Download Grade 5 Math - Ritu Chopra

Synopsis – Grade 5 Math
Chapter 1: Number System
 Reading and writing numbers according to Indian number system
The place value chart in the Indian Number System can be represented as follows:
Periods
Place 
Lakhs
Ten
lakhs
TL
Lakhs
L
Thousands
Ten
Thousands Hundreds
thousands
T Th
Th
H
ones
Tens
Ones
T
O
For example: 1732256 can be placed in the place value chart as follows:
TL
1



L
7
T Th
3
Th
2
H
2
T
5
O
6
All the digits in the same period are read together and the name of the period (except
ones) is read together.
The periods in numerals are separated by commas.
Example: 1732256 can be read and written in the Indian Number System as follows:
Seventeen lakh thirty-two thousand two hundred fifty-six and 17, 32,256 respectively
The expanded form of the number is obtained by multiplying the digits in the number
with the place value and separating them with ‘+’ sign.
Expanded form of 17, 32,256 = 10, 00,000 + 7, 00,000 + 30,000 + 2,000 + 200 + 50 +
6
 Reading and writing numbers according to International system
The place value chart in the International System can be written as follows:
Th
Ones
T Th
Tens
H Th
ones
Hundreds
M
Ten
Thousands
TM
Hundred
thousands
Ten
millions
HM
Thousands
Millions
Hundred
millions
Place 
Millions
Thousands
Periods 
H
T
O
In the International System, millions comes after thousands.
For example: 1732256 can be placed in the place value chart as follows:
HM
TM
M
1
H Th
7
T Th
3
Th
2
H
2
T
5
O
6



All the digits in the same period are read together and the name of the period (except
the ones) is read together.
The periods in the numerals are separated by commas.
Example: 1732256 can be read in the International System as one million, seven
hundred thirty-two thousands, two hundred fifty six.
1732256 can be written in the International System as 1,732,256.
The expanded form of the number is obtained by multiplying the digits in the number
with the place value and separating them with ‘+’ sign.
Expanded form of 1,732,256 = 1,000,000 + 7, 00,000 + 30,000 + 2,000 + 200 + 50 +
6
 Representing numbers on abacus
Abacus is a calculating tool. It consists of beads that can slide on vertical spikes.
Numbers can be represented on an abacus by inserting suitable number of beads in the
spikes.
The number represented on the following abacus is 4, 37,365.
 Representing numbers on number line
To draw a number line, we take a line and mark a point on it, labelling it as 0. Then, we
mark the points to the right of zero at equal intervals and label them as 1, 2, 3 …, as
follows:
On the number line, we can say that out of any two whole numbers, the number to the
right of the other number is greater.
 Comparing and ordering numbers
 Rule 1: If numbers with different number of digits are given, then the number that
has more digits is greater.
Example: 2, 56,325 > 97,325
 Rule 2: If numbers having the same number of digits are given, the number with the
greater digit at the leftmost place is greater.


If the leftmost digits are the same, then we compare the next digit to the right and
continue until the digits are different.
Example: 6258 < 6289
Arranging numbers in the ascending order means arranging the numbers from smaller
to greater
325 < 437 < 567 < 892 << 1023
 325, 437, 567, 892, 1023 are in the ascending order
Arranging numbers in the descending order means arranging the numbers from
greater to smaller
1023 > 892 > 576 > 437 > 325
 1023, 892, 567, 437, 325 are in the descending order
 Formation of numbers from given digits
 The greatest and smallest numbers, without repetition, can be formed using any
number of digits by arranging them in the descending and ascending order
respectively.
For example, if we form a 4-digit number from the digits, 9, 2, 7 and 3, then
Greatest number = 9732
Smallest number = 2379
 When repetition of digits is allowed, then the greatest number can be formed by
writing the greatest digit as many times as the number is required. Similarly, the
smallest number can be formed by writing the smallest digit as many times as
required.
Greatest number (with repetition) = 9999
Smallest number (with repetition) = 2222
 Roman numerals
Roman numeral is another system for writing numerals. In this system, symbols are used
for representing numbers.
The symbols are I, V, X, L, C, D and M.
Number
Roman
numeral

1
5
10
50
100
500
1000
I
V
X
L
C
D
M
Rules for writing Roman numerals
1. Any symbol cannot be repeated more than 3 times.
2. If a smaller symbol is written to the left of a bigger symbol, then subtract it from
the bigger symbol.
3. If a smaller symbol is written to the right of a bigger symbol, then add it to the
bigger symbol.
4. V, L and D cannot be written to the left of a bigger symbol.
5. V, L and D cannot be repeated, whereas I, X, C and M can be repeated
Chapter 2: Arithmetic Operations on Numbers
 Addition facts
 If zero is added to a number, then its value remains the same.
23 + 0 = 23
 Changing the order of the numbers to be added does not alter the sum of the numbers.
25 + 35 = 60
25 + 35 = 60
 Grouping of numbers in any order does not alter the sum of the numbers.
(35 + 15) + 45 = 50 + 45 = 95
35 + (15 + 45) = 35 + 60 = 95
 The numbers being added are called addends and the result of addition of addends is
called the sum.
Th
H
T O
5

3
5
6 2
3 8
5
9
0 0
Here, 5362 and 538 are addends and 5900 is the sum.
 Addition of numbers
 Smaller numbers can easily be added horizontally.
12 + 62 = 74
 Larger numbers are added in vertical columns according to the place value chart.
For example: 526 + 3124 + 2576 = ?
Th H T O
5 2 6
3
 2
1
5
2 4
7 6
6
2
2
6
526 + 3124 + 2576 = 6226
 Subtraction of numbers
 Smaller numbers can easily be subtracted horizontally.
32 – 23 = 9
 Larger numbers can be subtracted vertically by arranging the numbers according to
their place values.
For example: 2376 – 938 = ?
Th H T
O
1 13 6 16
2
3
7
6

9
3
8
1
4
3
8
2376 – 938 = 1438
 Subtraction facts
 The value of any number does not change when 0 is subtracted from it.
 The number from which the other number is subtracted is called the minuend.
 The number that is subtracted is called the subtrahend.
 The number obtained on subtraction is called the difference.
 Finding missing numbers in addition and subtraction
 To find the missing addend in an addition, the given addend is to be subtracted from
the sum.
For example: The sum of two numbers is 1235. If one number is 835, find the other
number.
Missing addend = 1235 – 835 = 400
 To find the missing minuend, the subtrahend is to be added to the difference.
 To find the missing subtrahend, the difference is subtracted from the minuend.
For example: Find the number from which 1230 must be subtracted in order to get
2337.
Here, difference = 2337 and subtrahend = 1230
 Minuend = Difference + subtrahend = 2337 + 1230 = 3567
 Multiplication facts
 The product of any number with ‘1’ is the number itself.
 The product of any number with ‘0’ is always zero.
 The product remains unaltered when the numbers are multiplied in any order.
2  7 = 7  2 = 14
(2  5)  6 = 2  (5  6) = 60
 Multiplication of numbers ending in zeros
Count the number of zeros in the numbers.
Multiply the numbers leaving the zeros.
Write the product followed by the counted number of zeros.
For example: Find the value of 160  240.
Number of zeros in the given numbers = 2
16  24 = 384
 160  240 = 38400
 Multiplication with one, two and three-digit numbers
 The value of 2356  4 can be found as follows:
T Th Th H T O
2
3
5 6

