers.
The order of grouping of rational numbers does not affect the
result.
3. Commutative property of multiplication of rational numbers.
The order of multiplication of rational numbers does not affect
the product.
4. Associative property of multiplication of rational numbers.
The grouping of rational numbers does not affect the product.
33
34
5
Distributive property of multiplication over addition and
subtraction.
Multiplication is distributed over addition and subtraction.
Non‑recurring or terminating decimals
To write a rational number as a decimal we can divide the
numerator by the denominator.
e.g. Express the rational number 1/2 as a decimal.
Solution: To express the rational number as a decimal we
can express the denominator as a power of 10.
Such as:
or
1x5 = 5 =
0.5
1/2 =
2x5
10
We can divide 1 by 2.
Since 1 is less than 2, we cannot perform division so we take 1
as 10 tenths.
1/2 = 10 tenths = 5 tenths = 0.5
2
We can see that the process of division has ended and there is
no remainder.
Such a decimal is called a ‘terminating decimal ’.
Recurring decimal or non-terminating decimal
Now express the following rational number by division:
1/3
Solution: 1 divided by 3.
1/3 = 0.333…
We can see that the process of division is never ending. The
remainder is not a zero.
35
1x5 = 5 =
0.5
1/2 =
2x5
10
10
2
36
Such a decimal is called a ‘non-terminating decimal ’.
All terminating and non-terminating decimals represent rational
numbers that can be written in the form n/d where n is an integer
and d is a positive integer.
Unit IV: Square Roots Pages 24-28
We have learnt that subtracting a number is the inverse of adding
that number, and that dividing by a non-zero number is the inverse
of multiplying by that number. The inverse of squaring a number
is finding the ‘square root’ of that number.
Explain the symbol used to denote the square root of a positive
number, which is called the ‘radical sign’. Often it is convenient
to use the + or – notations with radicals.
An expression written beneath the radical sign is called the
‘radicand ’.
It is interesting to note that zero has only one square root and
that is zero itself.
The values of certain square roots can be seen at a glance; e.g. the
square root of 49 is 7. We may be able to find other square roots
by expressing them as a product of square roots familiar to us.
e.g. To find the square root of 144.
Solution: Factorizing 144
= 9 x 16
The square root of 9 is 3 and that of 16 is 4.
Therefore, the square root of 144 is 3 x 4 = 12.
To find square roots by simple factors
We keep dividing the number by prime numbers till we get zero
as the remainder.
Then by pairing off the factors and selecting one factor from each
group we get a set of factors that when multiplied together give
us the required square root.
37
38
e.g. Find the square root of 324.
Solution:
2324
2162
381
327
39
33
1
Grouping two equal factors:
324 = 2 x 2 x 3 x 3 x 3 x 3
Selecting one factor from each pair we get:
2 x 3 x 3 = 18.
The square root of 324 is 18.
Square root of a fraction
We find the square roots of the numerator and the denominator
separately, then write them as a fraction.
e.g. Find the square root of 25/36.
Solution:
√25/36
The square root of 25 is 5 and that of 36 is 6.
Writing the square roots as a fraction, we get
5/6.
The square root of 100 and its powers
Give the pupils a tip to find out the square root of 100 and its
powers. Give them the following example:
e.g.10,0002 = 100,000,000
Then tell them to convert two zeros into one:
1 00 00 00 00
39
2324
2162
381
327
39
33
1
324 = 2 x 2 x 3 x 3 x 3 x 3
√25/36
10,0002 = 100,000,000
1 00 00 00 00
40
Now, each pair should be reduced to one zero to find out its
square root.
Therefore, the square root of 100,000,000 is 10,000.
The square root of a decimal number
The square root of 100 and its powers can be shown by finding
their prime factors.
e.g. Find the square root of 100.
Solution: The prime factors of 100 are 10 x 10.
So, the square root of 100 is 10.
We can use two methods for finding the square root of a decimal
number.
First, we convert the decimal into a fraction and then we find the
square root.
e.g. Find the square root of 0.16.
By changing it into a decimal fraction we get:
16/100.
Finding the square root of the numerator and the denominator
we get:
4/10.
Changing it back into a decimal number we get:
4/10 = 0.4.
A decimal fraction is a perfect square if it can be converted
into a perfect square fraction.
