Prime Math 6 - Final rabi_final

Kinds of Sets
Cardinal number of a set
A = {a, e, i, o, u}

Set A contains five elements. So, we say the cardinal number of set A is 5. It is
denoted by n(A) = 5.

Thus, the number of distinct elements of a set is called its cardinal number.

For example

B = {cow, cat, god, goat} \ n(B) = 4

C = {1, 2}, \ n(C) = 2

Empty set
Consider the sets

A = {x : x is a boy student of St. Mary's School }

B = {x : x is a girl with three eyes}

C = {x : x is a girl of height 9 feet}

All the set A, B and C contain no element because there is no boy student in St.
Mary's School, there is no girl with three eyes and there is no girl with height 9 feet.

Thus, a set containing no element is called an empty set or a null set or a void set.
It is denoted by { } or f.

Note: {f} is not a null set because it is a set containing an element.

Singleton set

Consider the sets

A = {2} B = {boy} C = {lion}

Each set A, B and C are containing only one element. Such sets are called singleton

set.

Thus, a set containing only one element is called a singleton set.

Finite set

A = {1, 2, 3, 4, 5}
The set A is containing five elements.
B = {x : x is a district of Nepal}
The set B is containing 75 elements.

Prime Mathematics Book - 6 95

The sets A and B are called finite sets because they contain finite number of elements.
Thus, a set containing finite number of elements is called a finite set.

Infinite set
Consider the sets
N = {1, 2, 3, …….}
The set N is the set of natural numbers containing infinite number of elements.
P = {2, 3, 5, 7, …..}
The set P is the set of prime numbers containing infinite number of elements. Such
sets are called infinite sets.
Thus, a set containing infinite number of elements is called infinite set.

Equal sets
Observe the following examples:
A = {g, o, d} and B = {d, o, g}
The sets A and B are containing the same elements. Such sets contain same and
equal number of elements. Such sets are called equal sets.
Thus, the sets containing the same elements are called equal sets.

Equivalent sets
Inspect the following sets.
A = {1, 2, 3, 4, 5} and B = {a, e, i, o, u}
The cardinal number of set A, n(A) = 5.
The cardinal number of set B, n(B) = 5.
As the sets A and B contain equal number of elements, they are called equivalent
sets.
Thus, the sets containing equal number of elements are called equivalent sets.

Note: The equal sets are equivalent sets but the equivalent sets are not necessarily equal
sets. The equivalent sets A and B are represented by A B.

96 Prime Mathematics Book - 6

Disjoint sets

Observe the following examples. The sets A and B have no common elements. Such
sets A and B are called disjoints.

AB

13 24 6
57 8 10

The set containing no common elements are called disjoint sets.

Overlapping sets
The set A and B have one common element 2. Such sets are called overlapping sets.
Thus, the set containing at least one common element are called overlapping sets.

AB

35 2 46 A B
78
abcd aei
Subset efg ou
..... z
All the elements of set B are in the set A.
So, set B is called the subset of set A. It is

denoted by B Ì A and read as B is a subset

of A or 'B is contained in A'.

If the set is not a subset, we use the symbol Ë. An empty set is a subset of any set.
If A = {1, 2, 3}, we can form the subsets as f, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},

{1,2,3}.

Consider a set B = {a,b}, its subsets are f, {a}, {b}, {a,b}

Universal set

Observe the following examples:

A = {1, 3, 5, 7, 9} B = {2, 3, 5, 8, 10}

C = {2, 3, 5, 7, 11} U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ………… , 20}

Here, the sets A, B and C are the sub-sets of the set U. the set U is called universal
set and it is denoted by U .

Thus, the set from which the elements are selected to form subset is called the
universal set.

Prime Mathematics Book - 6 97

Exercise 5.3

1. State whether the following are finite or infinite sets:
(a) {2, 4, 6, ………………….. , 20}
(b) {vowels of English alphabets}
(c) {days of a week}
(d) {stars in the sky}
(e) {months of a year}
(f) {set of points in a line segment}
(g) {1, 2, 3, . . . . . . }
(h) {1, 2, 3, 4, 5}

2. State whether the following sets are equal or equivalent sets.
(a) A = {letters of the word "tap"}, B = {letters of the word "pat"}
(b) A = {a, e, i, o, u}, B = {3, 5, 7, 9, 11}
(c) A = {x : x is an even prime number}, B = {2}
(d) A = {letters of the word "set"}, B = {letters of the word "god"}
(e) A = {a, b, c, d}, B = {2, 4, 6, 8}
(f) A = {x : x is a prime number less than 9}, B = {2, 3, 5, 7}
(g) A = {letters of the word "bat"}, B = {letters of the word "tab"}

(h) A = {x : x Î N; x < 6}, B = {1, 2, 3, 4, 5}

(i) A = {1, 3, 5, 7, 9}, B = {x : x is a natural odd number less than 10}
(j) A = {letter of the word "tub"}, B = {letter of the word "but"}
(k) A = {2, 4, 6, 8, 10}, B = {1, 3, 5, 7, 9}

3. State the whether the following are empty or singleton sets.
(a) {A boy with four hands.}
(b) {Banana}
(c) {Men who have travelled to the sun}
(d) {A prime number less than 2}
(e) A = { o }
(f) The set of deepest ocean of the world
(g) The set of man who discovered India
(h) A = {x : x is a whole number between 1 and 2}
(i) B = {x : x is a boy in a girls school}
(j) The set of highest mountain of the world

98 Prime Mathematics Book - 6

4. State whether the following pairs of sets are overlapping or disjoint.

(a) A = {1, 3, 5, 7}, B = {2, 4, 6, 8} (b) A = {a, b, c, d}, B = {b, c, d, e}

(c) A = {p, q, s, t}, B = {m, n, p, r} (d) A = {2, 4, 5}, B = {1, 3, 9, 11}

(e) A B (f) A B

5. State whether the following are the subsets of the set
A= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(a) C = {2, 4, 6, 8} (b) B = {2, 3, 5, 7, 11}

6. In the folowing pair of sets, which one is universal set of subset ?
(a) W = {whole number less than 16} and
A = {even number less than 16}
(b) C = {0, 1, 2, .......... 20} and
A = {prime numbers less than 20

7. Write the possible subsets of the following sets
(a) {1}, (b) {a,b}

Unit Revision Test I

1. Write the elements of the following sets and name the sets

(a) The set of English months of a year starting from first letter J.

(b) The set of five fruits name

Î Ï2. Use the correct symbol or in the gaps.

(a) 5 ………………. {set of prime numbers}

(b) 4 ………………. {set of odd numbers}

3. Write the following sets in the Roaster method.

(a) The set of first four cube numbers.

(b) The set of letter of the word cabbage.

4. State whether the following sets are equal or equivalent sets

(a) A = {4, 5, 6} and B = {p, q, r}

(b) A = {letters of the word ‘pot’} and B = {letters of the ‘top’}

5. Write the possible subsets of the following sets

(a) {5, 6} (b) {a, b, c}

6. Write one example of the following sets.

(a) Empty set (b) Singleton set (c) Disjoint set (d) Overlapping set

Prime Mathematics Book - 6 99

Estimated periods – 28

Objectives

At the end of this unit, the students will be able to:

define whole numbers and perforn the operations involved with brackets.
apply divisibility rule.

know prime and composite numbers their factors and multiples.
find the squares and square roots and cubes and cub roots of whole numbers.

find the H.C.F. and L.C.M. of whole numbers.

define integers. 120

be acquainted with rational numbers.