4
9

4
4
2356  4 = 98952
The value of 2356  42 can be found as follows:
Multiplier = 42 = 4 tens and 2 ones
Multiply 2356 with 2 ones = 4712
Multiply 2356 with 4 tens = 94240
Adding the partial products, 4712 + 94240 = 98952
T-Th Th H T O
2
3
9
4
4
7 1 2
2 4 0
9
8
9 5


2
5 6
4 2
2
2356  42 = 98952
The value of 2356  423 can be found as follows:
Multiplier = 423 = 4 hundreds, 2 tens and 3 ones
Multiply 2356 with 3 ones = 7068
Multiply 2356 with 2 tens =47120
Multiply 2356 with 4 hundreds = 942400
Adding the partial products, 7068 + 47120 + 942400 = 996588
L T Th Th H T O
2 3
 4
5
2
6
3
7
6
8
0
0
0
9
4
4
7
2
1 2
4 0
9
9
6
5
8 8
2356  423 = 996588
 Division facts
 When any number is divided by 1, the quotient is the number itself.



A non-zero number divided by itself gives the quotient as 1.
Zero divided by any non-zero number, gives the quotient as 0.
0  30 = 0
Any number cannot be divided by zero.
 Division by numbers ending in zeros
 If the dividend also ends is zeros:
When divided by 10, remove 1 zero
500  10 = 50
When divided by 100, remove 2 zeros
5,00,000  100 = 5,000
Perform similar operations if the divisors are 1000, 10000 … and so on.
 If the dividend does not end in zeros:
a) When a number is divided by 10, the remainder is the digit at ones place, while
the quotient is obtained by removing the digit at ones place.
In the division 2756  10, remainder = 6 and quotient = 275
b) When a number is divided by 100, the remainder is formed by the digits at the
tens and ones places, while the quotient is obtained by removing these two digits.
In the division 2756  100; remainder = 56 and quotient = 27
 Division by one, two and three-digit numbers
 The number 5236 can be divided by 7 as follows:

 Remainder = 0 and quotient = 748
The number 5236 can be divided by 72 as follows:

Remainder = 52 and quotient = 72
The number 5236 can be divided by 723 as follows:
Th H T O
7
723 5236
5061
175

Remainder = 175 and quotient = 7
Observation: Divisor  quotient + remainder = 723  7 + 175
= 5061 + 175
= 5236
= Dividend
This observation is used whether the division is correct or not.\
 Estimation
Estimation gives a rough idea of a calculation. It is to guess and to come close to the
correct answer.
 Rounding method
In this method, we consider the digit just to the right of the digit to be rounded. If this
digit is less than 5, then change all digits to the right of the rounding digit to zero. If it
is 5 or greater than 5, then add one to the rounding digit and change all digits to the
right of the rounding digit to zero.
7300 + 1684
Rounding 7300 and 1684 to nearest thousand,
7000
 2000
9000

Front end method – In this method, we focus to compute the digits at the highest
place. The estimate can be made closer to accurate by computing the digits at next
place.
Hence, the estimated quotient is 1340.
4324 + 8729
4000 + 8000 = 12000
300 + 700 = 1000
12000 + 1000 = 13000
Hence, the estimated sum is 13000.
5678 × 7
5000 × 7 = 35000
600 × 7 = 4200
Estimated product = 35000 + 4200 = 39200


2815 – 1403
2000 – 1000 = 1000
800 – 400 = 400
1000 + 400 = 1400
Hence, the estimated difference is 1400.
Estimation of sum by grouping of nice numbers
Grouping focuses on organising numbers into groups of nice numbers i.e., the
numbers that can easily be added mentally.
2439 + 4183 + 3609 + 5780
2000 + 4000 + 3000 + 5000 = 14000
14000 + 1000 + 1000 = 16000
Estimating sum by clustering
When given numbers centre around a specific number, we use this method.
51 + 47 + 53 + 56+ 42
All numbers cluster around 50.
Hence, estimated sum = 50 × 5 = 250
Chapter 3: Factors and Multiples
 Multiples of a number
The multiples of a given number are the products obtained on multiplying the number
with any number.
Example: Find the multiples of 12.
12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48 …, and so on
Hence, the numbers 12, 24, 36, 48, etc. are the multiples of 12.
 Zero is a multiple of every number.
0 × 1 = 0, 0 × 2 = 0, 0 × 3 = 0, 0 × 4 = 0 …
 Every number is a multiple of 1.
1 × 1 = 1, 1 × 2 = 2, 0 × 3 = 0, 0 × 4 = 0 …
 Every multiple of a number is greater than or equal to that number.
For example, the multiple of 6 is 6, 12, 18 …
Here, the multiples of 6 are greater than or equal to 6.
 Factors of a number
When a number exactly divides another number, the divisor is called a factor of the
dividend and the dividend is called a multiple of the divisor.
6
12 72
72
0
Since 12 exactly divide 72, 12 is a factor of 72.
Here, 72 is a multiple of 12.