Word Problems
Read the word problems carefully and discuss them thoroughly
before attempting to solve them.
41
10,000
100,000,000
42
Unit V: Percentage Pages 30-37
Discuss the term ‘percentage’ as applied to the marks that pupils
get in their tests and exams.
When a pupil gets 6/10 marks in a test he has actually got 60
marks out of 100 and that means 60%.
The ratio of a number can be expressed as a per cent. The
word ‘per cent ’ (usually denoted by %) means ‘hundredths’ or
‘divided by 100 ’.
e.g. 29/100 is called 29 per cent and is written as 29%.
To express per cent as a common fraction
A percentage can be changed into a common fraction or a decimal
fraction.
e.g. To express 20% as a common fraction:
20% = 20/100 = 1/5.
To convert a per cent to a decimal
When finding a per cent of a number, it is often convenient to
express the per cent as a decimal.
e.g. 15% = 15/100 = 0.15
There are two types of fractions
Fractions whose denominators are divisors of 100 are multiplied
by 100 and then reduced to their lowest terms.
e.g. 1/10 = 1/10 x 100 = 10%.
Fractions whose denominators are not divisors of 100 are also
multiplied by 100 and then reduced to their lowest terms.
e.g. 1/15 = 1/15 x 100 = 20/3 = 6 2 = 6.66%
3
43
2
3
44
To express a decimal fraction as a percentage
Change the decimal fraction into a common fraction and then
multiply by 100.
e.g.0.19
Changing it into a common fraction:
0.19 = 19/100
Multiplying by 100:
19/100 x 100
= 19%
Practical applications of percentage
Discuss the word problems and explain what the statements imply
before attempting to solve them.
Profit and loss
Discuss the meaning of ‘profit’ and ‘loss’ in terms of business
transactions.
When something is bought for a lower price and sold at a higher
price, it is called ‘profit’.
When something is bought for a higher price and sold for a lower
price, it is called ‘loss’.
Profit or loss percentage
Profit or loss can be calculated in terms of percentage:
Loss % = loss/cost price x 100
Profit % = profit/cost price x 100
Selling price when cost price and profit or loss per cent are
given
From the profit or loss per cent we can calculate the selling price.
e.g.Ahmed bought a radio for Rs 640. At what price must he sell
it to gain 15%?
45
0.19
0.19 = 19/100
100
19/100 x 100
= 19%
46
A gain of 15% means that when the cost price is Rs 100 the
selling price should be Rs 100+15.
When the cost price is Rs 640, the selling price should be:
115/100 x 640 = Rs 736.
Cost price, when selling price and profit or loss per cent are
given
From the profit or loss per cent we can calculate the cost price.
e.g. By selling a calculator for Rs 475, a shopkeeper lost 5%.
What is the cost price of the calculator?
A loss of 5% means that when the cost price is Rs100 the selling
price is Rs 100 – 5 = Rs 95.
When the selling price is Rs 475, the cost price would be:
100/95 x 475 = Rs 500.
Unit VI: Ratio and Proportion Pages 38-45
This is a continuation of the unit as given in Book 6.
A ratio is a relation that one quantity bears to another quantity
of the same kind with regard to their magnitudes.
The comparison is made by considering what multiple, part or
parts the first quantity is of the second.
If we say that the ratio of two quantities is ‘5 is to 6’ we mean
that the magnitude of the two quantities has been compared,
that is if one quantity has the magnitude 5, then the other has a
magnitude of 6.
A ratio is an abstract number given by a fraction. The numerator
denotes the magnitude of the first quantity and the denominator
gives the magnitude of the second quantity.
If the quantities to be compared are in different units then it is
essential to express them in the same unit to find their ratio.
47
115/100 x 640 = Rs. 736
100/95 x 475 Rs. 500
48
e.g. If two sticks of length 2m and 40 cm are to be compared; then
the lengths expressed to the same units are 200 cm and 40 cm
respectively and the ratio of their lengths is 200 : 40 i.e. 5 : 1.
The colon inserted between the two quantities denotes the ratio
and is read as ‘5 is to 1’.
A ratio remains unaltered when both the terms are multiplied or
divided by the same number.
A ratio should be expressed in its simplest form.