2 60

(+6) 2 30
(-4) (+2)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 15

\ (-4) + (+6) = +2 3 5

\ 120 = 2 × 2 × 2 × 3 × 5

Teaching Materials Activities

Multiplication tables, tables of squares It is better to:
and cubes e.t.c. clearify the students about different
numbers with the help of closure
property.
discuss operations involved with the
numbers.

Whole Numbers Estimated periods – 22

To count a collection of objects, we use the numbers 1, 2, 3, …….. . Such counting
numbers are called natural numbers. The set of natural numbers is represented by N.

N = {1, 2, 3, 4, …….}

From the above example, it is clear that 1 is the smallest natural number. For every
natural number there is another natural number which is just one more. This natural
number is called successor. There are no largest natural number. So, the set of
natural number is infinite.

Let’s take the natural numbers 3 and 4. The sum of these two numbers 3 and 4 is 7
which is also a natural number. Thus, the sum of two natural numbers is again a
natural number. But if we subtract 4 from 4, the difference is 0 which is not a natural
number. Thus, the difference of two natural numbers may not be a natural number.
If we include 0 to the set of natural number, the set is called whole number. The
set of whole numbers is denoted by W

W = {0, 1, 2, 3, ……}

The whole number can be represented in a number line as

01234

Note: - 0 is the smallest whole number.
- There is no largest whole number.
- The set of whole numbers is infinite.
- ForÐeDve^ry®w\hole number there is a number greater than the previous one by one
more.
Therefore, every whole number has its successor.

Order of operation:

The four basic operations in mathematics are addition, subtraction, multiplication
and division. If the mathematical sum involves different operations, then the result
depends in the order in which operations are performed.

Prime Mathematics Book - 6 101

Lets see an example. 6×7+8
6×7+8

= 42 + 8 = 6 × 15

= 50 = 90

Now, the question arise that which one is correct?
To avoid the confusion, we have to perform the operations in the following order.

B First perform the operations in brackets.
O Perform 'of' which is ‘X’
D Perform division
M Perform multiplication
A Perform Addition
S Perform subtractions
This rule is known as 'BODMAS' rule.

Example: Simplify:

6+4-5 [First perform 'addition' then perform 'subtraction'.]
Solution: 6 + 4 - 5

= 10 - 5
=5
Example: Simplify:

6-5+4
If the operation addition and subtraction involves in a sum. It is not necessary
to perform addition first according to BODMAS rule. Perform in the order in
which they appear from left to right.

Solution: 6 - 5 + 4

=1+4 [Perform subtraction first]

= 5 [Perform addition]

Example: Simplify:

7 x 10 ÷ 2
Solution: 7 × 10 ÷ 2

=7×5 [Perform division first]

= 35 [Perform multiplication]

102 Prime Mathematics Book - 6

Also, 7 × 10 ÷ 2

If the sum involves operation of multiplication and division, it is not necessary
to perform division first according to the BODMAS rule. We can perform the
operations in order in which they appear from left to right.

7 × 10 ÷ 2

= 70 ÷ 2 [Perform multiplication]

= 35 [Perform division]

Simplifications involving brackets:
When any operation is to be performed first, we use brackets. There are three types
of brackets.

( ) are called small brackets or round brackets.
{ } are called curly brackets or braces.
[ ] are called square brackets or large brackets.
If a sum includes all the three types of brackets
i. First perform for small brackets.
ii. Then perform for braces and
iii. Then finally for the large brackets.

Example: simplify [Operate inside the brackets]
Solution: 2 + 5 × (6 - 4) [Multiply]
2 + 5 × (6 - 4) [Add]
Example: =2+5×2
Solution: = 2 + 10
= 12

Simplify

4 + 2 [5 ÷ 5 + {2 + 6 + (4 ÷ 2)}]

4 + 2 [5 ÷ 5 + {2 + 6 + (4 ÷ 2)}]

= 4 + 2 [5 ÷ 5 + {2 + 6 +2}] [Operation inside the small bracket.]

= 4 + 2 [5 ÷ 5 + 10] [Operation inside the braces]

= 4 + 2 [1 + 10] [Brackets stand for multiplication]
= 4 + 2 × 11
= 4 + 22 = 26

Prime Mathematics Book - 6 103

Exercise 6.1

1. Simplify: (b) 12 ÷ 6 + 5 × 2 - 7
(a) 4 × 2 ÷ 2 - 1 + 5 (d) 200 - 50 ÷ 5 × 4 + 7
(c) 5 - 7 - 1 + 8 × 9
(b) 11 + 8 ÷ 4 + 2 of 6
2. Simplify: (d) 200 ÷ 5 + 4 × 6 - 5

(a) 99 ÷ 11 + 5 × 8 - 6 (b) 5 + 60 ÷ 4 - 7 of 5
(c) 60 ÷ 12 + 9 ÷ 3 × 5 - 7 (d) 90 + 30 ÷ 6 - 7 × 5

3. Simplify: (b) 10 ÷ 5 of 2 + 6 - 3 of 5
(a) 12 × 5 + 25 ÷ 5 + 4 (d) 200 ÷ 50 - 6 of 7 + 8
(c) 15 × 5 ÷ 5 + 20 - 4
(b) 14 ÷ 2 - [17{14 - (15 ÷ 3)}]
4. Simplify: (d) 3 - [12 ÷ {25 + 4(7 - 8 - 6)}]
(f) 14 - [15 of 2 - {84 ÷ 6 (8 - 3 + 2)}]
(a) 50 of 5 ÷ 10 + 3 - 7
(c) 60 ÷ 12 + 5 of 8 - 5

5. Simplify:
(a) 3 - {12 ÷ 3 + (-4)(-2) + 5}
(c) 14 of 2 [70 ÷ 5{12 - (3 + 2)}]
(e) 30 ÷ {25 - 3(10 - 3 + 2)}]

104 Prime Mathematics Book - 6

Odd numbers
Observe the examples.

2) 9 (4 2) 13 (6 2) 19 (9
-8 -12 -18
1 1 1

In above examples, the remainder is 1 in each case. The natural numbers 9, 13 and
19 are not exactly divisible by 2. Such numbers are called odd numbers. Some
examples of odd numbers are 11, 15, 17, etc.

Even numbers
Observe the examples.

2) 14 (7 2) 18 (9 2) 30 (15

-14 -18 -30

00 0

In the above examples, the natural numbers 14, 18 and 30 are exactly divisible by
2. Such numbers are called even numbers. Some other examples of even numbers
are 4, 8, 12, etc.

Prime numbers:
Observe the examples.

1) 7 (7 7) 7 (1

-7 -7
00

The natural number 7 is exactly divisible by 1 and by itself only. Such natural numbers
are called prime numbers. Some other examples are 2, 3, 5, etc. 1 is not c o n s i d e r e d
as a prime number. 2 is the only even number which is a prime number.

Composite n u m b e r s
Observe the example

1) 6 (6 6) 6 (1 2) 6 (3 3) 6 (2
-6 -6 -6 -6
0 0 0 0

Prime Mathematics Book - 6 105

The natural number 6 is exactly divisible by 1, by itself and by other numbers 2 and
3. Such natural numbers are called composite numbers. Some other examples are
4,8,9,10,12 etc.