1 is a factor of every number. E.g., 5 = 1 × 5, 19 = 1 × 19
Every number is a factor of itself. For example: 20 = 20 × 1. This shows that 20 is a
factor of 20.
Every factor of a number is less than or equal to the number.
For example, factors of 12 are 1, 2, 3, 4, 6 and 12. Here, 1, 2, 3, 4, 6 and 12 are less
than or equal to 12. The greatest factor of a number is the number itself.
Numbers that are multiples of 2 are called even numbers. For example, 12, 16, 22
…are even numbers.
Numbers that are not multiples of 2 are called odd numbers. For example 3, 5, 7…
are odd numbers.
 Common multiples and LCM
Multiples of 4: 4
8 12 16 20 24 28 32 36 …
Multiples of 6: 6 12 18 24 30 36 42 48 54 …
The common multiples of 4 and 6 are 12, 24, 36 …
The lowest common multiple (LCM) of two or more given numbers is the least of their
common multiples.
LCM of 4 and 6 is 12.
 Common factors and HCF
The factors common to the given numbers are called common factors.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
The factors of 32 are 1, 2, 4, 8, 16, 32
The common factors of 24 and 32 are 1, 2, 4 and 8.
The highest common factor (H.C.F.) of two or more given numbers is the highest of their
common factors.
The HCF of 24 and 32 is 8.
 Common divisibility tests
 A number is divisible by 2 if the digit at ones place is either 0, 2, 4, 6 or 8.
For example, the numbers 9218, 6054, 932 are divisible by 2.
 A number is divisible by 3 if the sum of the digits in the number is divisible by 3.
For example, the sum of the digits of 9528 is 9 + 5 + 2 + 8 = 24, which is a multiple
of 3. Hence, 9528 is divisible by 3.
 A number is divisible by 6 if it is divisible by both 2 and 3. For example, 39612 is
divisible by 2. The sum of the digits of 39612 is 3 + 9 + 6 + 1 + 2 = 21, which is a
multiple of 3. Hence, 39612 is divisible by 3. Now, 39612 is divisible by both 2 and
3. Hence, it is divisible by 6.
 A number is divisible by 5 if it ends in 0 or 5.
For example, 5235 is divisible by 5 because it ends in 5.
 A number is divisible by 9 if the sum of the digits in the number is divisible by 9.
For example, the sum of the digits in the number 9567 is 9 + 5 + 6 + 7 = 27, which is
divisible by 9. Hence, 9567 is divisible by 9.
 A number is divisible by 10 if it ends in 0.
For example, 120 is divisible by 10 because 120 ends in 0.
 Prime and composite numbers
 Prime numbers are numbers having exactly two factors i.e.,1 and the number itself.
For example, the factors of 23 are 1 and 23 only. Hence, 23 is a prime number.
Some of the prime numbers are 2, 3, 5, 7, 11, 13…
 Composite numbers are numbers having more than two factors. For example, the
factors of 20 are 1, 2, 4, 5, 10 and 20. Since 20 have more than two factors, it is a
composite number.
 Prime factorisation
When a number is factorised into a set of factors that are a set of prime numbers only, the
method followed is called prime factorisation.

Prime factorisation by factor tree method
The number 96 can be prime factorised by factor tree method as follows:
Here, we consider 96 as the base of the tree.
96 is expressed as 96 = 2 × 48
Factorise the composite factors completely till the prime factors are reached.
96 can be prime factorised as follows:
96 = 2 × 2 × 2 × 2 × 2 × 3

Prime factorisation by division method
To find the prime factorisation of a number, we need to divide it by prime numbers
that are factors of the given number, till we get 1.
Example: Find the prime factorisation of 24.
We can proceed as follows:
2
2
2
3
24
12
6
3
1
Hence, the prime factorisation of 24 is given by 24 = 2 × 2 × 2 × 3.
 Finding H.C.F. and L.C.M. by prime factors
 The H.C.F. of two or more numbers is the product of the common prime factors of
the numbers.
To find the H.C.F. of given numbers, we first need to prime factorise the given
numbers.
For example, the H.C.F. of 24 and 42 can be found as follows:
2
2
7
28
14
7
1
28 = 2 × 2 × 7
42 = 2 × 3 × 7
H.C.F. of 28 and 42 = 2 × 7 = 14
2
3
7
42
21
7
1


L.C.M. is the product of the prime factors counted the maximum number of times
they occur in the prime factorization of any number.
28 = 2 × 2 × 7
42 = 2 × 3 × 7
L.C.M. of 28 and 42 = 2 × 2 × 3 × 7 = 84
Here, the maximum number of times 2 occurs is two and each of the other primes
occurs once.
When the H.C.F. of a pair of numbers is 1, the pair of numbers is called co-prime
Consider 28 and 33:
1
H.C.F. of 28 and 33 = 1
 28 and 33 are co-prime.
28,
33
28,
33
Chapter 4: Fractions
 A fraction is a number representing a part of a whole.
The whole may be a single object or a group of objects.
Example:
Numbers such as half, one-third, three-fifths etc. are called fractional numbers.
1 1 3
For example, , ,
etc. are called fractions
2 3 5
 Fractions that have the same denominator belong to the same fraction family.
1 2 3
For example, , ,
etc. belong to the fourth fraction family.
4 4 4
 The smallest member of a fraction family with numerator as 1 is called a unit fraction.
1
1 1 1
For example,
is a unit fraction of the fourth fraction family. , ,
are some other
4
6 7 15
examples of unit fractions.
 Like and unlike fractions
 Fractions that have the same denominator are called like fractions.
2 6 8
For example, , ,
etc. are like fractions.
7 7 7
 Fractions that have different denominators are called unlike fractions.
2 3 5
For example, , ,
etc. are unlike fractions
5 7 9
 Fractions can be categorized into three types: proper, improper and mixed.
 Proper fractions are those fractions in which the numerator is less than the
denominator. These fractions are always less than 1.
17
Example:
is a proper fraction since the numerator, 17, is less than the
24
denominator, 24.
 Improper fractions are those fractions in which the numerator is greater than the
denominator. These fractions are always greater than 1.
15
is an improper fraction since the numerator (15) > denominator (7).
7
A mixed fraction or mixed number is a combination of a whole number and a part.
5
5
Example: 9   9
is a mixed fraction.
13
13
Example:

 Conversion of mixed fractions into improper fractions and vice-versa
 A mixed fraction can be converted into an improper fraction as follows:
(Whole  Denominator)  Numerator
Denominator
2 (3  7)  2 21  2 23
Example: 3 


7
7
7
7
 To convert an improper fraction into a mixed fraction, first of all, the quotient and
remainder are obtained by just dividing the numerator by the denominator. Then, the
mixed fraction corresponding to the given improper fraction is written as follows:
Remainder
Quotient
Divisor( Denominator)
23
Example: To find the mixed fraction corresponding to the improper fraction , first
7
of all, 23 is divided by 7.
3
7 23
21
2
Here, divisor = 7, quotient = 3 and remainder = 2
23
2
 3
7
7
 A fraction obtained on multiplying or dividing both the numerator and denominator by
the same non-zero number is called equivalent fraction. Such fractions express the same
value of the part of a whole.
2 2  2 4 2 2  3 6 2 2  5 10
Example: 
 , 
 , 

3 3  2 6 3 3  3 9 3 3  5 15
 A fraction can be reduced to its simplest form or lowest form by dividing both numerator
and denominator by their H.C.F.
49
Example:
can be converted into its lowest form as follows:
91
The H.C.F. of 49 and 91 is 7.
49 49  7 7
 

91 91  7 13
Therefore,
7
49
is the simplest form of
.
13
91
 Comparing and ordering of fractions
 If two or more fractions have the same denominator, then greater the numerator,
greater is the fraction.
2 5 3
6
Example: Among the fractions , ,
and , 2 < 3 < 5 < 6.
7 7 7
7
2 3 5 6
   
7 7 7 7
 If two or more fractions have the same numerator, then smaller the denominator,
greater is the fraction.
17 17
17
Example: Among the fractions
and , 3 < 5 < 11.
,
5 3
11
17 17 17



