1
Number System
Numbers are everywhere and play a very imporant role in our daily life. In every part of our life, we use
numbers.
e.g., weight, length, scores of a game, time-table, marks in exam etc are shown only through the numbers.
Digit
To represent a number, we use ten different basic symbols of mathematics, 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
These symbols are called digits.
Remember that it is not necessary to use all the digits to represent any number. Any one or more digits can
be used to represent a number. e.g., 11, 321, 333, 4978, 5521 etc are some examples of numbers.
How to write a number?
To write a number, we put the digits given in the number from right to left at the ones, thousands, lacs and
crores places.
We can write any number in words using the table given below. Let us write 592371221 in words.
Crores
Ten crore Crore
108
5
107
9
Lacs
Ten Lac
Lac
106
2
105
3
Thousands
Ten Thousand
Thousand
10 4
7
103
1
Ones
Hundred
Ten
102
2
101
2
One
100
1
We can write the above number in words as
“Fifty nine crore twenty three lac seventy one thousand two hundred and twenty one.”
Face value and Place value of the digits is a number
Face Value The face value of a digit in a number is the value of the digit itself. Here, the place of the
digit is not considered. It is known as real value also.
e.g., In the number 92347, the face value of 9 is 9, face value of 2 is 2, face value of 3 is 3 and so on.
Place Value The place value of a digit in a number changes according to the change of its place. In a
number, place value of
unit digit = (unit digit) × 100
ten digit = (ten digit) × 101
hundred digit = (hundred digit) × 102
thousand digit = (thousand digit) × 103 and so on.
The place value of number is called the local value of the number also.
e.g., in the number 28397, the place value of 8 is
8 × 103 = 8000
4
Mathematics Number System
Classification of Numbers
Numbers
Real Numbers
Imaginary Numbers
Rational Numbers
Natural Numbers
Irrational Numbers
Even/Odd
Numbers
Integers
Whole Numbers
Fractions
Co-prime or
Relatively Prime
Numbers
Prime/Composite
Numbers
Real Numbers The numbers which on squaring give positive result, are called real numbers. All the
positive, negative, rational and irrational numbers are real numbers.
Imaginary Numbers All non-real numbers are called imaginary numbers. On squaring these
numbers, we get negative values. e.g., − 2, is an imaginary number.
Natural Numbers All counting numbers are called natural numbers and they start from 1.
e.g., 1, 2, 3, 4,… are natural numbers.
These are denoted by N.
Note 0 is not a natural number.
Whole Numbers All natural numbers including zero are called whole numbers. These are denoted by W.
e.g., 0, 1, 2, 3,… are whole numbers.
Integers All positive and negative whole numbers are called integers. These are denoted by I.
e.g., …… − 3, − 2, − 1, 0, 1, 2, 3, … are integers.
Note
¢
¢
¢
I + = 1, 2, 3, 4, … are positive integers.
I − = − 1, − 2, − 3, − 4, … are negative integers.
0 (zero) is neither positive nor negative.
Prime Numbers The numbers other than ‘1’ which have exactly two factors ‘1’ and ‘itself’ are called
prime numbers.
e.g., 2, 3, 5, 7, 11, 13, 17, … are prime numbers.
Note
¢
¢
‘2’ is the least and only even prime number.
‘1’ is not the prime number.
Test of Prime Numbers
Suppose, we are to test the number P whether it is prime or not. We can check it using the following steps
Step I Find a whole square number just greater than P say x2 .
Step II Find the square root of x2 i.e., x.
Step III Take all the prime numbers less than or equal to x.
Step IV Check whether P is exactly divisible by any one more or number taken in Step III or not.
Step V If yes, then P is not prime and if not, then P is prime.
e.g., let P = 193
Now 196 is the whole square number just greater than 193.
CTET Success Master 5
196 > 193
14 > 193
Prime numbers upto 14 are 2, 3, 5, 7, 11 and 13. Since no one of these numbers divides 193
exactly, thus, 193 is a prime number.
Composite Numbers The numbers other than ‘1’ which are not prime, are called composite numbers.
e.g., 4, 6, 8, 9, 10, 12,… are composite numbers.
Note ‘1’ is not a composite number.