Square numbers:

Look at the examples

2 × 2 = 4, 5 × 5 = 25, 8 × 8 = 64

The natural numbers like 4, 25 and 64 are called square numbers. Thus, any natural
number which is a product of two same numbers is called a square number. Some
other examples are 1, 9, 16, 36, etc.

Cube numbers: 2 × 2 × 2 = 8, 6 × 6 × 6 = 216
Look at the examples:
1 × 1 × 1 = 1,

The natural numbers like 1, 8 and 216 are called cube numbers. Thus, any natural
number which is a product of three same numbers is called a cube number. Some
of the other examples are 27, 64, 125, etc.

Test of divisibility:

Divisibility test of 2:

Look at the examples.

20, 104, 256, 482, 698, 980

The digit in the unit place of the above numbers are 0, 4, 6, 2, 8. They are exactly
divisible by 2. Thus, if the digit in unit place is either 0 or even number, the number
is exactly divisible by 2.

Divisibility test of 3:

If the sum of the digits involved in a natural number is exactly divisible by 3, then
the natural number is also exactly divisible by 3.

312, 6723, 927, 840

3 + 1 + 2 = 6, 6 + 7 + 2 = 15, 9 + 2 + 7 = 18, 8 + 4 + 0 = 12

The sum of the digits used in the above numbers are exactly divisible by 3. So, the
given numbers are also exactly divisible by 3.

106 Prime Mathematics Book - 6

Divisibility test of 4:
If the number formed by the last two digits of the number is divisible by 4, the
number is divisible by 4. Observe the example:
160, 948, 1032, 2416, 324, 5112, 920
In the above examples, the number formed by the last two digits are exactly divisible
by 4. So, these numbers are exactly divisible by 4.

Divisibility test of 5:
If the ones digit is either 0 or 5, the number is exactly divisible by 5.
Look at the examples,
150, 265, 375, 480, 1920, 5430, 8945
As the ones digit of the given numbers are either 0 or 5, the number are exactly
divisible by 5.

Divisibility test of 6:
The numbers exactly divisible by 2 as well as by 3 are also exactly divisible by 6.
Look at the examples.
534, 1056, 2862, 3168, 6132
The above numbers are exactly divisible by 2 as well as by 3. So, they are exactly
divisible by 6 also.

Divisibility test of 7:
If the difference between two times the units digit and the number formed by the
remaining digits is divisible by 7, then the whole number is also divisible by 7.

Look at the example.
In 245, the digit in unit place is 5. Two times 5 is 10. Let's subtract the two times
of unit digit place from the number formed by the remaining digits.
24 - 10 = 14
The difference 14 is divisible by 7. So, 245 is divisible by 7.

Prime Mathematics Book - 6 107

Divisibility test of 8:

If the number formed by the last three digits is divisible by 8, then the number is
exactly divisible by 8.
Look at the example: 7248.
The number formed by the last three digits is 248.

8 ) 248 (31
-24
8
-8
0

As the number 248 formed by the last three digits of 7248 is exactly divisible by 8.
So, the number 7248 is also divisible by 8.

Divisibility test of 9:
If the sum of the digits of a number is divisible by 9, the number is divisible by 9.

Observe the example: 234
The sum of the digits of 234 = 2 + 3 + 4 = 9. It is divisible by 9. So, 234 is also divisible
by 9.

Divisibility test of 10:
If the ones digit of a number is 0, the number is divisible by 10.
Observe
3140, 5670, 8940, 9850, 1230
The unit digit of the above numbers is 0 so they are divisible by 10.

Divisibility test of 11:
If the difference of the sum of the alternate digits is a multiple of 11, then the
number is divisible by 11.
Observe;

8371
The up-up digit are 8 and 7. Their sum = 8 + 7 = 15. The down-up digits are 3 and
1. Their sum 3 + 1 = 4.
15 - 4 = 11. The number 11 is divisible by 11. Therefore, 83 71 is divisible
by 11.

108 Prime Mathematics Book - 6

Exercise 6.2

1. Write the odd numbers between 250 and 260.
2. Write the even numbers between 100 and 120.
3. Write the prime numbers between 20 and 30.
4. Write the composite numbers between 60 and 70.
5. Write the first five square numbers.
6. Write the first six cube numbers.
7. Separate the odd numbers and even numbers from the following numbers.

214, 310, 413, 618, 759, 1041, 3052, 8045
8. Separate the prime numbers and composite numbers from the following

numbers.

17, 21, 31, 91, 13, 52, 19, 60, 23, 29, 58, 99

9. Identify the square numbers and cube numbers.

16, 49, 27, 324, 64, 729, 1000, 144, 2744, 125.

10. Which of the following numbers are divisible by 2?

(a) 1058 (b) 395 (c) 4050 (d) 692
3000
(e) 5064 (f) 1386 (g) 2931 (h)
5409
11. Which of the following numbers are divisible by 3? 1983

(a) 357 (b) 1200 (c) 3421 (d)

(e) 1206 (f) 1234 (g) 3957 (h)

Prime Mathematics Book - 6 109

Factors and Multiples:

Look at the examples.

(a) 1 ) 3 2 ( 3 2 (b) 2 ) 3 2 ( 1 6 (c) 4 ) 3 2 ( 8 (d) 8 ) 3 2 ( 4
-32 -32
-3 -2 0 0

02 12

-2 -12

00

(e) 1 6 ) 3 2 ( 2 (f) 3 2 ) 3 2 ( 1
-32 -32
0 0

The numbers 1, 2, 4, 8, 16 and 32 divide 32 leaving no remainder. So, these numbers
are factor of 32. So, the factors of any number are those numbers which divide the
number without leaving any remainder.

Similarly, the factors of 50 are 1, 2, 5, 10, 25 and 50.

Example: Write down the factors of 25.
Solution: Here, 1 × 25 = 25

5 × 5 = 25
25 × 1 = 25
\ The factors of 25 are 1, 5 and 20.

Prime factors 2 × 12 = 24
Let’s find factor of 24. 4 × 6 = 24
1 × 24 = 24 8 × 3 = 24]
3 × 8 = 24 24 × 1 = 24
6 × 4 = 24
12 × 2 = 24

Thus, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. Among these numbers 2 and
3 are prime numbers. They are called prime factors of 24.

110 Prime Mathematics Book - 6

Example: Find the prime factors of 40.
Solution:
The factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40. Among these factors
2 and 5 are prime numbers. So, 2 and 5 are the prime factors of 40.

Multiplies:

Look at the examples.

4 × 1 = 4, 4 × 2 = 8, 4 × 3 = 12, 4 × 4 = 16,

4 × 5 = 20, 4 × 6 = 24, 4 × 7 = 28, 4 × 8 = 32.

The numbers 4, 8, 12, 16, 20, 24, 28 and 32 are multiples of 4. Thus, to obtain the
multiples of any number, we have to multiply it by a natural number.

Example: Find the first five multiples of 6.
Solution: 6 × 1 = 6

6 × 2 = 12
6 × 3 = 18
6 × 4 = 24
And 6 × 5 = 36
\ Thus, the first five multiples of 6 are 6, 12, 18, 24 and 30.