3
5 11
 To compare two unlike fractions (without the same numerator), first of all, these
fractions are converted into their equivalent fractions of the same denominator, which
is the L.C.M. of the denominators of the fractions. The like fractions are then
obtained, which can be easily compared.
 Addition and subtraction of fractions
 The addition of two fractions with the same denominators can be performed by just
adding the numerators and retaining the common denominator of the fractions.
17 3 17  3 20 20  5 4
Example:





25 25
25
25 25  5 5
 The subtraction of two like fractions can be performed by just subtracting the
numerators and retaining the common denominator of the fractions.
31 4 31  4 27 27  3 9
Example:
 



15 15
15
15 15  3 5
 To perform the addition and subtraction of unlike fractions, first of all, they are
converted into their equivalent fractions with the denominator as the L.C.M. of their
denominators. Then, addition or subtraction can be easily performed.
5
7
Example: Subtract from .
8
6
H.C.F. of 6 and 8 = 24
7 7  4 28


6 6  4 24
5 5  3 15


8 8  3 24
7 5 28 15 28  15 13
Thus,  



6 8 24 24
24
24

To add or subtract mixed fractions, first of all, they are converted into improper
fractions. Then, they can be easily added or subtracted.
Example:
2
4 37 31
7 3 

5
9 5 9
37  9 31 5


 L.C.M. of 5 and 9 is 45.
5 9 9 5
333 155


45 45
333  155

45
488

45
38
 10
45
Chapter 5: Decimals
 The fractions in which the denominators are 10, 100, 1000 etc. are called decimal
fractions.
Decimals are used for separating whole numbers from decimal fractions.
58
2.58  2 
100
58
In 2.58, 2 is the whole number and
is the decimal fraction.
100
Fractional number
1 tenth
9 tenth
1 hundredth
99 hundredth
1 thousandth
999 thousandth
Common fraction
1
10
9
10
1
100
99
100
1
1000
999
1000
Decimal fraction
0.1
0.9
0.01
0.99
0.001
0.999
 Conversion of decimals into fractions and vice-versa
 For converting a given decimal into common fraction, consider the given decimal
without the decimal point as the numerator and take the denominator as the number
formed by placing as many zeros to the right of 1 as the number of decimal places.
92
923
9234
Example: 9.2  , 9.23 
, 9.234 
10
100
1000
 For converting common fractions into decimals, place the decimal point in the
numerator after as many digits as there are zeros in the denominator.
2536
235
23
Example:
 2.536,
 2.35,
 2.3
1000
100
10
 Place value chart for decimals
Ten thousands
Thousands
Hundreds
Tens
Ones
Decimal point
10000
1000
100
10
1
.
Tenths
Hundredths
Thousandths

The decimal 235.523 can be represented in the place value chart as follows:
Hundreds
Tens
Ones
Decimal point
Tenths
Hundredths
Thousandths


1
10
1
100
1
1000
2
3
5
.
5
2
3
The decimal 235.523 can be read as two hundred thirty-five point five two three or
two hundred thirty-five five tenths two hundredths and three thousandths.
The decimal number 235.523 can be written in expanded form as follows:
235.523 = 2 hundreds + 3 tens + 5 ones + 5 tenths + 2 hundredths + 3 thousandths
5
2
3
 200  30  5  

10 100 1000
 Decimals can be represented using figures as well as the number line
 Representation of decimals using figures
Example: Decimal 0.7 can be represented using a figure as follows:

Here, the figure is divided into 10 equal parts, out of which 7 are shaded.
7
The shaded portion of the figure represents
or 0.7 .
10
Representation of decimals on number line
Example:
Here, the gap between consecutive whole numbers is divided into ten equal parts to
represent tenths.
0.2, 0.5, 1.1, 1.2, 1.6 are represented by the marked points.
 Comparing and ordering of decimal fractions
 The comparison starts with the whole part. If the whole parts are equal, then the tenth
part is compared and so on.
 Arranging the decimals in the ascending order means arranging them from the
smallest to the greatest
 Arranging the decimals in the descending order means arranging them from the
greatest to the smallest
Example: The decimals 1.235, 2.356, 2.536 and 3.528 can be arranged in the
ascending order As follows:
In the whole parts, 3 > 2 > 1
 3.528 > (2.356, 2.536) > 1.235
Now, consider 2.356 and 2.536.
The whole parts are equal. The digit at the tenths place of 2.536 i.e., 5 is greater than
the digit at the tenths place of 2.356 i.e., 3.
 2.536 > 2.356
Hence, 3.528 > 2.536 > 2.356 > 1.235
 Addition and subtraction of decimals
1. Write the numbers according to the place value chart such that the decimal point falls
on the decimal line.
2. Adding zeros to the extreme right of the decimal column makes no difference to the
value of the number.
3. Now, the addition or subtraction of decimals is exactly the same as the addition or
subtraction of whole numbers.
3
1
1
9
9
0
4
3
0
One
hundredths
14
10
One
thousandths
One tenths
Decimal line
One
Ten
Hundred
Example: Find the value of 194.506 – 93.52.
4
.
.
.
194.506 – 93.52 = 100.986
5
5
9
0
2
8
6
0
6
Chapter 6: Unitary Method
 Unitary method is the method of carrying out calculation for finding the value of required
number of units by first finding the value of one unit.
Example:
If the cost of 5 pens is Rs 30, then find the cost of 8 pens.
Solution:
Firstly, calculate the cost of 1 pen and then calculate the cost of 8 pens.
Cost of 5 pens = Rs 30
Cost of 1 pen = Rs 30  5 = Rs 6
Cost of 8 pens = Rs 6 × 8 = Rs 48
Chapter 7: Metric System
 Table of metric measures
Thousand
1000
Hundred
100
Ten
10
One
1
Tenth
1
10
Hundred
1
100
Thousandth
1
1000
Metric
unit
Kilo
( 1000)
Hecto
(100)
Deca
( 10)
Basic
unit
Deci
( 10)
Centi
( 100)
Milli
( 1000)
Length
Kilometre
Mass
Kilogram
Kilolitre
Deci
metre
Deci
gram
Deci
litre
Centi
metre
Centigra
m
Capacit
y
Hectometr Decametr Metr
e
e
e
Hecto
Decagram Gram
gram
Hecto
Decalitre Litre
litre
Centilitre
Millimetre
Milligram
Millilitre
 Metric conversions
Measures of length
10 millimetres
=
10 centimetres
=
10 decimetres
=
10 metres
=
10 decametres
=
10 hectometres
=
Measures of mass
Measures of capacity
10
1
10
1
=
=
milligrams
centigram millilitres
centilitre
10
1
10
1
=
=
centigrams
decigram centilitres
decilitre
10
10
1 metre
=
1 gram
=
1 litre
decigrams
decilitres
1
1
1
10 grams =
10 litres =
decametre
decagram
decalitre
1
10
1
10
1
=
=
hectometre decagrams
hectogram decilitres
hectolitre
10
1
10
1
1 kilometre
=
=
hectograms
kilogram hectolitres
kilolitre
1
centimetre
1
decimetre
 Conversion of length
 Smaller units to bigger units
Example: 8653 mm can be converted into bigger units as follows:
10 8653 mm
10 865 cm 3 mm
10 86 dm 5 cm
8 m 6 dm