To find the ratio of two quantities of the same kind:
Express the measures in the same unit and then divide them.
The ratio 9/6 can be expressed in its simplest form as 3/2.
Algebraic method for direct proportion
An equation which states that two ratios are equal is called
a proportion.
We can write proportions in several different ways.
e.g. 2 : 3 = 4 : 6
We read it as: ‘2 is to 3 as 4 is to 6’.
x/12 = 9/4
We read it as ‘x divided by 12 is equal to 9 divided by 4’.
In the proportion 2 : 3 = 4 : 6
2 and 6 are called the ‘extremes’ and 3 and 4 are called the
‘means’.
We can use the multiplication property of equalities to show that
in any proportion:
49
50
The product of the extremes is equal to the product of the
means.
e.g. If 15 pens cost Rs 45, find the cost of 25 pens?
Solution:Let x be the cost of 25 pens.
Hence the proportion:
15 : 45 : : 25 : x
15/45 = 25/x
Product of the extremes: 15 multiplied by x.
Product of the means: 45 x 25.
Product of the extremes = product of the means
15x = 45 x 25
x = 45 x 25
15
x = Rs 75.
Proportional division
Proportional division means dividing a given quantity in a
specified ratio.
The number in the ratio is considered to be the total number of
units that the given quantity is to be divided into. From this we
find the quantity per unit.
e.g. Divide Rs
Solution:
5,000 in the ratio 2:3:5.
The sum of the ratios is 2 + 3 + 5 = 10.
Quantity per unit is Rs 5,000 divided by 10.
One unit is equal to Rs 500.
According to the given ratio:
2 units will be equal to 500 x 2 = Rs 1,000;
3 units will be equal to 500 x 3 = Rs 1,500; and
5 units will be equal to 500 x 5 = Rs 2,500.
This is the required proportional division.
Quick reckoning
This is a simpler method of calculating proportions. Teachers
should explain the formula to the pupils so that they can retain it
better.
51
45 x 25
15
52
e.g. A’s share = the ratio of A x total quantity.
total ratio
Continued ratio
Three quantities of the same kind are said to be in a
continued proportion when the ratio of the first to the second
is equal to the ratio of the second to the third.
e.g. In the proportion 1 : 3 = 3 : 9;
3 constitutes the second as well as the third terms of the
proportion, i.e. the means are the same. In this case, 3 is called
the ‘mean proportional’ between 1 and 9. Also, the three numbers
1, 3 and 9 are said to be in continued proportion. The third
quantity is called the ‘third proportional’.
To convert ratios into continued ratios:
Consider the proportion 2 : 3 = 5 : 7
This can be written as:
A:B=2:3
B:C=5:7
To change the ratios into continued ratios; we multiply the first
ratio by 5 and the second by 3 to make the second proportional
B the same.
2 : 3 = 10 : 15 (multiplying by 5)
5 : 7 = 15 : 21 (multiplying by 3)
Now B is the same in both the ratios.
So the continued proportion will be A : B : C.
10 : 15 : 21
We can divide a given quantity into the given ratio by
changing the ratios into continued ratio and then dividing.
53
A:B=2:3
B:C=5:7
54
e.g. Divide Rs 7,000 amongst A, B, C in the following ratios:
A:B=2:3
B:C=4:5
Solution: For changing the ratios into continued ratio, multiply
the first ratio by 4 and the second ratio by 3.
We get 8 : 12, and 12 : 15.
So the continued ratio is:
A: B: C
8 : 12 : 15
The sum of the ratios is 8 + 12 + 15 = 35.
Hence, the shares are:
A’s share is 8/35 x 7,000 = Rs 1,600,
B’s share is 12/35 x 7,000 = Rs 2,400 and
C’s share is 15/35 x 7,000 = Rs 3,000.
Partnership
In a business partnership, the profit or loss depends upon the
amount of money invested (capital) and the period for which each
partner has invested his share.
If the period of investment is the same for all the partners, the
profit is divided amongst them in the same way as for proportional
division.
But if the period of investment is different for all the partners, we
have to calculate the ratio of the partners in continued proportion.
e.g. The investments of A, B and C in a business are Rs 2,500 for
6 months, Rs 4,000 for 3 months and Rs 5,000 for 4 months.