Exercise 6.3

1. Write the factors of the following numbers.

(a) 20 (b) 40 (c) 60 (d) 75

2. Write the first five multiplies of the following numbers.

(a) 8 (b) 12 (c) 15 (d) 22

(e) 25 (f) 30 (g) 16

3. Find the prime factors of the following numbers.

(a) 50 (b) 75 (c) 110 (d) 150

Prime Mathematics Book - 6 111

Prime factorization:
Look at the example:
Factor tree method

120

2 60 i. 2 is a prime and 60 is a composite
2 30 which can be further factroized.

ii. 30 is a composite

2 15 iii. 15 is a composite

3 5 iii. 3 and 5 both are prime

\ 120 = 2 × 2 × 2 × 3 × 5

The process of factorizing a composite number in the form of product of prime
factors is called prime factorization. The method shown above is called factor tree
method. Observe this example.

Division method i) Divide by the possible smallest prime number.
ii) Divide till we get a prime number.
2 120
2 60
2 30
3 15
5

\ 120 = 2 × 2 × 2 × 3 × 5
The method shown above is called division method.

112 Prime Mathematics Book - 6

Exercise 6.4

1. Factorize the following numbers by using division method:

(a) 132 (b) 60 (c) 576 (d) 425

(e) 925 (f) 1215

2. Factorize the following numbers by using factor tree method.

(a) 84 (b) 90 (c) 140 (d) 180

Squares and square roots, cubes and cube roots

Squares and square roots

Observe the following.

1×1=1 1 is called square of 1

2×2=4 4 is called square of 2

3×3=9 9 is called square of 3

4 × 4 = 16 16 is called square of 4

Thus, when two identical numbers are multiplied then the square of that number
is obtained. Therefore, 1, 4, 9 and 16 are called square numbers.
Let’s, find the prime factorization of 18.

2 18
39
3

\ 18 = 2 × 3 × 3

2 has no pair. So, 18 is not a square number.
Lets see whether 81 is a square number or not.

Prime Mathematics Book - 6 113

3 81
3 27
39
3

\ 81 = 3 × 3 × 3 × 3

There is no number without pair. So, 81 is a square number.

Square roots

6 × 6 = 36. Here 36 is a square of 6 and 6 is a square root of 36.

Similarly,

12 × 12 = 144, Here, 144 is a square of 12 and 12 is a square root of 144.

Now, 4 × 4 = 42 It is read as four squared.

5 × 5 = 52 It is read as five squared.

6 × 6 = 62 It is read as six squared.

The above method is also a method to express the square number

Let’s find the square root of 324.

2 324
2 162
3 81
3 27
39
3

\ 324 = 2 × 2 × 3 × 3 × 3 × 3

Square root of 324 = 324
= 2×2×3×3×3×3
= 2×3×3
= 18

From a group of two numbers, one number is taken out.

114 Prime Mathematics Book - 6

Cubes:

See the example:

1 × 1 × 1 = 1; 1 is called cube of 1.

2 × 2 × 2 = 8; 8 is called cube of 2.

3 × 3 × 3 = 27; 27 is called cube of 3 and so on.

Thus, when three identical numbers are multiplied, we obtain the cube number of
that number. The numbers 1, 8 and 27 are called cube numbers.

Also, they can be represented as

1 × 1 × 1 = 13; It is read as one cubed.

2 × 2 × 2 = 23; It is read as two cubed.

3 × 3 × 3 = 33; It is read as three cubed.

Example 1: Find the cube number of 13.
Solution: The cube of 13 = 13 × 13 × 13 = 133 = 2197

Example 2: Is 250 a cube number?
Solution: Here,

2 250
5 125
5 25
5

\ 250 = 2 × 5 × 5 × 5
As 2 is a single number, 250 is not a cube number.

Cube root:
We know,
2 × 2 × 2 = 23 = 8
8 is the cube of 2 and 2 is cube root of 8. 23 is read as two cubed. It is also written
as 3 8 = 2.

Prime Mathematics Book - 6 115

Similarly,
3 × 3 × 3 = 33 = 27. So 27 is the cube of 3 and 3 is cube root of 27.
\ 3 27 = 3

Example: Find the cube root of 64.
Solution:
Here,
From a group of three twos, one
2 64 two is taken out.
2 32
2 16
28
24
2

\ 64 = 2 × 2 × 2 × 2 × 2 × 2

3 64 = 3 2 × 2 × 2 × 2 × 2 × 2
=2×2

=4

Exercise 6.5

1. Evaluate: (b) 102 (c) 122 (d) 142
(a) 82 (b) 103 (c) 133 (d) 153
(e) 162 (d) 15

2. Evaluate:
(a) 83
(e) 183

3. Find the square of the following numbers.

(a) 9 (b) 11 (c) 13

(e) 17

116 Prime Mathematics Book - 6

4. Find the cube of the following numbers.

(a) 9 (b) 11 (c) 12 (d) 14

(e) 16

5. Find the square root of the following number by prime factor method.

(a) 196 (b) 256 (c) 576 (d) 625
(e) 900 (f) 1225

6. Find the cube root of the following numbers by prime factor method.

(a) 512 (b) 1728 (c) 3375 (d) 4096

7. Find the least number when multiplied to the following numbers make them
a square number.

(a) 18 (b) 48 (c) 98 (d) 45

8. If the area of a square room is 196m2, find the length of the room.

9. A certain number of people collected Rs. 11025 for get a together. Each
person contributed as many rupees as there were people. Find the number
of people.

10. An instructor wanted to arrange 1225 students into a square form making
equal number of rows and columns. Find the number of students in the
front row.

Prime Mathematics Book - 6 117

Highest common factor (H.C.F.)

See the examples.
4=2×2
6=2×3

The common factor of 4 and 6 is 2. Thus, a common factor of two or more numbers
is a number which divides each of them exactly.
Let’s take another example:

Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

The common factor of 18 and 24 are 1, 2, 3 and 6 in which 6 is the highest common
factor of 18 and 24. Thus, the H.C.F. of two or more numbers is the highest number
which divides each of them exactly.
There are mainly three methods for finding the highest common factor of numbers.

i. Set of factors method
ii. Prime factor method
iii. Division method

Set of factors method:
Let’s find the H.C.F. of 15 and 20.
Set of factors of 15 = F(15) = {1, 3, 5, 15}
Set of factors of 20 = F(20) = {1, 2, 4, 5, 10, 20}
Set of common factors of 15 and 20 = {1, 5}
Here, the highest common factor is 5. So, the H.C.F. of 15 and 20 is 5.

Prime factor method:
Let’s find the H.C.F. of 20 and 25.

20 = 2 × 2 × 5
25 = 5 × 5
\ H.C.F. = 5

118 Prime Mathematics Book - 6

Lets find H.C.F. of 20, 30 and 40.

2 20 2 30 2 40
2 10 2 15 2 20
5 5 2 10
5

20 = 2 × 2 × 5
30 = 2 × 3 × 5
40 = 2 × 2 × 2 × 5

\ H.C.F. = 2 × 5 = 10

Let’s find H.C.F. of 15 and 16.

3 15 2 16
5 28
24
2

15 = 3 × 5
16 = 2 × 2 × 2 × 2
They do not have common prime factor. So, their H.C.F. = 1.
Such numbers which do not have common prime factor are called co-prime numbers.