8653 mm = 6 m 6 dm 5 cm 3 mm
Bigger units to smaller units
Example: 7 m 8 dm 9 cm 2 mm can be converted into millimetres as follows:
1 m = 1000 mm
1 dm = 100 mm
1 cm = 10 mm
7 m 8 dm 9 cm 2 mm = (7  1000 + 8  100 + 9  10 + 2) mm
= (7000 + 800 + 90 + 2) mm
= 7892 mm
Similarly, conversions of weights or capacities from smaller to bigger units or bigger
to smaller units can also be performed.
 Arithmetic operation on metric units
Arithmetic operations on metric units are performed just as on whole numbers.
1. 5 dag 5 dg 8 cg and 2 dag 1 g 4 dg 5 cg can be added as follows:
dag g dg cg
5
1
0
1
5
8

2
1 4 5
7
2 0 3
 5 dag 5 dg 8 cg + 2 dag 1 g 4 dg 5 cg = 7 dag 2 g 3 cg
2. 7 gm 5 cg 2 mg can be divided by 4 as follows:
gm dg cg mg
4 7 0
5
2
1 7 6 3
 7 gm 5 cg 2 mg  4 = 1 gm 7 dg 6 cg 3 mg
Similarly, subtraction and multiplication can be performed on metric units.
 Inter-conversion of metric units involving decimals
 To convert from higher units to lower units, multiply the higher unit with the
difference between the two units.
Example: 2.38 dl can be converted into ml as follows:
1 dl = 100 ml
2.38 dl = 2.38  100 l = 238 ml
 To convert a lower unit into higher unit, first convert the lower unit into a decimal
fraction and then change it into decimal number.
Example: 3253 dag can be converted into kg as follows:
3253
3253 dag 
(100 dag = 1 kg)
100
= 32.53 kg
 More conversions of metric units

Arrange the numbers in the correct place values. Then, the decimal point should lie to
the right of the place corresponding to the given unit. For the length 4.523 dam, the
decimal should lie between deca and metre.
Example: 4.523 dam can be converted into mm as follows:
Kilo
Hecto
Deca
Metre
Deci
Centi
Milli
4.523
4.
5
2
3
0
45230 mm
 4.523 dam = 45230 mm
 If there is no decimal, then place the last digit of the number in the corresponding
column and if there are not enough digits, then place as many zeros at the front.
Example: 2437 m can be converted into km as follows:
Kilo
Hecto
2437 m
2
4
 2437 m = 2.437 km
Deca
3
Metre
7
Deci
Centi
Milli
2.437 km
 For addition or subtraction on metric units given in decimal form, first convert the
decimals into same metric units and then add or subtract as in the case of whole numbers.
Example: Find the value of 5.326 kg + 3.254 kg + 7.328 kg.
11
5. 3 2 6 kg
3. 2 5 4 kg
 7. 3 2 8 kg
15. 9 0 8 kg
5.326 kg + 3.254 kg + 7.328 kg = 15.908 kg
Chapter 8: Money
 Money is an object that is usually accepted as the payment for goods or services.
 In India, the units of measuring money are rupees and paise.
1
Rs 1 = 100 paise or 1 paise = Re
100
 While we read the amount of money given in decimals, the number in rupees column is
read together and the number in paise column is read together.
Example: Rs 26.93 can be read in words as twenty-six rupees and ninety-three paise.
Similarly, an amount of money given in words can be written in decimals.
Example: Five rupees and eight paise can be written in decimals as follows:
Five rupees and eight paise
= Rs 5 + 8p
1
= Rs 5 + Re (8  100)
(1 p = Re
)
100
= Rs 5 + Re 0.08
= Rs 5.08
 The amounts of money are added or subtracted just as decimals are added or subtracted.
Example: Rs 93.75 and 26.35 can be added as follows:
Rupees Paise
93
26
75
35
120 10
Rs 93.75 + Rs 26.35 = Rs 120.10
 To multiply an amount of money with a whole number, we are required to follow the
given steps.
Step 1: Write the given amount of money in decimals.
Step 2: Multiply the obtained decimal with the given whole number as the numbers are
multiplied.
Step 3: In the product, put the decimal point after the second digit from the right.
Example: Rs 19.63 can be multiplied with 8 as follows:
19.63

8
157.04
 8  Rs 19.63 = Rs 157.04
 To divide a given amount of money by a whole number, we are required to follow the
given steps.
Step 1: Write the given amount of money in decimals.
Step 2: Divide the obtained decimal by the given whole number taking the decimal as an
ordinary number.
Step 3: In the quotient, put a decimal point after 2 digits from the right.
Example: Rs 23.04 can be divided by 12 as follows:
1.92
12 23.04
12
110
108
24
24
0
 Rs 23.04  12 = Rs 1.92
Chapter 9: Time
 Clock
The measurement of time is an important aspect of daily life.
The time is read from a clock. The face of a clock is divided into 12 equal big divisions
and the space between two consecutive big divisions is divided into 5 equal parts.
It consists of a minute-hand (long hand) and an hour hand (short hand) at the centre.
Sometimes, it consists of a second hand at centre.
 Units of time
Some of the units used while measuring time are second, minute, hour, day etc. The
conversions between these units are given as follows:
1 minute = 60 seconds
1 hour = 60 minutes
1 day = 24 hours
 Reading time from a clock
The time in a clock is read by looking at the positions of its hands.
Example:
The time shown in the given clock can be read as follows:
In the given clock, it is observed that the hour hand is between 3 and 4.
Hence, the time is between 3 o’clock and 4 o’clock.
It is also observed that the minute hand is at 9.
This means that the minute hand has moved 9 × 5 = 45 small divisions.
Thus, the time shown in the clock is 3:45 or quarter to 4 o’clock.
 Writing or expressing time



The time between 12 midnight to 12 noon is denoted by A.M.
The time between 12 noon to 12 midnight is denotes by P.M.
The time of a day is expressed in terms of 24-hour clock. If the time is in between 12
noon and 12 midnight, then the time is expressed in 24-hour clock by adding 12:00.00
hours to the given time.
Example:
The times 9:31 A.M. and 9:31 P.M. can be expressed in 24-hours clock As follows:
9:31 A.M. = 9:31 hours
9:31 P.M. = (9:31 + 12:00) hours = 21:31 hours
 Arithmetic operations on time
Time is added or subtracted as the decimals are added by keeping in the mind that 1 hour
= 60 minutes and 1 minute = 60 seconds.
Example: 9 hours 41 minutes 32 seconds can be subtracted from 4:25:17 P.M. as follows:
4:25:17 P.M. = 16:25:17 hours
Hours Minutes Seconds
60
15
24
60