Calculate their shares if the profit is Rs 4,700.
Solution: The ratio of their investments is:
55
A’s investment is Rs 2,500 x 6 = 15,000,
B’s investment is Rs 4,000 x 3 = 12,000 and
C’s investment is Rs 5,000 x 4 = 20,000.
A:B=2:3
B:C=4:5
A: B: C
8 : 12 : 15
56
Ratio of investments is:
A: B: C
15,000: 12,000: 20,000
Reducing the ratio to its simplest form:
15: 12: 20
The sum of ratios is 15 + 12 + 20 = 47.
The profit is Rs 4,700.
A’s share is 15/47 x 4,700 = Rs 1,500,
B’s share is 12/47 x 4,700 = Rs 1,200 and
C’s share is 20/47 x 4,700 = Rs 2,000.
Unit VII: Algebra Pages 46-63
Algebraic expressions
A variable is a symbol that is used to represent one or more
numbers.
e.g. a, b, c . . . . . . are ‘variables’.
To solve problems using algebra, we must often translate word
phrases of numbers into numerical or variable expressions.
An algebraic expression is a combination of numbers and
variables connected by one or more symbols such as ‘+’ or ‘–’.
Coefficient, base, exponents
In the expression: 2 a3, 2 is called the ‘coefficient’, ‘a’ is the base
and 3 is the exponent.
Polynomial expressions
An algebraic expression having one or more variables whose
exponents are positive integers, is called a polynomial expression.
e.g. 8x, 8x + 9, 8x2 + 2x + 1
A ‘monomial’ is considered a polynomial of one term, which is an
expression that is either a numeral,
57
58
a variable, or a product of a numeral and one or more variables. A
numeral such as 7, is called a ‘constant monomial’ or a ‘constant’.
e.g. 7, a, 3c, 8x2y
Binomial expressions have two terms:
e.g. 4x + 9, 6b2 – 7a
Trinomial expressions have three terms:
e.g. x2 – 3x – 2, 5b2 + 3ab – a2
Degree of polynomials
The ‘degree of a monomial in a variable’ is the number of times
that variable occurs as a factor in the monomial.
e.g.3xy2z3 is of degree 1 in x, 2 in y and 3 in z.
The degree of any non-zero constant monomial is ‘0’ (It has no
degree).
The ‘degree of a polynomial’ is the greatest of the degrees of its
terms.
e.g. 4x3 + 5x2 + 7.
The greatest degree of the variable x is 3. So the degree of the
polynomial is 3.
Ascending and descending order
It is often helpful to rearrange the terms of a polynomial so that
their degrees in a particular variable are in either increasing or
decreasing order.
e.g. Consider the expression:
59
– 4x3 + 3x2 + 3x5
60
The lowest power of x is 2, and the highest power of x is 5.
To arrange the terms in ascending order we start from the lowest
power i.e. 3x2 – 4x3 + 3x5.
To arrange the terms in descending order we start from the
highest power i.e. 3x5 – 4x3 + 3x2.
Exponential expressions
The number 9 can be written as:
3 x 3 or 32 and is called a ‘power’ of 3 (It is read as: ‘3 to the
second power’ or ‘three squared’).
The numbers 27 and 81 are also powers of 3.
27 can be written as 3 x 3 x 3 or 33 (It is read as: ‘three to the
third power’ or ‘three-cubed’).
In the expression a3, a is called the ‘base’ and the small raised
symbol 3 is called the ‘exponent’.
The exponent indicates the number of times the base occurs as
a factor.
Addition of algebraic expressions
The terms that have the same variables and exponents are called
‘like terms’. The coefficients of ‘like terms’ may be different.
e.g. a, 2a, 3a
The terms that have different variables and exponents are called
‘unlike terms’, even if their coefficients are the same.
e.g. 2a, 2a2, 2a3
Note: Only ‘like terms’ can be added or subtracted.
To add two polynomials, we write the sum and simplify by adding
‘like terms’.
61
62
e.g. 2x + 5x = 7x.
e.g. 2a + 3b and 2b + a.
Write in a vertical form:
2a + 3b
a + 2b
3a + 5b
Addition of positive and negative terms
e.g.3a – 6a + 9a – 4a
First, group the terms with similar signs:
3a + 9a – 6a – 4a
Then, add ‘like terms’:
12a – 10a = 2a.