Exercise 6.6

1. Find the H.C.F. of the following numbers by set of factor method.

(a) 6, 8 (b) 10, 15 (c) 20, 30

(d) 12, 15, 30 (e) 24, 36, 60 (f) 25, 45, 75

Prime Mathematics Book - 6 119

2. Find the H.C.F. of the following numbers by prime factor method.

(a) 8, 12 (b) 15, 25 (c) 30, 40

(d) 24, 32 (e) 18, 30, 36 (f) 24, 42, 60

(g) 24, 36, 48

3. Find the highest number which div ide 45 and 60 exactly.

4. Find the largest number which divides 30, 50 and 60 exactly.

5. Find the greatest number of students to whom 120 oranges and 180 mangoes
can be divided equally?

6. What is the maximum number of students to whom 75 copies and 100
pencils can be distributed equally?

Lowest common multiple (L.C.M.)
Let’s take the numbers 4 and 6.
Set of multiplies of 4 = {4, 8, 12, 16, 20, 24, ……}
Set of multiplies of 6 = {6, 12, 18, 24, 30, …….}
Set of common multiplies of 4 and 6 = {12, 24, …….}

Here, the least common multiple of 4 and 6 is 12. Therefore, 12 is the L.C.M. of 4
and 6. Thus, the lowest common multiple of two or more numbers is the smallest
number which is exactly dividable by each of the given numbers.
There are mainly three methods to find L.C.M.
1. Set of multiples method
2. Prime factor method
3. Common division method

120 Prime Mathematics Book - 6

1. Set of multiplies method:
Let’s find the L.C.M of 5 and 8 by set of multiples method.
Set of multiples of 5 = {5, 10, 15, 20, 25, 30, 35, 40, 45, ……..}
Set of multiples of 8 = {8, 16, 24, 32, 40, 48, 56, ……….}
Set of common multiples of 5 and 8 = {40, ………….}
Thus, the L.C.M. of 5 and 8 is 40.
Let’s find the L.C.M of 2 and 4.
Set of multiples of 2 = {2, 4, 6, 8, 10, 12, ………..}
Set of multiples of 4 = {4, 8, 12, 16, 20, 24, ……………}
Set of common multiples of 2 and 4 = {4, 8, 12, ………..}
\ The L.C.M. of 2 and 4 is 4.

2. Prime factor method
Let’s find the L.C.M. of 6 and 15 by prime factor method.

2 6 3 15
35

6=2×3 First common prime factor and then the
15 = 3 × 5 remaining prime factors.

L.C.M = 3 × 2 × 5
= 30

Thus, 30 is the L.C.M. of 6 and 15.
Let’s find the L.C.M of 15, 20 and 24.

3 15 2 20 2 24
5 2 10 2 12
5 26
3
15 = 3 × 5
20 = 2 × 2 × 5 i. First find common prime factor of all the
24 = 2 × 2 × 2 × 3 three numbers.

L.C.M. = 2 × 2 × 2 × 3 × 5 ii. Find common prime factor from two numbers.

= 120 iii. Write the remaining prime factors.

Prime Mathematics Book - 6 121

3. Division method
Lets find the L.C.M of 10 and 15 by division method.

5 10, 15 [Both are divisible by 5.]
2, 3 [2 and 3 both are not division by the same number.]

\ L.C.M. of 10 and 15 = 5 × 2 × 3 = 30

Let’s find the L.C.M of 9, 12 and 15 by division method.

3 9, 12, 15 i. All three division by 3
3, 4, 5
ii. No any two are divisible by a single number so
division work is stopped.

\ L.C.M. = 3 × 3 × 4 × 5

= 180

Relation between H.C.F. and L.C.M.

Let's take two numbers 10 and 15.

2 10 3 15
5 5

10 = 2 × 5
15 = 3 × 5

\ H.C.F. of 10 and 15 = 5
\ L.C.M. of 10 and 15 = 5 × 2 × 3 = 30
\ H.C.F. × L.C.M. = 5 × 30 = 150
\ Product of two numbers = 10 × 15 = 150

H.C.F. × L.C.M. = product of two number

H.C.F. = Product of two number
L.C.M

L.C.M. = Product of two number
H.C.F

122 Prime Mathematics Book - 6

Exercise 6.7 (A)

1. Find the L.C.M. of the following set of number by set of multiples method.

(a) 2 and 3 (b) 3 and 4 (b) 4 and 5

(c) 6 and 7 (d) 6 and 9 (e) 2, 4 and 6

(f) 4, 8, 12 (g) 7, 14, 21 (h) 3, 6, 9

2. Find the L.C.M. of the following numbers by prime factor method.

(a) 18, 24 (b) 24, 30 (c) 15, 20

(d) 30, 40 (e) 12, 36, 72 (f) 10, 15, 20

(g) 32, 48, 64

3. Find the L.C.M. of the following numbers by division method.

(a) 10, 12 (b) 6, 8 (c) 9, 12

(d) 11, 33 (e) 22, 32 (f) 15, 20

(g) 7, 28, 35 (h) 6, 8, 12 (i) 8, 18, 27

4. Find the least number which is exactly divisible by 15, 20 and 25?

5. Three bells ring at the intervals of 6, 8 and 12 minutes respectively. If they
all ring at 2:00, at what time will they ring together a g a i n ?

6. Three bells ring at the intervals of 10, 15 and 20 minutes respectively.If
they all ring at 9:00 am at once, at what time will they ring together again?

7. Three measuring sticks are of 12 ft, 18 ft and 27 ft long. Find the shortest
length that can be measured exactly by any one of these sticks.

Prime Mathematics Book - 6 123

Exercise 6.7 (B)

1. Find the L.C.M of the following numbers using their H.C.F.

(a) 9, 12 (b) 10, 15 (c) 12, 18

(d) 20, 30 (e) 18, 24 (f) 36, 84

(g) 64, 72 (h) 96, 192 (i) 40, 48

(j) 45, 105

2. (a) The H.C.F. of 48 and 64 is 16, find their L.C.M.
(b) The H.C.F of 32 and 40 is 8, find their L.C.M.

3. (a) The L.C.M. of 24 and 30 is 120, find their H.C.F.
(b) He L.C.M. of 40 and 50 is 200, find their H.C.F.

4. (a) The product of two numbers is 216 and their H.C.F is 6. Find the L.C.M.

(b) If the product of two numbers is 320 and their H.C.F is 4, find the L.C.M.

5. (a) The H.C.F. and L.C.M. of two numbers are 9 and 108 respectively. If one
of the numbers is 27, find the other.

(b) The H.C.F. and L.C.M of two numbers are 4 and 168 respectively. If one
of the numbers is 24, find the other.

124 Prime Mathematics Book - 6

Integers Estimated periods – 2

The height of Mount Everest is 8848m above the sea level. The depth of Mariana
Trench is 10920 m below sea level. In some places, the temperature in winter season
is below 0oc. To represent such things distinctly, we need some other measures along
with whole numbers i.e. the opposite of them. Hence, the negative numbers are
introduced. Thus, -1 is the opposite of +1, -2 is the opposite of +2 and so on.

Zero is considered as neither negative nor positive number.

The set of natural numbers with zero and their negative numbers is called integers.
The set of integer numbers is denoted by I.

I = {……., -3, -2, -1, 0, 1, 2, 3, ……..}

The natural number 1, 2, 3, ………………….. are called positive integers and the negative
number ……., -3, -2, -1 are called negative integers.

The set of integers can be represented in a number line as

-4 -3 -2 -1 01 234 5
Negative integers
zero Positive integers

i. In this number line, the point O is called point of reference. The positive
integers are kept on the right of the point O and the negative integers are kept
on the left of the point O.

ii. The points are kept at equal intervals. It means that the distance between O
and -1, similarly the distance between -2 and -3, -3 and -4, 1 and 2 and so on
are equal.

iii. The arrow on both the ends of a number line indicates that the number line
continues to infinity in both directions.

iv. Zero is lesser than every positive integer.
e.g. 0 < 1 < 2 < 3 and so on.

v. Zero is greater than every negative integer.
e.g. 0 > -1 > -2 > -3 and so on.

