UNIT 1:NUMBER SYSTEM AND
INDICES
NUMBER SYSTEM
A number system is a way of
expressing numbers through
writing. It is a mathematical notation
for consistently expressing
numbers from a particular set using digits or other symbols.
Natural Numbers (N)
The counting numbers 1, 2,3, 4,5.....arecalled naturalnumbers.
They are also called positive integers. Natural numbers denoted
by N'.
N= {1,2,3A.....]
Whole Nnumbers (W)
Allnatural numbers together with 0 are called whole numbers
which is denoted by 'W.
W= {0, 1, 2, 3...]
Integers (Z)
The set of all positive numbers,negative numbers together with
Ois called the set of integers and is denoted by 'Z'.
Z={.... 3, -2, -1 , 0, 1, 2, 3 ...
Even Numbers
Numbers which are divisible by 2 are even numbers
0,2,4,6,8.... -2,-4,-6,.... are even numbers
nh even number is 2n
Example:3rd even number= 2 x3 =6
10th even number = 2 x 10 = 20
(1) 66 is 33rdeven number
(ii) 128 is 64heven number
Since 2n = 66 ’n= 33
Since 2n = 128’n=64
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RUSINESS MATHEMAT
Odd Numbers
Numbers which are not divisible by 2 are
1, 3, 5, 7......are odd numbers
odd numbers.
nh odd number is 2n -1
Example: 10th odd number = 2(10) - 1=19
66th odd number = 2(66) - 1 =131
(i) 257 is odd number;
(ii) 995 is odd
2n -1= 257
number;
2n -1= 995
Here n = 257 +1 = 129
+1 = 498
Here n= 995
2
2
Prime Numbers
A number p > 1 is said to be a
prime number, if it has no othe
devisors except 1 and itself.
" 2is a prime number, since only
divisors of 2 are 1 and 2.
" 13 is a prime number, since only
divisors of 13 are l and 13.
" 15 is not a prime
number, since divisors of 15 are 1, 3, 5 and
15.
Set of all prime numbers
denoted by P.
P= (2, 3, 5,7, 11, 13, 17....
Note:
" 2is the only even prime
number.
prime numbers.
" There are infinitely many
Composite Numbers
The numberswhich are notprime
numbers arecalled composite
numbers. Composite numbers has divisors/
factors other than 1
and itself.
NUMBER SYSTEM AND INDICES
Example: 4, 6, 8, 9, 10, 12 ...
0 and 1are neither prime nor
composite numbers.
Q. Is 51 a prime number.
Ans.: No. The divisors of 51 are 1, 3, 17 and 51.
Rational Numbers
A
number which can be expressed in the form of
q? where 'p'
and 'q' are integers and denominator q 0 is called rational number.
17 -9
Example:;g4, -9, 0...
All natural numbers, integers, are rational numbers.
Set of all rational nåmbers denoted by 'Q'
Decimal expansion of rational numbers either terminates or
repeat after sometimes.
Example:
50
=
111
=14, terminates
0.45045045.. = 0.450,
10
61.6666=16
6
11=0.545454 =0.54.
(450 recurring)
(6 recurring)
(54 recurring)
Conversion of a terminating or recurring decimal into
P
form.
Example: Express the following decimals as rational numbers.
i) 0.555 i) 0.32 ii) 0.333.... iv) 1.32
Solution
555
(i) 0.555 = 1000
(i) 0.32= 32
100
(iii) Let x =0.333.....
)
v) 5.915915....
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BUSINESS MATHEMATIC
Multiply by 10 on boh sides
10x =3.3333..
(ii)
)
Subtracting (i) - ()
9x =3
iv) Let x= 1.323232......
()
Multiply by 100 on both sides
100x = 132.3232..
(i)
Subtracting (i) - ()
99x = 131
131
99
v) Let x= 5.915915.
()
Multiply by 1000 on both sides
1000x = 5915.915915..... . (ü)
Subtracting (i) - (1)
999x = 5910
X=
5910
999
Irrational Numbers
A number which cannot be expressed in the form of
fraction
where p and q are integers is called irrational number.
Eg: 2,/3, 5,. . .
Set of irrational numbers denoted by I
NUMBER SYSTEM AND INDICES
5
Decimal expansion of irrational numbers neither
become periodic.
terminates nor
Example: I =3.142857...
J2= 1.414213...
V10 =3.16227766...
Note:
If p is a prime number, then p is irrational.