Addition of mixed expressions
Mixed expressions can be added by associating the ‘like terms’
horizontally or vertically.
e.g. Add: 5x2y + 3x2 – 8 + 4x2y + 2x2 + 9.
Solution:(i) Associating the terms horizontally:
5x2y + 4x2y + 3x2 + 2x2 – 8 + 9
9x2y + 5x2 + 1.
(ii) Associating the terms vertically:
5x2y + 3x2 – 8
4x2y + 2x2 + 9
9x2y + 5x2 + 1
Subtraction of algebraic expressions
Subtracting polynomials is very much like subtracting real
numbers. To subtract a number, you add the opposite of that
number.
63
2x + 5x = 7x.
2a + 3b and 2b + a.
2a + 3b
a + 2b
3a + 5b
3a – 6a + 9a – 4a
3a + 9a – 6a – 4a
12a – 10a = 2a
5x2y + 3x2 – 8 + 4x2y + 2x2 + 9
5x2y + 4x2y + 3x2 + 2x2 – 8 + 9
9x2y + 5x2 + 1
5x2y + 3x2 – 8
4x2y + 2x2 + 9
9x2y + 5x2 + 1
64
To subtract a polynomial, you add the opposite of ‘each’ term of
the polynomial that you are subtracting and then simplify.
e.g. Subtract: –5a2 + 2ab + 3b2 – 4 from 7a2 + 6ab – b2 – 9.
Solution:
(i) Associating the terms horizontally:
(7a2 + 6ab – b2 – 9) – (–5a2 + 2ab + 3b2 – 4)
= 7a2 + 6ab – b2 – 9 + 5a2 – 2ab – 3b2 + 4
= (7 + 5) a2 + (6 – 2) ab + (–1 – 3) b2 + (–9 + 4)
= 12a2 + 4ab – 4b2 – 5.
(ii) Associating the terms vertically:
7a2 + 6ab – b2 – 9
– 5a2 + 2ab + 3b2 – 4
+
–
–
+
(changing to the opposite)
12a2 + 4ab – 4b2 – 5(adding)
Multiplication of polynomials
When we multiply two monomials, we use the rule of exponents
along with the commutative and associative properties for
multiplication.
e.g. (i) Multiply 2x2 by 3.
Solution: 2x2 x 3 = 6x2.
(ii) Multiply –5x2 by –3.
Solution: –5x2 x –3 = + 15x2.
Multiplying polynomials
When we multiply two powers having the same base, we add the
exponents.
e.g.(i) x2 x x5 = x2+5 = x7.
e.g.(ii) 3x3y4 x –7xy5
= (3 x –7) (x3 x x) (y4 x y5)
= (–21) (x3+1) (y4+5)
= –21x4y 9.
65
–5a2 + 2ab + 3b2 – 4 7a2 + 6ab – b2 – 9
(7a2 + 6ab – b2 – 9) – (–5a2 + 2ab + 3b2 – 4)
= 7a2 + 6ab – b2 – 9 + 5a2 – 2ab – 3b2 + 4
= (7 + 5) a2 + (6 – 2) ab + (–1 – 3) b2 + (–9 + 4)
= 12a2 + 4ab – 4b2 – 5
7a2 + 6ab – b2 – 9
– 5a2 + 2ab + 3b2 – 4
+
–
–
+
12a2 + 4ab – 4b2 – 5
66
Multiplying a polynomial by a monomial
We use the distributive property and the rules of exponents to
multiply.
e.g. Multiply 3a2 – 4a + 3 by 4a.
We can multiply horizontally or vertically.
(i)4a (3a2 – 4a + 3)
= 12a3 – 16a2 + 12a
(ii)3a2 – 4a + 3
4a
12a3 – 16a2 + 12a
(multiplying horizontally)
(multiplying vertically)
Multiplying two polynomials
We can use the distributive property of multiplication to multiply
two polynomials.
e.g. Multiply 4x + 3 by 3x + 4.
(i) Multiplying horizontally:
(4x + 3) (3x + 4)
= 4x (3x + 4) + 3 (3x + 4)
= 12x2 + 16x + 9x + 12
= 12x2 + 25x + 12.