16
9
25
41
17
32
6
43
45
6:43:45 hours = 6:43:45 A.M.
 4:25:17 P.M. – 9 hours 41 minutes 32 seconds = 6:43:45 A.M.
 Time duration
The time duration of an activity is calculated by subtracting the time at which the activity
started from the time at which the activity is ended.
 Leap year
A year is said to be a leap year if any of the following conditions is satisfied.
 If the year ends in 00 and is divisible by 400.
 If the years does not end in 00 and is divisible by 4
 In a leap year, there are 29 days in the month of February. There are 28 days in the
month of February if the year is not a leap year.
 Number of days in different months
 There are 31 days in the months of January, March, May, July, August, October and
December.
 There are 30 days in the months of April, June, September and November.
 Starting time and finishing time
Starting time is the time when an activity begins. Finishing time is the time when an
activity ends. Time taken is the total time spent in the activity.
Time taken = Finishing Time – Starting Time
 Starting date and finishing date
Starting date is the date on which an activity starts. Finishing date is the date when an
activity ends. Duration is the number of days spent on the activity.
Duration = Finishing date – Starting date
Chapter 10: Geometry
 The length of a line segment can be measured using a ruler and a pencil. To do so, we are
required to follow these steps.
1. Place the scale in such a manner that the left end-point of the line segment coincides
with the zero mark of the ruler and the line segment coincides with the edge of the
ruler.
2. Read the position of the other end-point of the line segment with respect to the scale.
This reading is the measure of the line segment.
Example:
The given figure shows the position of a scale while measuring a line segment XY.
Find the length of XY.
Solution:
It can be observed in the given figure that the length of the given line segment XY is 5.6
cm.
 Angles and its types
 Angles can be measured in degree units.
 An angle is said to be acute if its measure is less than 90.

An angle is said to be a right angle if its measure is 90.

An angle is said to be obtuse if its measure is between 90 and180.

An angle is said to be straight if its measure is 180.

An angle is said to be reflex if its measure is between 180 and 360.
 Measurement of an angle
 An angle is measured and drawn by using a protractor.
 A protractor is divided into 180 small divisions. Each small division of a protractor
measures 1. There are two scales in a protractor – an outer scale and an inner scale.
In each scale, 0, 10, 20, … 180 are marked and there are 10 equal divisions between
these marks.
 To measure an angle using a protractor, we are required to follow these steps.
1. Place the protractor on the given angle so that the vertex of the angle coincides
with the centre point of the protractor and one arm of the angle coincides with the
zero mark of the protractor.
2. Start counting 0 from this arm and see the reading of the remaining arm of the
angle in the protractor. This reading gives the measure of the given angle in
degree unit.
Example:
The given figure shows the position of a protractor while measuring an angle PQR.
Find the measure of PQR.
Solution:
It can be observed in the given figure that the measure of PQR is 118.

To draw an angle of the given measure, we are required to follow these steps.
1. Draw a line segment.
2. Place the protractor on this line segment such that the centre of the protractor
coincides with one end-point of the line segment and the zero mark of the
protractor coincides with this line segment.
3. Mark a point on the outer edge of the protractor that represents the given angle.
4. Join this point and the point at which the centre of the protractor was placed.
In this way, an angle of given measure can be drawn.
 Perpendicular lines
 A line is said to be perpendicular to another line if the measure of the angle between
the lines is 90.

If a line l is perpendicular to another line m, then it is represented as l  m.
 To draw a perpendicular m to a given line l through a point P on l using a set-square, we
are required to follow these steps.
1. Place a ruler with one of its edges along line l. Take a point P on it. Hold this firmly.
2. Place a set-square with one of its edges along the already aligned edge of the ruler
such that the right-angled corner is in contact with the ruler at P.
3. Hold the set-square firmly and draw a line m through P.
4. l is perpendicular to m through point P.
 If a curve does not intersect itself, then it is called a simple curve.
 If a simple closed curve is made of line segments only, then it is called a polygon.
 A polygon of three sides is called a triangle.
 Classification of triangles
 A triangle can be classified as an equilateral triangle or an isosceles triangle or a
scalene triangle on the basis of the length of its sides.
1. Equilateral triangle: All the sides are of equal length.
2. Isosceles triangle: Two sides are of equal length.
3. Scalene triangle: All the sides are of different length.

A triangle can be classified as acute-angled or obtuse-angled or right-angled on the
basis of the measures of the angles.
1. Acute-angled triangle: If all the angles are acute.
2. Obtuse-angled triangle: If one of the angles is obtuse.
3. Right-angled triangle: If an angle is right.
 A quadrilateral is a polygon of four sides. It may be a parallelogram or a rhombus or a
rectangle or a square or a trapezium or simply a general quadrilateral.
 Parallelogram: It is a quadrilateral whose opposite sides are of equal length and
opposite angles are of equal measure.

Rhombus: It is a parallelogram all of whose sides are of equal length.

Rectangle: It is a parallelogram the measure of all of whose interior angles is 90.

In a rectangle, the diagonals are equal in length.
Square: It is a rectangle whose sides are of equal length.

In a square, the diagonals are equal in length.
Trapezium: It is a quadrilateral in which exactly one pair of opposite sides is parallel.
 Polygons of five and six sides are called pentagon and hexagon respectively.
 A circle is a collection of points that are at a fixed distance from a fixed point. The fixed
distance is the radius and the fixed point is the called centre of the circle.


The line segment joining any two points on the circle is called a chord.
The chord through the centre of the circle is called the diameter of the circle. The
relation between the diameter and radius of a circle is given by diameter = 2  radius.
 To draw a circle of given radius, we are required to follow these steps.