Even Numbers The numbers which are divisible by 2, are called even numbers.
e.g., 2, 4, 6, 8, 10, … are even numbers.
Odd Numbers The numbers which are not divisible by 2, are called odd numbers.
e.g., 1, 3, 5, 7, 9, … are odd numbers.
p
Rational Numbers The numbers which can be expressed in the form of , where p, q ∈ I and q ≠ 0, are
q
called rational numbers. These are denoted by Q.
3 4 2
e.g., , , ,… are rational numbers.
5 9 7
p
Irrational Numbers The numbers which cannot be expressed in the form of , are called irrational
q
numbers.
e.g., 2, 3, 5,… are irrational numbers.
Co-prime Numbers Two natural numbers are called co-primes if they don’t have any common factor
other than ‘1’.
e.g., (2, 5), (7, 15), (9, 13) are co-prime numbers.
Example 1. What is the difference of five digit greatest number and six digit smallest number?
(1) 1
(2) 9
(3) 11
(4) 99
Sol. (1) 5-digit greatest number = 99999
6-digit smallest number = 100000
Required difference = 100000 − 99999 = 1
Example 2. In the number 98739246, find the sum of place values of both the 9’s.
(1) 89991000
(2) 90009000
(3) 9009000
(4) 899100
Sol. (2) Place value of first 9 = 9 × 107 = 90000000
Place value of second 9 = 9 × 103 = 9000
Required sum = 90000000 + 9000 = 90009000
Example 3. A girl wants to check the number 187, whether it is divisible or not. What should she do?
(1) First check the divisibility
(2) First check its non-divisibility
(3) First check divisibility and then non-divisibility
(4) Check any one i.e., divisibility or non-divisibility
Sol. (4) Here, last option is better i.e., the girl is free to check the divisibility or non-divisibility of the number
about which she knows better. Thus, only one checking is enough.
Example 4. Which of the following is a real number?
(1)
2
3
(2) −
4
5
(3) 8
(4) All of these
6
Mathematics Number System
2
3
Sol. (4) On squaring , −
4
and 8, we get the positive numbers. Hence, all of these are real numbers.
5
Example 5. Find the sum of face values of all the 8’s in the number 87388243.
(1) 8
(2) 24
(3) 80088000
(4) 888
Sol. (2) We know that the face value of a digit is the digit itself. Hence, required sum = 8 + 8 + 8 = 24
Operations of Numbers
Addition When two or more numbers are combined together, it is called addition. It is denoted by the
sign ‘+’.
e.g.,
32 + 29 + 78 = 139
193
. + 275
. = 468
.
Subtraction When one or more numbers are taken out from a larger number, it is called subtraction. It
is denoted by the sign ‘−’.
e.g.,
182 − 13 − 56 = 113
1763
. + 12 − 1328
. = 555
.
Multiplication When two numbers, say x and y are multiplied together, then x is added y times or y is
added x times. It is denoted by the sign ‘×’.
e.g.,
3 × 4 = 3 + 3 + 3 + 3 = 4 + 4 + 4 = 12
Division on Numbers
In a sum of division, we have for quantities.
(i) Dividend
(ii) Divisor
(iii) Quotient
These quantities are inter-related by the following relations
1. Dividend = Divisor × Quotient/Remainder
2. Divisor = (Dividend − Remainder)/Quotient
(iv) Remainder
3. Quotient = (Dividend − Remainder)/Divisor
Example 1. In a sum of division, the quotient is 110, the remainder is 250, the divisor is equal to the sum of
Sol.
the quotient and remainder. What is the dividend?
(1) 3940
(2) 39850
(3) 40000
(2) Divisor = (110 + 250) = 360
Dividend = (360 × 110) + 250 = 39850
Hence, the dividend is 39850.
(4) 3890
Test of Divisibility
A number is divisible by
2 if unit digit of the number is either even or 0 (zero).
3 if the sum of the digits is divisible by 3.
4 if the number made by last two digits of the number is divisible by 4.
5 if unit digit is either 5 or 0.
6 if it is divisible by both 2 and 3.
7 if the differnce between twice the unit digit and the number formed by other digits is either 0 or a
multiple of 7.
e.g., 658 is divisible by 7 as
65 − 2 × 8 = 65 − 16 = 49 = divisible by 7
8 if the number made by last three digits of the number is divisible by 8.