Real numbers
Set of rational and irrationals together called Real numbers,
denoted by R'.
NcZcQcR
Note:
1. Sum of first 'n' natural numbers is n(n+1)
2
2. Sum of first 'n' even numbers is n (n+1)
3. Sum of first 'n' odd numbers is n²
ILLUSTRATIONS
1. Find the sumn of first 64 natural numnbers
Solution: Sum of first n natural numbers is
n(n+1)
2
Here n=64
.". Sum= 64(64+1)
2
64(65) =32(65) = 2080
2
2, Find the sum of 1+ 2+3+
+ 835
Solution: Sum of first n natural number
Here n= 835
835(836) =835 (418) = 3, 49, 030
2
n(n+1)
2
BUSINESS MATHEMATI
6
:1+ 2+3 +....... .+835 = 3,49,030
3. Find the sun of 35 + 36+ 37t ...+ 110
Solution:
We can write
35 +36+ 37+...+110 = (1+2+..+110 ) -(1+2+.. +34)
=sum of first 110 natural number-sum of first 34 natural numbe
110{111) 34(35)
2
2
= 6,105 -595 =5,510
.35+36 + 37+...+110 = 5,510
4.Check whether 820 is sum of first natural numbers?
n(n+1)
Solution: Sum of first n natural number
2
Suppose 820 is sum of first n natural numbers
Then, 820
=
820 x 2
=
1640
n(n+1)
2
=
’1640
for some value of n
n (n+1)
n (n+1)
40(41) = n(n+1)
n=40
..820 is sum of first 40 natural numbers.
5. Find the sum of 1 + 3 + 5 +....+ 91
Solution: Sum of first n odd numbers is n'
Here
2n -1 = 91
n= l+1
2
-= 46
.:.Sum of first 46 odd number = 462
46 x 46 = 2,116
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NUMBER SYSTEM AND INDICES
6. Find the sum of first 75 odd numbers.
Solution: Sum of first n odd numbers is n?
Here n = 75
Sum of first 75 odd number = 752
=75 x75 = 5,625
7. Find the sum of 31+33+........+ 99
Solution: We can write
31+33+...... . 99 = (1+3+5+...... 99)-(1+3+5+....+29)
From 1 to 99 there are 99 +1 = 50 odd numbers
2
From 1 to 29 there are
29 +1 =15 odd numbers
2
.:.31 +27+.......+ 99 = 50
- 152
= 2500 - 225 = 2,275
8. Find sum of first 500even numbers
Solution: Sum of first n even number is n (n +1)
Here n =' 500
Sum = 500 (500+1)
= 500 (501) = 2,50,500
.:. Sum of first 500 even. numbers is 2,50,500
9. Find 2+4+ 6+.
+448
Solution:Sum of first n even number is n (n +1)
Here
2n = 448
n
=
448
2
= 224
Sum = 224(225) =50,400
2+4+6+
+448 = 50,400
BUSINESS MATHEMATICS
8
10.Find the sum 52 + 54 +......+ 540
Solution: Sum of first n even numbers is n (n +1)
We can write
52 + 54 +...+ 540 = (2 + 4+ 6+...+ 540) - (2 + 4+ 6+ ...+50)
= 270 even numbers
From 2 to 540 there are 540
2
From 2 to 50, there are 50 = 25 even numbers
Sum = 270 (271) - 25(26)
= 73,170 - 650 = 72,520
Sum 52+34 +......+ 540 = 72,520
Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of
primes and this decomposition is unique, apart from the order in
which the prime factors occurs.
Example:
i)The factors of 45 are 9 and 5
.:. 45 = 3x3 x5= 32 x 5!
ii) The prime factors of 72
72 = 12 x6= 4 x3 x3 x2 =2x 2 x3 x3 x 2
72 = 28 x32
ii) The prime factors of 28
28 = 4 x7
2x7
This is called prime factorisation or canonical
representation.
NUMBER SYSTEM AND INDICES
9
11.Write prime factorisation of 460.
Solution: Dividing 460 by smallest prime factor until we
prime umber
get
2)460
2230
5115
23
.". 460 = 2x2 x5x 23
= 2x 5' x 23
12.Write prime factorisation of 33,075
Solution: Dividing 33,075 by smallest prime factor until we get
prime number.
333075
3/11025
33675
51225
5|245
749
7
.:. 33075
=3x3 x3 x5x5x7x7
= 3 x5 x72
Least Common Multiple (LCM)
The least/smallest number which is exactly divisible by each one
of the given numbers is called their Least common multiple or LCM.