Insert a pencil in a compass.
Take the distance between the tip of the pencil and the tip of the compass equal to the
radius of the circle.
By keeping the tip of the compass at a fixed point, rotate the pencil smoothly. Hence,
a circle of the given radius is obtained.
Chapter 11: Patterns
 Important divisibility rules
 A number is divisible by 3, if sum of its digits is divisible by 3. For example, sum of
digits of 33 is 3 + 3 = 6 and 6 is divisible by 3. Hence, 33 is divisible by 3.
 A number is divisible by 9, if sum of its digits is divisible by 9. For example, sum of
digits of 45 is 4 + 5 = 9 and 9 is divisible by 9. Hence, 45 is divisible by 9.
 While dividing a number by 10 or 100, the quotient can be obtained by removing the
ones digit or tens digit from the dividend and the digits so removed will be the
remainder.
For example, 41263 ÷ 10
Quotient = 4126, remainder = 3
For example – 367596 ÷ 100
Quotient = 3875, remainder = 96
 Square numbers
When a number is multiplied by itself, we obtain a square number.
For example: 9 × 9 = 81; 81 is a square number.
 All square numbers can be represented as a collection of beads or pebbles arranged in
a square shape i.e., the numbers of beads or pebbles in each column and row are
same.
For example: 9 can be arranged as:



The first 10 square numbers are:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
The difference between any 2 consecutive square numbers is always an odd number.
The sum of first n odd numbers is the square of n.
Let us take first 6 odd numbers.
1, 3, 5, 7, 9, 11
1 + 3 + 5 + 7 + 9 + 11 = 36 = (6)2
Hence, the sum of first 6 odd numbers is the square of 6.
 Triangular numbers
Some numbers can be represented as a collection of beads arranged in a triangular shape.
In this type of representation, the uppermost row contains 1 bead, second row will have 2
beads, 3rd will have 3 beads, and so on. Thus, each row contains 1 more bead than the
number of beads represented in the row just above it or simply we can say that the nth
row contains n number of beads in it.
The numbers so arranged are called triangular numbers.
 The number of beads in each side of such triangle is same and also same as the
number of rows used to represent the triangle.

The first 10 triangular numbers are:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55 _ _ _

Addition of two consecutive triangular numbers always gives a square number.

A triangular number can never end with 2, 4, 7, or 9.
 Pascal’s triangle
In Pascal’s triangle, each number is the sum of the numbers above it.
Pascal’s triangle is drawn below and sum of each horizontal row is written against it.
 Border Patterns
 A border pattern has unit, which is repeated alongside it. This unit is the cell of the
pattern and is repeated by translations.

Apart from translation, H, V, and R are the three other symmetries, which transform
the strip into itself.

A border pattern can be formed by following any of the symmetries, H, V, and R, and
their combinations as follows:
 Tiling patterns
Tiling is the pattern obtained when a collection of individual tiles are fit together without
any gaps or overlaps. Tiling is done to fill flat surfaces such as floors, table tops etc.
Example:

Two tiles are congruent if they are of the same shape and size.

Example: The pair of tiles
and
is congruent.
A tiling, like the one shown below with only one tile as the generating set, is called
monohedral tiling.

A tiling with two tiles in its generating set is called a dihedral tiling.
Example: The tiling shown below is a dihedral tiling.
Here, the tiling is generated by the tiles
and
.
Chapter 12: Symmetry
 Symmetrical figures and line of symmetry
 A figure is said to be symmetrical if it is in an evenly-balanced proportion.
 A figure has line symmetry if a line can be drawn dividing the figure into two
identical parts in such a way that the obtained two parts exactly overlap each other
when they are folded along the line. The line is called the line of symmetry.
Example:

In this figure, line l divides the figure into two identical parts and the obtained two
parts exactly overlap each other when they are folded along line l. This line l is
known as the line of symmetry. Hence, this figure has a line of symmetry.
A figure may have no line of symmetry, only one line of symmetry or multiple lines
of symmetry.
Example:
1) A scalene triangle and a parallelogram have no line of symmetry.
2) An isosceles triangle and the letters of the English alphabet such as A, B etc.
show one line of symmetry.
3) An equilateral triangle, a square, a circle etc., show multiple lines of symmetry.
 Relation between line of symmetry and mirror reflection
The line of symmetry is closely related to mirror reflection. When we deal with mirror
reflection, we have to take into account that the object and the image are symmetrical
with reference to the mirror line. There is no change in the length and the angle of the
object and the corresponding length and angle of the image, with respect to the mirror
line; only the left-and-right alignment changes.
Example:
Draw the reflection of the given shape considering the dotted line as the mirror line.
The reflection of the given shape can be drawn considering the dotted line as the mirror
line as follows:
 Applications of line of symmetry
The concept of symmetry is widely used in the field of technology, architecture,
geometrical reasoning, designing etc.
 Rotational symmetry
Some figures have a rotational symmetry in it. When such figures are rotated by some
angle, we obtain exactly the same figure again.
 Symmetry in 3-D shapes
Some 3-D figures have a plane of symmetry i.e., this plane divides the 3-D shape into 2
halves and if we put a mirror in place of plane, then the two halves will be the mirror
images of each other. This plane is called as the plane of symmetry.

There are nine planes of symmetry of a cube and these nine planes can be shown as:

A sphere has infinitely many planes of symmetry.
Chapter 13: Three-Dimensional Shapes
 We see certain shapes in our day-to-day life that are not flat. Some of these shapes are
solids.
 Solids and their attributes
 An object that has a fixed shape and size is called a solid.
 The corners of a solid figure are called its vertices; the line segments joining the
vertices are known as its edges and its flat surfaces are called faces.
Solid
Cube
Cuboid
Figure
Properties
6 faces
12 edges
8 vertices (corners)
6 faces
12 edges
8 vertices
Cylinder
2 flat faces (circles)
1 curved face
Cone
1 flat surface
1 curved surface
1 vertex
Triangular
pyramid
4 faces
6 edges
4 vertices
Square
pyramid
5 faces
8 edges
5 vertices
 Nets of different solid figures
Net of a solid figure is a two-dimensional imagination of all its faces attached at the
edges on a point. When this net is again folded in a meaningful way, it again gives the
original 3-D figure.
 Net of a cube
o 11 nets of a cube are as follows:
o Net of a cube can have four squares aligned together at maximum. The net that
consists of 5 or 6 squares in a row cannot be the net of a cube.
o If in a net of a cube, 4 squares are in a row, then the remaining two will be on
opposite sides i.e., one is on one side and the other one on the other side.