9 if sum of the digits of the number is divisible by 9.
CTET Success Master 7
10 if unit digit of the number is 0.
11 if the difference of the sum of digits at even places and the sum of digits at odd places is 0 or a
multiple of 11.
12 if the number is divisible by 3 and 4.
13 if the sum of four times the unit digit and the number formed by remaining digit is divisible by 13.
Sometimes, we are to apply this rule more than one time according at our need. If the number has 5
or 6 or more digit, then we apply this rule until we get a 2 or 3 digit number.
e.g., let us take the number 113945.
113945 → 11394 + 20 = 11414
→ 1141 + 16 = 1157 → 115 + 28 = 143
which is divisible by 13.
14 if the number is divisible by both 2 and 7.
15 if the number is divisible by both 3 and 5.
16 if the number made by last four digits of the number is divisible by 16.
18 if the number is divisible by both 2 and 9.
21 if the number is divisible by both 7 and 3.
25 if the number made by last two digits is divisible by 25.
125 if the number made by last three digits is divisible by 125.
Note As we have shown in the divisibility rules of 14, 15, 18, 21, etc, it can be concluded that if we have to check
the divisibility by any number, then we can check the divisibility by their prime factors.
If a number is made by writing a digit 6 times, then the number is divisible by 7, 11 and 13.
e.g., 888888 is divisible by 7, 11 and 13.
If a number is made by writing a 2-digit number three times, then the number is divisible by 7 and 13.
e.g., 939393 is divisible by both 7 and 13.
l If a number is made by repeating a 3-digit number two times, then the number is divisible by 7, 11 and
13.
e.g., 973973 is divisible by 7, 11 and 13.
Example 2. What is not true about 480249?
(1) It is divisible by 9
Sol. (2)
(1)
(2)
(3)
(4)
(2) It is divisible by 4
(3) It is divisible by 3
(4) It is divisible by 11
Sum of digits = 4 + 8 + 0 + 2 + 4 + 9 = 27, which is divisible by 9.
Last two digits is 49, which is not divisible by 4.
Number 480249 is divisible by 9, so it is divisible by 3.
Sum of even digits = 8 + 2 + 9 = 19
Sum of odd digits = 4 + 0 + 4 = 8
Difference = 19 − 8 = 11, which is a multiple of 11, so given number is divisible by 11.
∴
Example 3. What should come in place of ‘–’ so that the number 8432_96?
Sol.
(1) 7
(2) 2
(3) 3
(4) 5
(2) A number is divisible 8 when the number made by last three 3 digits is divisible by 8. Only 296 is
divisible by 8. So ‘–’ must be replaced by 2.
Example 4. 983 − 32 .17 − 185 + 18 = ?
Sol.
(1) 783.83
(2) 764.13
(3) 693.33
(1) 983 − 32 .17 − 185 + 18
= (983 + 18) − (3217
. + 185)
= 1001 − 21717
. = 78383
.
(4) None of these
8
Mathematics Number System
Roman Numbers
The numbers which we normally use (e.g., 1, 2, 3,…) are called ‘Arabic Numbers’. Sometimes, we use the
another system for writing numbers, called Roman system.
Roman numbers are occasionally used to denote the class (in which a student study) and position of a
candidate, in faces of clocks, in page numbering etc.