Example: LCM of 4 and 6 is 12
Here4| 12 (4 divides 12) and 6|12 and 12 is the smallest number
which is divisible by both 4 and 6.
T'o find the LCM by Prime factorization method
(Canonical
express each number as the product of primes
First
Representation)
Then LCM =Product of all the prime factors with highest powers.
BUSINESS MATHEMATIC
10
13.Find the LCM of 15, 20 and 30
Solution :
15 =3 x5.
20 - )
30 =2 3)x 3
LCM = 22x 3 x 5!
= 4x 3 x 5=60
:LCM of 15, 20 and 30 = 60
14.Find the LCM of 45, 36 and 64
Solution:
45 3)
36 = 2' x3)
64 =2)
LCM = 26x 3²x 51
= 64 x 9 x 5= 2,880
:LCM of 45, 36 and 64
2,880
To find LCM by Division Method
Write the given numbers in ahorizontal line, separating them
by commas.
" Divide by any one of the prime numbers 2, 3, 5, 7 etc. which
will exactly divide at-least any two of the numbers.
Write the quotient and undivided numbers in a line below the
first line.
" Repeat the process until we get a line of numbers which are
prime to one another (co-primes).
The product of all the divisors and the numbers on the last line
will be the required LCM.
NUMBER SYSTEM AND
11
INDICES
15 Find the LCM of 25, 40 and 60
Solution:
5 25, 40, 60
25, 8, 12
2|5, 4, 6
5, 2, 3
(common divisor is 5)
(commondivisor for 8 and 12 is 2)
(5,2,3 are co-primes)
LCM =5x2 x 2x5x2x3 = 600
LCM of 25, 40 and 60 = 600
16.Find the LCM of 225 and 135
Solution:
3| 225,135
3 75, 45
5 25, 15
5, 3
:.LCM =3 x3 x 5 x5x3
675
LCM of 225, 135 = 675
17.Find the LCM of 32 and 64
Solution: Here 32 divides 64 ie., 32| 64
.. LCM of 32 and 64 = 64
Note:
" Ifxis a multiple of y then LÇM (1,y) = 1
z) =x
" Ifx is multiple of y and z then LCM of (, y and
i.e., if the number
If given numbers are co-prime to each other
have no common factors, then LCM =
For example 5, 8, 9 are co-primes
:.LCM (5,8, 9) =5 x 8x 9- 360
Product of numbers.
BUSINESS MATHEMAT
12
Highest Common Factor : (HCF) or (Greatest
Common Divi:
GCD) or (Greatest common measure)
An integer d' is called as the GCD or HCF of 2
"b' (both of them are not zero) if,
integers 'a'
i)dla and d|b (d divides a and d divides b)
ii) Every other common divisor of 'a' and 'b' divides 'd
i.e., xa, xb ’ xld
HCF of 'a' and 'b' is denoted by d = (a, b)
Example: HCF of 21 and 63 is 21 and is denoted by (21, 63) =1
HCF of 15 and 25 is 5 and is denoted by (15, 25) =5
Note:
HCF of 2 numbers is a unique positive
"If (a, b) = d then
integer.
(-a, b) = (a, -b) =(-a, b) = d, a
positive integer
Example :(15, 25) =(-15, 25) = (15, -25) =(15, -25) = 5
To find HCF by Prime
Factorization Method
Express each number as the product of primes.
Then HCF = Product of
common factors with least powers.
18.Find the HCF of 24, 28, 40
Solution: 24 = 8x3 = 2 x3!
28 = 4x 7=2 x7
40 =8 x5= 2 x5!
HCF = Product of common
factors with least powwer.
Here common factor is 2 with
least power 2
:. HCF of (24, 28, 40) = 2? = 4
NUMBER SYSTEM AND INDICES
13
19.Find the HCF of 72, 114 and 198
Solution:
272
2|36
39
399
19
333
114 =2x3x 19
198 = 2x3 x3 x 11
3
72 =2x 2 x 2x3x3
72 = 2 x 32
2198
2114
3\57
2|18
11
114 =2' x3' x 19' 198 = 2' x3x 11!
Here common factors are 2 and 3 with the least power 1and 1.
.:. HCF of (72, 114, 198) =2' x 3' =6
To find HCF by Division method
Let 'a' and 'b' are any 2 integers,
Without loss of generality we can assume b > a >0.