(ii) Multiplying vertically:
4x + 3
3x + 4
12x2 + 9x
+ 16x + 12
2
12x + 25x + 12
Division of polynomials
When we divide two monomials, we use the laws of exponents.
When the base of the divisor and the dividend is the same then
the power of the quotient is the difference of their exponents.
67
4a
3a2 – 4a + 3
4a (3a2 – 4a + 3)
= 12a3 – 16a2 + 12a
3a2 – 4a + 3
4a
12a3 – 16a2 + 12a
(4x + 3) (3x + 4)
= 4x (3x + 4) + 3 (3x + 4)
= 12x2 + 16x + 9x + 12
= 12x2 + 25x + 12.
4x
3x
12x2
12x2
+3
+4
+9x
+16x + 12
+25x + 12
68
e.g. Divide 5x2 by x.
Solution:5x2/x= 5x2-1 = 5x.
Note: any number raised to the power ‘0’ is 1.
e.g.5x/x
= 5x1-1 = 5x0 = 5 x 1 = 5.
e.g.15y3/3y
= 15/3 x y3-1 = 5y2.
Dividing a polynomial by a monomial
2a2x – 10a3x2 + 12a5 x3 ÷ 2ax
Divide each term of the polynomial by the monomial:
2
3 2
5 3
= 2a x – 10a x + 12a x
2ax
2ax
2ax
Add the quotients:
= a – 5a2x + 6a4x2.
Substitution
The meaning of substitution is to ‘replace’ or ‘put in place of’.
Changing the numeral by which a number is named in our
expression does not change the value of the expression.
Replacing each variable in an expression by a given value and
simplifying the result is called ‘evaluating the expression’ or
'finding the value of the expression.’
e.g. Evaluate the expression:
a + b, when a = 5 and b = –2.
Solution:
a+b
= 5 + (–2) (putting the values of a & b)
=5–2
= 3.
69
5x/x
= 5x1-1 = 5x0 = 5 x 1 = 5.
15y3/3y
= 15/3 x y3-1 = 5y2.
2a2x – 10a3x2 + 12a5 x3 ÷ 2ax
2
3 2
5 3
= 2a x – 10a x + 12a x
2ax
2ax
2ax
= a – 5a2x + 6a4x2.
a
b
a b
a b
a b
70
Algebraic sentences
An algebraic sentence is formed by placing =, >, <, ≠ signs
between two numerical or variable expressions.
Sentences containing variables such as 5x – 1 = 9, and 3 + c =
c + 3, are called ‘open sentences’.
When we replace each variable in an open sentence by a given
number, we obtain a ‘true’ sentence or a ‘false’ sentence.
e.g. 5 + 2 = 7 is a true sentence.
5 > 10 is a false sentence.
3+
=10 is an open sentence.
The last sentence is an open sentence because it is impossible
to judge whether it is a true or a false sentence, unless the blank
space is replaced by a number. If we replace the blank space by
7, the sentence becomes a true sentence.
If we replace the blank space by 6, the sentence becomes a false
sentence.
Equations
A polynomial equation has polynomials on both sides and where
both the sides have an equality sign in the middle.
A simple linear equation
An equation is formed by placing an ‘is equal to’ sign between two
numerical variable expressions called the ‘sides’ of the equation.
In an equation, the sign of equality between the sides shows that
the two sides are equal.
e.g. x – 5 = 7.
An equation in which the exponent of the variable is 1 is called a
linear equation.
71
3 + c = c + 3 5x–1=9
x–5=7
72
Equivalent equations
Equations having the same solution are called equivalent
equations.
Transforming equations
To solve an equation we usually try to change or ‘transform’ it into
a simple equivalent equation, whose solution can be seen at a
glance. This transformation into a simple equivalent equation can
be done by substitution, addition, or subtraction.
Addition and subtraction properties of an equality
1.
If the same number is added to equal numbers, the sums are
equal.
2. If the same number is subtracted from equal numbers, the
differences are equal.
We can use these properties to solve some equations.
e.g.Solve x – 5 = 7.
x – 5 + 5 = 7 + 5 (5 is added to both sides)
x = 12.
e.g.Solve x + 10 = 20.
x + 10 – 10 = 20 – 10 (10 is subtracted from both sides)
x = 10.