Prime Mathematics Book - 6 125

vi. Every positive integer is greater than the negative integer.
e.g. 3 > -2, 4 > -4, 6 > - 3

vii. An integer on a number line is greater than the integer on its left.
e.g. 4 < 3, -2 < -3, 0 < -1, 5 > 4

Similarly, an integer on a number line is lesser than the integer on its right.
e.g. 3 < 4, -3 < -2, -1 < 0.

viii. The greater the number, the lesser is its opposite.
Thus, 8 > 5 but -8 < -5

ix. The lesser the number, the greater in its opposite.
e.g. 8 < 9 but -8 > -9

Exercise 6.8

1. Draw a number line and answer the following questions:

(a) Which is greater -4 or -6?
(b) Which is smaller -9 or 0?
(c) On which side of point of reference in the line, number smaller than zero

are kept?
(d) On which side of point of reference, the number greater than zero lies?

2. Draw a number line and with its help write all integers between.

(a) -3 and 3 (b) 0 and -5 (c) -4 and 0

(d) -2 and 1 (e) -3 and 5 (f) -4 and 6

3. Draw a number line and taking its help find which number is greater in each
pair.

(a) -2 or -4 (b) 0 or -3 (c) 3 or 5

(d) -3 or -5 (e) -7 or -12 (f) 0 or -4

4. With the help of number line, write the number lying 4 units to the left of

(a) 0 (b) 2 (c) -1

(d) -3 (e) 3 (f) -4

126 Prime Mathematics Book - 6

5. Insert the symbol < or > between the number.

(a) -3 _____ -5 (b) 0 _____ -2 (c) -4 _____ -6

(d) 3 _____ -4 (e) -5 _____ 6 (f) 6 _____ -7

6. Write true or false against each statement.

(a) -3 > 0 (b) 0 < -4 (c) -5 < 0
(f) -7 > -12
(d) -6 > 6 (e) 6 > -7

7. Arrange the following integers in ascending order.

(a) 0, -3, 4, -7, 8 (b) -3, 2, -6, 0, 4

(c) -6, 7, -10, 1, -20 (d) -2, 8, -30, -1, -25

(e) -4, -10, -8, -2, -5

8. Arrange the following integer in descending order.

(a) -2, 0, - 5, 6, 5 (b) -1, -4, 4, 9, -10

(c) -200, -50, 10, 20, -40 (d) -2, 40, -40, 30, 0

(e) -10, 10, -20, 20, -30 (f) 40, 50, -2, -45, -100

Prime Mathematics Book - 6 127

Addition of integers

Absolute value of an integer
The absolute value of -2 is 2. Similarly, the absolute of -4 is 4.
The absolute value of -6 is expressed as I-6I. It is equal to 6. Hence, the absolute
value of an integer is always positive. Thus, the absolute value of an integer is the
numerical value of the integer regardless of its sign. If the absolute value of an
integer is 7 then integer may be +7 or -7.

Addition of integers:
On a number line, as we go towards the right, the values go on increasing and when
we go towards the left, the value go on decreasing. So, we represent the positive
direction when we go towards the right and negative direction when we go towards
the left.

1. When two integers with same sign (either positive or negative) are added,
their absolute value are added and the sum is assigned the same sign.
For example:
i. To find the sum of (+3) and (+4).

(+7)
(+3) (+4)

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
\ (+3) + (+4) = +7

ii. To find the sum of (-2) and (-3).
(-5)

(-3) (-2)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
\ (-2) + (-3) = -5

128 Prime Mathematics Book - 6

2. When the integers are of different sign, find the their absolute value and
subtract the smaller value from the larger value. Assign the sign of the larger
value.

i. To find the sum of (+3) and (-5).
(-5)

(-2) (+3)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
\ (+3) + (-5) = -2

ii. To find the sum of (-4) and (+6).

(+6)
(-4) (+2)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
\ (-4) + (+6) = +2

Examples 1: Add (+210) and (+320)
Solution: ½+210½ = 210 and ½+320½ = 320

210 + 320 = 530
As both the number are positive, the answer + 530.

Examples 2: Add (-625) and (-325)
Solution: Here, ½-625½ = 625 and ½-325½ = 325

625 + 325 = 950
As the both the integers are negative, the result is -950.

Examples 3: Add (+250) and (-575).
Solution: ½+250½ = 250 and ½-575½ = 575

As the integers are of different sign, subtract the smaller value from
the large value.
575 - 250 = 325
As the larger value 575 has (-)ve sign, so the answer is -325.

Prime Mathematics Book - 6 129

Exercise 6.9

1. Write the absolute value of each of the following.

(a) ½+4½ (b) ½-5½ (c) ½-8½ (d) ½-270½

(e) ½0½ (f) ½+120½ (g) ½+250½ (h) ½-350½

2. Write the negative of the following numbers.

(a) -5 (b) +6 (c) 0 (d) +12
(h) -1
(e) +8 (f) -15 (g) -18

3. Add: (b) (-23) + (+57) (c) (-81) + (+129)
(a) (-40) + (+50) (e) (+20) + (+30) (f) (+15) + (+45)
(d) (-45) + (+100) (h) (+25) + (+35) (i) (-81) + (-23)
(g) (+45) + (+65) (k) (-51) + (-41) (l) (-88) + (-22)
(j) (-91) + (-21) (n) (-42) + (+20) (o) (-101) + (+51)
(m) (-31) + (+10)
(p) (-93) + (+40)

4. Add: (b) (+42) + (-24) + (-36)
(a) (-15) + (+20) + (+25) (d) (-17) + (+24) + (-34)
(c) (+32) + (-42) + (+27)

130 Prime Mathematics Book - 6

Rational Numbers Estimated periods – 2

You have learnt about the following sets of numbers.

Natural/ counting numbers (N) = {1, 2, 3, ........... }

Whole numbers (w) = {0, 1, 2, 3, ..........}
Integers (z) = {......-3, -2, -1, 0, 1, 2, 3 ......}

Note: The letter z stands for German word zahlen meaning “+ count”.

Let’s observe the following properties considering two intergers 2 and 6..

Sum of intergers is closed.
i.e. 2 + 6 = 8 (sum of two intergers is also an interger.)

2 - 6 = -4 (difference of two intergers is also an interger.)

2 × 6 = 12 (product of two intergers is also an integer.)

6 ÷ 2 = 3 (quotient is also an integer.)

2÷6= 1 (quotient in this case is not an integer.)
3

Thus, when an integer is divided by another integer, the quotient is not always an

integer. This fact indicates the existance of another sets of number which includes

such non integral numbers. The set of such number is called a rational numbers.

Any number which can be expressed in p form, where p and q are integers and
q

q = 0 are called rational numbers. The set of rational numbers is denoted by ‘Q’.

The sets of rational numbers is a wider set which includes the sets of natural numbers
(N), whole numbers (W) and integers (Z).

e.g. Q = {........, -3, - 3 , -1, - 1 , 0, 1, 3 , 3 , 2,.....}
2 2 4 4 2

Converting a rational number into decimal, we may get a termainal or non

termainal but reccuring decimals. Thus, termainating and non-terminating but
p
reccuring decimals can be converted to q form and are rational number.