" On dividing 'b' by 'a', let 'q,' be the quotient and 'r,' be the
remainder
" Ifr,-0, then 'a' divides 'b', thus HCF
" If r, 0, divide abý r,. Let 'q.' be the quotient and r, be the
remainder, If r, =0 thenr, is the HCF of 'a' and 'b'.
If r,#0, divide r, by r, to get the quotient q, and remainder r,.
If r, = 0 then HCF = I,.
by the
If r, #0, continue this process of dividing each divisor
remainder till the remainder becomes zero.
'a' and 'b'
Ihe last non-zero remainder is the HCF of
" This is known as EUCLID'S ALGORITHM.
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BUSINESS MATHEMATI
20.Find the HCE of 63 and 35
Solution:
Here 35 <63,divide 63 by
35
35)63 (1
35
28) 35 (1
28
7)28 (4
28
The last non zero
28+0.: divide 35 by 28
7+0 divide 28 by 7
remainder is 7
:.HCF (63, 35) =7
21.Find the HCF of 195 and 495
Solution: Here 195 <495; divide 495 by
195) 495 (2
390
105) 195 (1
105
90) 105 (1
90
15)90(6
90
The last non zero
remainder
0
is 15
.:.HCF (195, 495) =15
22.Find the HCF of 72, 108 and 234
Solution: First find HCF of 72 and 108
Here 72< 108. Divide 108
72) 108 (1
72
36) 72 (2
72
by 72
195
VUMBER SYSTEM
15
AND INDICES
HCF (72, 108) = 36 (Last non zero
Divide 234 by 36
reminder)
36) 234 (6
216
18) 36 (2
0
HCF (36, 234) = 18
.:: HCF (72, 108, 234) = 18
Note :
"If a|b then HCF (a,b)=a
"If a|b and a<c then HCF (a,b,c)=a
If aandb are co-primes then HCF (a,b)=1
" If HCF of a, b, c= d Then HCF (b-a) -HCF(C-b)= HCF (a-c) =d
" IfHCF of (a, b) = dthen a = nd and b= md, where m and n are
integers.
23.Find the greatest number which divides 320, 173, and 404
leaving the same reminder.
Solution: Let h be the required number, and let x be the common
remainder.
Then h
= HCF (320-x, 173-x, 404-x)
404-x-173+x)
HCF (320-x-173+x, 404-1-320+x,
= HCF (147, 84, 231)
First find HCF of 147 and 84
84 < 147, so divide 147 by 84
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BUSINESS MATHEMAN
84) 147 (1
84
63) 84 (1
63
21) 63(3
63
0
HCF (84, 147) = 21
Further 21 divides 231 ’ HCF (21,231) = 21
:. HCF (84, 147, 231) = 21
The required number is 21
To find out the LCM and
C
e
If ~ are
HCF of fractions
the proper fractions, then
LCM = LCM of
numerators
HCFof denominators
(b,d, f)
HCF = HCF of numerators (a, c, e)
LCM of denominators (b, d, f)
(a,c, e)
24.Find the LCM and HCE of ð 32 16
Solution:Here numerators are 8, 32, 16 and
9, 81, 27
LCM of numerators (8, 32, 16) =32,
and 16
denominators are
since 32 is the multiple of 8
HCF of numerators (8, 32, 16) = 8,
since 8 is factor of 32 and 16.
LCM of denominators (9, 81, 27) = 81,
since 81 is the multiple of
9 and 27
HCF of denominators(9, 81, 27) = 9,
since 9 is factor of 81 and
27.
UMBER SYSTEM AND INDICES
LCM= LCM of numerators
HCF of denominators
17
32
HCE =HCF of numerators
LCM of denominators
8
81
25.Find the LCM of 6 5 8
714' 21
Solution: LCM of numerator (6, 5, 8)
26, 5, 8
3, 5, 4
LCM =2x3x5X4 =120
HCF of 7, 14, 21 =7
IM
(:.7 is the factor of 14 and 21)
LCMof numerators
HCF of denominators
26.Find HCE of 96 12
Solution: HCF of numerator (9, 6, 12) = 3
LCM of denominator (16, 11, 25) = 16 x 11 x 25 = 4400
(:.16, 11 and 25 are co-primes)
HCF =. HCF of numerators
LCM of denominators
4400
Application of LCM and HCF
LCM is used to find out the minimum/least common
timings,
Commnon capacity, and common length etc. of two or more
different values.