Net of a cylinder
Net of cylinder consists of three parts:
o One circle giving the base
o Another circle giving the top
o A rectangle giving the curved surface
Here, the circumference of these circles is exactly same as one of the sides of rectangle.
 Net of a cone
Net of a cone consists of two parts:
o A circle that gives the base
o A sector that gives the curved surface
 Drawing three-dimensional shapes on paper
 A three-dimensional shape can be drawn on paper using lines or can be made using
two-dimensional objects.
 A three-dimensional drawing of an object is called isometric drawing. The isometric
drawing of a solid is drawn on isometric dot paper.
For example, the isometric drawing of a cube is drawn as follows:
 Views of an object
There are three possible views of an object. They are elevation view, plan view and side
view.
The given solid, when viewed from the given directions, gives
as the elevation view
as the plan view
as the side view
 Perspective is an illusion of a three-dimensional nature on a two-dimensional surface,
mostly by giving the illusion of depth. It is one of the techniques for representing threedimensional objects on a two-dimensional surface.
 Drawing perspective
Let us draw the perspective of an L-shaped wooden block whose front view is as follows:
Take a paper and draw a horizontal line at the top of the page and mark a point on the
horizontal line. Then, draw the given shape below that line as follows:
Now, join 5 corners of the given shape to the point.
Between such joining lines, draw the lines that are parallel to the sides of the front view
of the given figure as follows:
Clearly, here we have to draw 4 such lines. Also, it can be noticed that these are either
horizontal lines or vertical lines. However, it is not true always. There may be some cases
when the lines are neither horizontal nor vertical. Now erase the dotted lines as shown
below:
This is the perspective view of the given shape.
Chapter 14: Perimeter, Area and Volume
 Perimeter
The perimeter of a closed figure is the length of the boundary of the figure.
Example: Find the perimeter of the given figure.
Solution:
Perimeter of the given figure = (2 + 7 + 4 + 1 + 1 + 3 + 1 + 3) cm = 22 cm
 Perimeter of a rectangle
 The perimeter of a rectangle is found by using the formula, perimeter = 2 (length +
breadth).
 When the perimeter and the length of a rectangle are known, its breadth can be found
1
by using the formula, breadth = perimeter – length.
2
 When the perimeter and the breadth of a rectangle are known, its length can be found
1
by using the formula, length = perimeter – breadth.
2
 Perimeter of a square
The perimeter of a square is found by using the formula, perimeter = 4 × side.
 Area
 The area of a closed figure is the amount of surface enclosed by the figure.
The units of area are square cm, square m etc.
 The concept of area is widely used in our daily life. For example, for finding the area
of the carpet required for covering the floor etc.
 We can estimate the area of a surface by drawing it on a square graph paper, where
every square measures 1 cm  1 cm. For this, we have to adopt the following
conventions.
o The area of 1 full square is taken as 1 square unit.
o The area of a region which is more than half the square is taken as 1 square unit.
1
o The area of half the square is taken as
square unit.
2
o We have to ignore the portions of area that are less than half-a-square.
Example: The given figure shows a shape drawn on a squared paper. Here, the
side of each small square is 1 cm.
Find the area enclosed by the given shape.
Solution:
The given figure covers 17 complete squares and 8 half squares.
Thus, area enclosed by the given figure = (17 + 8  2) sq. cm = 21 sq. cm

Area of a rectangle
The area of a rectangle is calculated using the formula, area = length × breadth.

Area of a square
The area of a square is calculated using the formula, area = side × side.
 Volume
Volume of a solid is how much three-dimensional space it occupies.
Suppose we have to find the volume of the figure shown below.
We can find that the given figure consists of 10 small cubes
Hence, the volume of this figure is 10 cm3.
.
Chapter 15: Data Handling
 Data can be represented in the form of tables, pictographs, and bar graphs.
 Data is formed so as to analyse the information regarding a particular object easily.
 Table graph
The title of a table indicates what the table is about whereas the column heading indicates
what the data is in the column.
Production of wheat and sugar in last 5 years in a village
Year
Wheat (in tons)
Sugar (in tons)
2004
502
480
2005
546
540
2006
588
581
2007
625
672
2008
682
740
 Pictographs
In pictographs, a key is used to represent the data. The title of pictograph indicates what
the pictograph is about.
Bikes sold in last 5 months at a particular showroom
Month Number of bikes
April
40
May
60
June
140
July
160
August 200
Bikes sold in last 5 months at a particular showroom
April
May
June
July
August
Each
stands for 20 bikes
 Bar Graphs
 Title of a bar graph, always given on top, indicates the information being represented
in the graph.
 There are two axes, vertical and horizontal, in each bar graph. The labels on each axis
represent the type of information plotted on them. One axis represents the group of
data and the other represents the values of data.
 Scale is the range of values being represented along both axes.
 Bars are rectangular blocks of uniform width with equal spacing between them.
Each bar represents only one value for the given data. Their base may be either on
horizontal axis or on vertical axis. On this basis, there are two types of bar graphs –
horizontal bar graph and vertical bar graph.
 Following steps are taken to draw a bar graph:
o Put title of bar graph on the top of graph.
o Draw two perpendicular lines as axes and give a suitable label for it.
o Choose a suitable width for each bar and a suitable spacing between 2 consecutive
bars.
o Choose a suitable scale to represent the value of each bar and draw the bars
according to the heights calculated.
Class
VI
VII
VIII
IX
X
Number of boys
35
40
30
40
50
Number of girls
15
20
35
10
40
Number of students in different classes of a school

Double bar graphs
Double bar graphs (double column graphs) are also drawn as bar graphs. It is the
collection of two sets of data on the same graph. It is helpful in comparing two sets of
data.
Example: The given data shows the revenue
A and B, in 6 months.
January
February
March
16
24
A 28
18
30
B 32
generated (in lakhs) by two companies,
April
22
31
May
38
26
June
20
26
Construct a double bar graph representing the given data. Also, in which particular
month is the revenue generated by company A more than that generated by company
B.
Solution: In the given data, the lowest value of observation is 16. Therefore,
choosing the scale as 1 unit = 4 lakhs and drawing bars of corresponding lengths for
each month, we obtain the bar graph as follows:
It can be seen that the bar for the month of May is higher for company A as compared
to company B. Therefore, company A generated more revenue in the month of May
as compared to company B.
 Pie charts
A pie chart or a circle graph shows the relationship between a whole and its parts.
For example: Consider the given pie chart which shows the favourite colours of the
class-VIII students of a school.
In this pie chart, the portion of the sector for the colour red is given by,
Number of students whose favourite colour is red
Total number of students
45

80
9

16
th
9
Therefore, the sector representing red colour is   part of the circle.
 16 
 Construction of pie charts
Example: Construct a pie chart for the following data which gives the brands of
laptop preferred by the people of a locality.
Brand A :
100
Brand B :
120
Brand C :
180
Solution:
The total number of people is 100 + 180 + 120 = 400.
We can form the following table to find the central angle of each sector:
Brand
of
Central angle
Number of people
Fraction
laptop
100 1
1

 360  90
A
100
400 4
4
120 3
3

 360  108
B
180
400 10 10
180 9
9

 360  162
C
120
400 20 20
Steps of construction:
(a) Draw a circle with any convenient radius. Let O be the centre of the circle and
OX be its radius.
(b) Draw the angle of the sector for brand A, which is 90. Using protractor, draw
XOY  90 .
(c) Now, draw the angle of the sectors for brands B and C.

Interpretation of pie charts
The given pie chart shows the foot wears preferred by the people of a locality.
From the above pie chart, we can infer that most people of the locality prefer wearing
leather foot wears. Also, we can infer that the least number of people prefer wearing
plastic foot wears.
Now, suppose the total number of people in the locality is 1000. Then, we can say
that the number of people who prefer wearing rubber foot wears is
30
1000  300
100