0.5 = 0.5 = 5 = 1
1.0 10 2

0.125 = 0.125 = 125 = 1
1.000 1000 8
p
Non terminating but reccuring decimals can be converted into q form as

Example:

0.333 ....... .

Let x = 0.333 = 0.3 ........... (i)

Multiply by .10, both sides
10x = 3.3 .............. (ii)

Prime Mathematics Book - 6 131

substracting (i).from.(ii)
10x - x = 3.3 - 0.3
9x = 3

\ x= 3 = 1
9 3
p
Example: To convert 0.323232 ........ to q form.

Solution: ..

Given non terminating but reccuring decimal = 0.323232 ...... = 0.32
..
Let x = 0.32 .............. (i)

Multiply both s.i.des by 100, we get
100x = 32.32 ............. (ii)

substracting (i) fr.o.m (ii).,.we get
100x - x = 32.32 - 0.32

99x = 32

\x = 32
99
Irrational numbers

Consider a number 2.

If we find the square root of 2 iteratively by using division method, we get

2 = 1.41421356237 ....... which neither terminal nor repeat same number

or group of numbers. such numbers can not be converted into p form and are called
q
irrational numbers. Roots of primes, their products or multiples are irrational number.

Thus, 2, 3, 5, 6, 7 ....... are irrational numbers.

Exercise 6.10

1. Write the following sets in listed form.
2.
3. (a) N, (b) w (c) Z.
4.
Write the following into p form.
132 q
(a) 0 (b) 5 (c) 0.25 (d) 0.0156625
p (e) 12
Write the following into q form. 0..25.
(a) 0.3. (b) 0.6. (c)

Write whether the following are rational or irrational.

(a) -3 (b) 0.35 (c) 3 (d) 0.8

Prime Mathematics Book - 6

Unit Revision Test

1. Simplify:

a) 120 ÷10 + 5 of 2 - 4

b) 70 ÷ 7 - [15 × 2 {16 of 2 + (14 - 8)}]

2. Write the acutual division say whether 3718 is divisible by 11 or not.

3. Write the prime numbers beteen 10 and 30.

4. Find the prime factors of 240.

5. Find the square root of 256 by prime factor method.

6. Find the cube root of 1728 by prime factor method.

7. Fidn the H.C.F. of: 24, 42, 60.

8. Find the L.C.M. of: 32, 48 and 64.

9. If L.C.M.. o.f 24 and 30 is 120, find their H.C.F.
p
10. Write 0.24 into q form.

Prime Mathematics Book - 6 133

Estimated periods – 15

Objectives

At the end of this unit, the students will be able to:
perform four fundamental operations on fractions.
simplify the problems on fractions.
convert fractions to decimals and decimals to fractions.
perform fundamental operations on decimal.
round off the decimals to desired decimal places.

3 - 1
4 6

-= -

= = = 7
12

Teaching Materials Activities

Chart of equivalent fractions, graphs and It is better to:
grids to illustrate tenths, hundredths and do revision with equivalent fractions, like
thousandths. and unlike fractions and comparison of
fractions.
clarify the idea of addition, subtraction,
multiplication and division of fraction
with the help of figures.

Fractions
Study the following for your revision
Equivalent fractions:

124
248

The fractions represented by the above figures are equivalent fractions. Whatever
be the number of division made, the shaded parts are equal and the wholes are also
equal.

Two or more fractions are said to be equivalent if they represent the equal portion
of the same whole.

Here, 1 = 1
2 2

2 = 2 = 1
4 2×1 2

4 = 2 2 × 2 2 = 1
8 × 2 × 2

Thus, the lowest terms of the equivalent fraction are same.
To find the fractions equivalent to the given fraction:
Let's consider a fraction 2

3

We get, 2 = 2×2 = 4
3 3×2 6

2 = 2×3 = 6
3 3×3 9

2 = 2 × 4 = 8
3 3 × 4 12

Being the equal portion of the same whole, thus 4 , 6 , 8 are the fractions equivalent
2 6 9 12
to the fraction 3 .

Prime Mathematics Book - 6 135

Again, Let's consider a fraction 12
18

We get 12 = 12 ÷ 2 = 6
18 18 ÷ 2 9

12 =1128 ÷3 =46
18 ÷3

12 =1182 ÷6 =32
18 ÷6

Bing equal portions of the same whole, thus, 2 , 4 , 6 are the fractions equivalent
12 3 6 9
to the given fraction 18 .

Thus, by multiplying or dividing the numerator and denominator of the given fraction

by the same natural numbers, we can obtain the fractions equivalent to the given

fraction.

There are infinite number of equivalent fractions to a fraction.

To check whether fractions are equivalent or not:

Let's consider two fractions 3 and 12
5 20

A. Converting the fractions into their lowest term: If the lowest terms of the

fractions are same, the fractions are equivalent, otherwise not. Here,

3 = 3
5 5

12 = 2 × 2 × 3 = 3
20 2 × 2 × 5 5

\ 3 = 12
5 20

B. Comparing the cross-product of numerators and denominators:

If the product of numerator of first and denominator of secord and the product
of numerator of secord and denominator of first fractions are equal, the fraction
are equivalent, otherwise not.

3 12
5 20

3 × 20 =60 and 12 × 5 = 60 are equal

\ 3 and 12 are equivalent fractions
5 20

136 Prime Mathematics Book - 6

Like and unlike fractions:

The fractions having same denominators are like fractions. e.g. 1 , 2 , 3 , 4 etc
5 5 5 5
where the total number of division of the whole are same thus, their unit fractions

are same.

Unit fractions 2 2 1
3 5 6
The fractions having different denominators are unlike fractions. e.g. , , etc.

where the total number of division of the wholes are different. Thus, unit fraction

of them are not equal.

2 21
3 56

Unit fractions ,, are not equal

To convert unlike fraction into like fractions:
2 3
Let's consider two unlike fractions 3 and 5 .

2 = 10
3 15

3 = 9
5 15

Not equal Equal

Prime Mathematics Book - 6 137

If we further divide the figures (wholes) into number of parts which is L.C.M. of the
denominators, unit parts will be equal and thus the unlike fractions convert into
like.

Here, L.C.M. of denominators 3 and 5 is 15

\ 2 = 2 × 5 = 10 - Find L.C.M. of D's
3 3 × 5 15
- Multiply N and D of each
3 = 3 × 3 = 9 fraction by numbers so as
5 5 × 3 15 to make D's = L.C.M.

\ 10 and 9 are like fractions.
15 15

Comparison of fractions:

A. If two fractions are like fractions, compare the numerators. The fraction having
greater number is greater fraction.

e.g. to compare 3 and 5
7 7

Two fractions 3 and 5 are like fractions
7 7
3 5
Here, 3 < 5 \ 7 < 7

B. If two fractions are unlike fraction, convert them to like fractions and compare
numerators as in (A).

e.g. to compare 5 and 7
6 9

L.C.M. of 6 and 9 is 18 3 6, 9
2, 3
\ 5 = 5 × 3 = 15
6 6 × 3 18 \ L.C.M of 6 and 9 = 3 × 2 × 3

7 = 7 × 2 = 14 = 18
9 9 × 2 18

Since, 15 > 14

15 > 14
18 18

i.e. 5 > 7
6 9

138 Prime Mathematics Book - 6

Reducing fractions to their lowest terms.