"HCF is used to split givern resources into smaller sections without
any wastage
" To equally distribute any number of sets of items into their
largest grouping.
18
BUSINESS
MATHEMA,
" To know highest number of people we can invite.
"To arrange things into rows or groups.
27.A, B and C start at thesame time in the same
direction.
around a circular stadium. A completes one round i
seconds, Bin 312 seconds and C in 168 seconds, all sta
the same point. After what time, will they again
meet a
starting point?
Solution: Here we need to find LCM of 120, 312 and 168
2 120, 312, 168
2 60, 156, 84
2 30, 78, 42
15, 39, 21
5, 13
7
LCM =2 x 2x 2x3 x5x13 x7=10,920 seconds = 0220
182 r
60
.:. A, B, Cwill again meet at
starting point after 182 minute
28.Three measuring rods are 50 cm, 75 cm, and 120 cm in
respectively. What will be the minimum length of clothlength
can be measured exact number of
times by using any of the
rods.
Solution:Here we need to find LCM of 50, 75 and 120
5 50, 75, 120.
2 10, 15, 24
5
3
5, 15, 12
1, 3, 12
1, 1, 4
LCM = 5x2x5 x3 x1x1x4=
. Minimum length of
600cm = 6m
cloth which can be measured =6m.
19
NUMBER SYSTEM AND INDICES
29.Find thà number which when divided hy 36. 40 and 48 leaves
the same remainder 5.
Solution: First we need to find LCM of 36, 40 and 48
2 36, 40, 48
2 18, 20, 24
2
9, 10,12
9, 5, 6
3, 5, 2
LCM =2x2x2x3 x3x5x 2= 720
Since 5 has to be the remainder, we have to add5 to 720
:.Required number = 720 + 5 = 725
30.The cost of a T.V is T60,000 and the cost of an AC is 25,000.
Find the least sum of money that a person must have in order
topurchase a whole numnber of TV or AC.
Solution : We need to find LCM of 60,000 and 25,000
5,000\60,000 25,000
12,
5,
LCM =5,000 x 12x5=3,00,000
.:.Least sum of money required = 3,00,000
31.Find the greatest number which divides 39, 48 and 90 leaving
remainders 6, 4 and 2 respectively.
Solution: Let h be the required number
Given h divides 39 and leaves reminder 6
’h divides 39 -6’his factor of 33
Similarly h divides 48- 4’his factor of 44
And h divides 90 - 2
his factor of 88
..h is HCF of 33, 44 and 88
33 = 111x3!
44 = 11lx 22
88 = 11'x 23
20
BUSINESS MATHEMAT
HCF of 33, 44, 88 is 11
.Required number is 11.
32.A rectangular courtyard of 3.33 meters
and 4.81
wide is paved exactly with square tiles oflong
the same mete
the largest size of the tile used for this
size. Fi
purpose.
Also find tnumber of tiles required.
Solution:Dimension of courtyard are 333 cm and 481 cm
The side of the required square tile =
HCF (333, 481)
Here 333 <481, divide 481 by 333
333) 481 (1
333
148) 333 (2
296
37) 148(4
148
HCF of 333 and 481= 37
..Side of the largest tile is 37 cm
Total number of tiles area of
x481 =117
coutryard33337 x37
area of each tile
Relation between LCM and HCF
If A and B are two
numbers, then the product of their LCM
and HCF is equal to the
product of two numbers.
LCM x HCF = A xB
HCF = AxB
LCM
LCM = AxB
HCF
For example : HCF (15, 10) =5
15 x 10 = 150
.LCM =AxB
HCF
150
530
21
NUMBER SYSTEM AND INDICES
Note
This relation may not be true when there are more than tWO
numbers.
For example, consider 20, 30 and 45
HCF (20, 30, 45) = 5
And LCM (20, 30, 45)=180
LCM x HCF =5 x 180= 900
20 x 30 x 45 = 27,000
Thus LCM xHCF
20x 30x 45
and their LCM is 36.
33.If the product of two numbers is 216
Findtheir HCF.
Solution : Let A and B are the two numbers
Given A x B=216and LCM=36
AxB
HCF = LCM
HCF =
216 =6
36
one of
and their LCM is 320. If
8
is
numbers
two
of
34.The HCF
Find the other number.
64.
is
numbers
the
the two numbers
Solution: Let A and B are
LCM =
Given A =64, HCF =8,
We have LCM x
HCF = A x B
320 x 8 =64 x B
’B=320x& = 40
64
320