Because errors may occur in transforming equations, we should
check our work by substituting the value of the variable found, so as
to show that the transformed equation satisfies the original equation.
e.g. x + 8 = 3.
x+8–8=3–8
x = –5.
73
x–5=7
x–5+5=7+5
x = 12
x + 10 = 20
10 x + 10 – 10 = 20 – 10
x = 10
x + 8 = 3.
x+8–8=3–8
x = –5.
74
Substituting the value of x = –5 in the given equation:
x + 8 = 3.
–5 + 8 = 3
3 = 3.
Multiplication and division properties of an equality
1. If equal numbers are multiplied by the same number, the
products are equal.
2. If equal numbers are divided by the same non-zero number,
the quotients are equal.
Transformation by multiplication
Multiply each side of a given equation by the same non-zero real
number.
e.g. x/2 = 14.
Multiplying both sides by 2:
(x/2)(2) = (14)(2)
x = 28.
Transformation by division
Divide each side of a given equation by the same non-zero real
number.
e.g.2x = 10.
Dividing both sides by 2:
2x/2 = 10/2
x = 5.
Using several transformations to solve an equation
We know that subtraction is the inverse of addition and that
division is the inverse of multiplication. In transforming equations,
we often use inverse operations.
75
x + 8 = 3.
–5 + 8 = 3
3 = 3.
x/2 = 14.
(x/2)(2) = (14)(2)
x = 28.
2x = 10.
2x/2 = 10/2
x = 5.
76
e.g. Solve 4y + 43 = 19.
Solution:4y + 43 = 19
Subtracting 43 from each side:
4y + 43 – 43 = 19 – 43
4y = – 24
Dividing each side by 4:
4y/4 = –24/4
y = –6.
To check:
Substituting the value of y = –6 in the given equation:
4y + 43 = 19
(4) (–6) + 43 = 19
–24 + 43 = 19
19 = 19
The following steps are usually helpful when we are solving an
equation in which all the variables are on the same side.
1. Simplify each side of the equation.
2. If there are indicated additions or subtractions, use the
inverse operations to undo them.
3. If there are indicated multiplications or divisions, use the
inverse operations to undo them.
Solving word problems
Follow the steps given below for solving word problems involving
linear equations:
1. Read the problem carefully a few times. Decide what
numbers are asked for and what information is given. Making
a sketch may be helpful.
2. Choose a variable and use it with the given facts to represent
the number(s) described in the problem.
3. Reread the problems. Then write an open sentence that
represents the relationship amongst the numbers in the problem.
4. Solve the open sentence and find the required numbers.
5. Check your results with the words of the problem. Give the
answer.
77
4y + 43 =
(4) (–6) + 43 =
–24 + 43 =
19 =
19
19
19
19
78
UNIT VIII: Geometry Pages 64-78
Parallel lines
Discuss parallel lines with examples of the railway line, the
horizontal and vertical sides of rectangular doors and windows.
Draw a square and rectangle on the board and explain the
definition of parallel lines.
Complementary and supplementary angles
Lay a large book flat on the desk and gradually lift its cover.
Explain the formation of angles by it. Discuss a right angle and
also complementary angles. Open the book and lay it flat on
the desk. Lift one page and hold it near the center. Discuss that
supplementary angles are equal to two right angles.
Angles of a triangle
Draw a triangle on the board and measure its angles. Explain that
the sum of the angles of a triangle is equal to 180°.
Construction of triangles
Following the steps of construction for the different types of
triangles, draw the triangles on the board and give the pupils
practice in constructing triangles.
Quadrilaterals
Discuss the various types of quadrilaterals. Following the steps of
construction for the different types of quadrilaterals, draw them on
the board. Give the pupils practice in constructing quadrilaterals.
Area
Discuss the meaning of area.
Use a squared paper to explain it. Help the pupils to find the area
of different geometrical shapes by using formulae.
79
80
Volume
Discuss the meaning of volume. Use solid figures to explain it.
Help the pupils find the volume of solid figures by using formulae.
Unit IX: Information Handling Pages 79-82
Discuss the meaning of statistics. Explain that we get numerical
information from various sources.
Numerical information is called ‘data’.
Ask the pupils to collect data regarding some information about
their class, their school, etc.
Discuss the different kinds of data.