Converting a given fraction to an equivalent fraction having no common factor
between its numerator and denominator (numerator and denominator prime to each
other) is called reduction of the fraction in its lowest term.

Reduction of fractions to their lowest term make the operation simpler.

Lets consider a fraction 48
120

To reduce it to its lowest terms.

A. In terms of prime factors, 48 = 2 × 2 × 2 × 2 × 3 2 48 2 120
In terms of prime factors, 120 = 2 × 2 × 2 × 3 × 5 2 24 2 60
2 12 2 30

26 3 15
3 5

Now, 48 = 2 × 2 × 2 × 2 × 3 (Cancel the common factors)
120 2 × 2 × 2 × 3 × 5

= 2
5

B. Dividing numerator and denominator by common primes one by one (cancellation)

48 = 24 (divide N and D by 2) = 12 (divide N and D by 2) = 6 (divide N and
120 60 30 15
2
D by 2) = 5 (divide N and D by 3)

or, 4824 12 6 2 = 2
120 60 30 15 5 5

C. Cancellation by H.C. F of N and D. 2 48, 120 or, 45) 120 (2
H.C.F of 48 and 120 is 2 × 2 × 2 × 3 = 24 96
So, dividing N and D by 24 2 24, 60 24) 48 (2
2 12, 30
48 2 = 2 3 6, 15 48
1205 5 ×

2, 5

Prime Mathematics Book - 6 139

Example 1: Write the first four fractions equivalent to 3
5
3
Solution: Here, given fraction = 5

We can take 3 = 3 × 2 = 6
5 5 × 2 10

3 = 3 × 3 = 9
5 5 × 3 15

3 = 3 × 4 = 12
5 5 × 4 20

and 3 = 3 × 5 = 15
5 5 × 5 25

\ 6 , 9 , 12 , 15 are four fractions equivalent to 3 .
10 15 20 25 5

Example 2: Find three equivalent fractions of 24
36
24
Solution: Here, given fraction = 36

We can take
24 24 ÷ 2 12
36 = 36 ÷ 2 = 18

24 = 24 ÷ 3 = 8
36 36 ÷ 3 12

24 = 24 ÷ 4 = 6
36 36 ÷ 4 9

\ 12 , 8 , 6 are the three fractions equivalent for 24
18 12 9 36

Example 3: Check whether the fraction 9 and 24 are equivalent or
not. 12 30

Solution: Given fractions are 9 and 24 3
12 30 9 3
or 124 = 4
Converting into the lowest terms

9 = 3×3 = 3 12 4 4
12 2×2×3 4 24 5
3015 =
And 24 = 2 × 2 × 2 × 3 = 4
30 2 × 3 × 5 5
5
3 4
Since 4 and 5 are not equal

\ 9 and 24 are not equivalent
12 30

140 Prime Mathematics Book - 6

Alternative method:

Here, 9 24
12 30

9 × 30 = 270 and 12 × 24 = 288

As 270 and 288 are not equal

So, 9 and 24 are not equivalent.
12 30

Example 4: Convert 3 and 5 into like fraction.
Solution: 4 6

Given fractions are 3 and 5 where
4 6

L.C.M. of denominators 4 and 6 is 12

Now, 3 = 3 × 3 = 9 2 4, 6
4 4 × 3 12 2, 3

5 = 5 × 2 = 10 L.C.M. = 2 × 2 × 3 = 12
6 6 × 2 12

So, 9 and 10 are like fractions.
12 12

Example 5: Compare the following fractions.

(a) 7 and 9 (b) 5 and 9
10 10 12 16

Solution: (a) Given fractions are 7 and 9 which are like fractions. Comparing
10 10

the numeritors, 7 < 9

\ 7 < 9
10 10

(b) Given fractions are 5 and 9 which are unlike fraction where
12 16
L.C.M of denominators 12 and 16 is 48.

Now, 5 = 5×4 = 20 4 12, 16
12 12 × 4 48
3, 4
and 9 = 9×3 = 27 L.C.M = 4 × 3 × 4
16 16 × 3 48
= 48
20 27 5 9
Since, 20 < 27, thus 48 < 48 i.e. 12 < 16

Prime Mathematics Book - 6 141

Example 6: Reduce 24 to its lowest term.
36
24
Solution: Given fraction = 36 2 24 2 36
2 12 2 18
= 2 × 2 × 2 × 3 26 39
2 × 2 × 3 × 3
3 3
= 2
3

Alternative method: Alternative method:

24 12 62 2 H.C.F of 24 and 36 is 12
3618 3
24 = = 24 = 24 ÷ 12 = 2
36 36 36 ÷ 12 3
9
3

Example 7: Reduce 2700 to its lowert term.
18000
2 2700, 18000
Solution: Here, 2700 2 1350, 9000
18000
H.C.F. N and D is 3 675, 4500
= 2700 ÷ 900 900 so, divided 3 225, 1500
18000 ÷ 900 by 900 5 75, 500

= 3 5 15, 100
20
3, 20

H.C.F = 2 × 2 × 3 × 3 × 5 × 5

= 900

Alternative method:

Here, 2700 3 27, 180
18000 3 9, 60

= 27 (Zeros cancelled) 3, 20
180 (Dividing N and
D by H.C.F.) H.C.F. = 3 × 3 = 9
= 27 ÷ 9
180 ÷ 9

= 3
20

142 Prime Mathematics Book - 6

Exercise 7.1 1/5 1/5
1. Study the table and complete the following:

1/5 1/5 1/5

1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10 1/10

(a) = 2 (b) 3 =10 (c) 2 = 4
10 5 10
5

(d) 10= 5 (e) 2 = 1 (f) = 6
5 10 10
5

2. Provide suitable numbers in the boxes to make equivalent fractions.

= ´3 3´ =
´5 5´

3
5

= ´3 3´ =
´5 5´

3. Write three equivalent fractions to each of the following.

(a) 3 (b) 6 (c) 9
4 7 11

(d) 20 (e) 35 (f) 100
21 42 125

4. Check whether the following are equivalent or not.

(a) 2 and 5 (b) 2 and 8 (c) 8 and 24 (d) 15 and 40
7 7 3 12 10 30 21 56

5. Convert the following pairs of fractions into like fractions.

(a) 1 and 3 (b) 3 and 4 (c) 2 and 2 (d) 3 and 5
2 5 8 5 5 3 4 8

Prime Mathematics Book - 6 143

6. Compare the following pair of fractions (use > or < or =).

(a) 7 and 11 (b) 17 and 7 (c) 4 and 6 (d) 8 and 7
15 15 21 21 5 7 15 12

7. Reduce the following fractions to their lowest terms.

(a) 9 (b) 27 (c) 20 (d) 72
12 45 60 84

8. (a) In an election, the candidate A got 7 and candidate B got 5 of the total
10 8
votes. Who got more votes?

(b) In a day, Netra can do 3 of a work and Rudra can do 7 of the work. Who
5 9
can do more work in a day?

Four basic operations on fractions

Addition and subtraction of fractions.

1. Addition and subtraction of like fractions.

To add or subtract like fractions we add or subtract numerators keeping
denominator as common.

e.g. To add 5 and 2 .
9 9

5 + 2 5 + 2 = 7
9 9 9 9 9

= 5+2
9

= 7
9

To subtract 3 from 6 6
7 7 7
6 3
7 - 7 ×××

= 6 - 3 3
9 7
=
= 3
7 3
7

144 Prime Mathematics Book - 6