Write ungrouped data on the board and explain the method of
arranging it into grouped data.
Classification
It is the method of arranging data into certain groups or classes
without losing valuable information.
Class interval
Data is classified on the basis of values or quantities and is
divided into a number of classes, each of which is called a ‘class
interval’.
Classification according to this method involves:
(i) number of classes and their sizes,
(ii) choice of class limits and
(iii) counting the numbers in each class.
Class limits
The maximum and minimum values within which a class interval
lies are known as the ‘class limits’.
81
82
Class size
The difference between two class limits is known as the ‘class
size’.
The difference between the maximum and the minimum number
of items of a class is known as the ‘range’ of the data.
The class size of the class intervals depends on the range of the
data and the number of classes.
e.g.
Marks obtained by 20 students in science:
42, 38, 37, 75, 45,
49, 52, 78, 84, 34,
41, 65, 73, 32 , 87,
34, 37, 56, 59, 89
The lowest score is 32 and the highest is 89.
So the range is 89 – 32 = 57.
Suppose we want to make 6 classes, the number nearest to 57,
which is divisible by 6 is 60.
Hence, the class size will be:
60/6 = 10.
Thus, the classes may be taken as:
30 – 39, 40 – 49, 50 – 59,
60 – 69, 70 – 79, 80 – 89.
Pie chart
Diagrams and graphs help in giving us an overall view of the data
under consideration. They present the facts in the form of pictures
by which data or information can easily be compared.
83
42, 38, 37, 75, 45,
49, 52, 78, 84, 34,
41, 65, 73, 32 , 87,
34, 37, 56, 59, 89
89 – 32 = 57
60/6 = 10
30 – 39, 40 – 49, 50 – 59,
60 – 69, 70 – 79, 80 – 89.
84
One kind of chart called the ‘pie chart’ is constructed by dividing
a circle into different sectors. Each sector corresponds to the
percentage of a category of data under consideration. The angle
of each sector is proportional to the number the sector represents.
To obtain the sectors in a pie chart we need to find the angles
at the center of the circle. Since the total area of the circle
corresponds to the total number of degrees in the circle, i.e. 360,
we can find the angle of the circle by dividing each item by the
total number of items and multiplying by 360°.
To draw a pie chart
First find out the angles of each sector. Then draw the radius
of the circle and with the help of a protractor, draw the required
angle. Using the arm of the angle drawn, draw the next angle
corresponding to the next sector and so on. Fill till all the sectors
have been represented.
Write the item in the sector it represents. Colour each sector in
a different shade.
e.g. Draw a pie chart to represent the monthly expenditure of a
family.
House rent
= 30%
Food
=50%
Education =10%
Health
= 2%
Savings
= 8%
85
30%=
50%=
10%=
2%=
8%=
86
(i) Adding all the items:
=100%
(ii) Finding the angles of each sector
(a) House rent
=
30/100 x 360° = 108°
(b) Food
=
50/100 x 360° = 180°
(c) Education
=
10/100 x 360° = 36°
(d) Health
=
2/100 x 360° = 7.2°
(e) Savings
=
8/100 x 360° = 28.8°
(a)
108°
draw the circle
draw the radius
(i)
(ii)
(i)
(b)
180°
(i)
(a)
108°
87
(a)
108°
(ii)
(b)
180°
(iii)
draw the
third sector
(v)
(a)
108°
(e)
28.8°
➝
(iv)
➝
➝
draw the
second sector
(iv)
(c)
36°
(iii)
(i)
➝
(ii)
(b)
180°
draw the first sector
(c)
36°
(ii)
(iii)
(d)
7.2°
then, draw the
last two sectors
(vi)
30/100 x 360° =
108° =
50/100 x 360° =
180° =
10/100 x 360° =
36° =
2/100 x 360°
=
7.2° =
8/100 x 360°
= 28.8° =
(a)
108°
(i)
(b)
180°
(i)
(a)
108°
(a)
108°
(c)
36°
(ii)
(b)
180°
(iii)
(a)
108°
(e)
28.8°
➝
(iv)
➝
➝
➝
(ii)
(b)
180°
(i)
(c)
36°
(ii)
(iii)
(d)
7.2°
88
NOTE
89
NOTE
90
NOTE
91