The letters used in Roman numbers are
l
l V=5
I=1
l X = 10
l L = 50
l C = 100
l D = 500
l M = 1000
Roman Numerals Chart
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
XIII
XIV
XV
XVI
XVII
XVIII
XIX
XX
XXI
XXII
XXIII
XXIV
XXV
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
XXVI
XXVII
XXVIII
XXIX
XXX
XXXI
XXXII
XXXIII
XXXIV
XXXV
XXXVI
XXXVII
XXXVIII
XXXIX
XL
XLI
XLII
XLIII
XLIV
XLV
XLVI
XLVII
XLVIII
XLIX
L
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
LI
LII
LIII
LIV
LV
LVI
LVII
LVIII
LIX
LX
LXI
LXII
LXIII
LXIV
LXV
LXVI
LXVII
LXVIII
LXIX
LXX
LXXI
LXXII
LXXIII
LXXIV
LXXV
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
LXXVI
LXXVII
LXXVII
LXXIX
LXXX
LXXXI
LXXXII
LXXXIII
LXXXIV
LXXXV
LXXXVI
LXXXVII
LXXXVIII
LXXXIX
XC
XCI
XCII
XCIII
XCIV
XCV
XCVI
XCVII
XCVIII
XCIX
C
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
DI
DL
DXXX
DCCVII
DCCCXC
MD
MDCCC
CM
501
550
530
707
890
1500
1800
900
Characteristic of Roman Numbers
1. In Roman system, a letter can be repeated one to two times and not more than two times. e.g.,
I = 1,
2 = II ,
3 = III but
4 ≠ IIII, is written as IV.
(No repeatition) (One time repeatition) (Two time repeatition)
Similarly 10 = X, 20 = XX, 30 = XXX and 40 = XV
In this case, XV means 50 minus 10.
2. In the repeatition of roman digits, the value of the roman number is equal to the sum of values of
roman digits separately.
e.g.,
III = 1 + 1 + 1 = 3
MM = 1000 + 1000 = 2000
CTET Success Master 9
3. V, L and D are not repeated.
4. I is subtracted by only V and X, X is subtracted by only L and C, C is subtracted by only D and M. V, L
and D are not subtracted.
5. If a small roman symbol is at the left position of any bigger symbol, then value of the number can be
determined by subtracting smaller symbol from the bigger symbol.
e.g.,
XC = C − X = 100 − 10 = 90
CD = D − C = 500 − 100 = 400
6. If a smaller roman symbol is at the right position of any bigger symbol, then value of the number can
be determined by adding both the symbols.
e.g.,
XIII = X + I + I + I
= 10 + 1 + 1 + 1 = 13
MD = M + D = 1000 + 500 = 1500
7. If bar sign (−) is put above any number, then the value of that number is increased 1000 times.
e.g., IV = 4000,
D = 500 × 1000 = 500000
Example 5. Write the given numbers in Roman system.
(1) 90
(5) 83
(2) 60
(6) 99
(3) 69
(7) 65
(4) 98
(8) 80
Sol.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
90 = 100 − 10 = XC
60 = 50 + 10 = LX
69 = 50 + 10 + 9 = LXIX
98 = 100 − 2 = (100 − 10) + 8 = XCVIII
83 = 50 + 30 + 3 = LXXXIII
99 = 100 − 1 = (100 − 10) + 9 = XCIX
65 = 50 + 10 + 5 = LXV
80 = 50 + 30 = LXXX
Commutative Property
In commutative property, we see the word commute which means exchange. For example, washing my
face and combing my hair hold commutative property because here, order does not matter. But reading a
maths lesson and then answering the review questions is not commutative. Here, order matters because
we have to read the lesson before knowing how to answer the review questions.
Commutative Property of Addition It states that changing the order of numbers which are to be added
does not change the sum
i.e.,
a + b=b+ a
e.g.,
8 + 3 = 11 = 3 + 8
Commutative Property of Multiplication It states that changing the order of number which are to
multiplied does not change the multiplication result.
i.e.,
ab = ba
e.g.,
5 × 4 = 20 = 4 × 5
Note Subtraction and multiplication are not commutative.
e.g.,
and
7 −2 =5 ≠2 −7
25 ÷ 5 = 5 ≠ 5 ÷ 25
10
Mathematics Number System
Additive Identity When we add 0 (zero) in any integer, we get the same integer as the result i.e., for any
integer a,
a + 0 = a =0 + a
Here, 0 is called additive identity.
Multiplication Identity When we multiply any integer by 1, we get the same integer as the result i.e.,
for any integer a.
a ×1= a =1× a
Here, 1 is called multiplicative identity.
Associative Rule For any three integers a, b and c associative rule is defined as
a + (b + c) = (a + b) + c (for addition)
a × (b × c) = (a × b) × c (for multiplication)
Negative Numbers The numbers which are less than zero, are called negative numbers.
Following are some properties of negative numbers
1. Multiplications of two negative numbers is positive.
e.g.,
(− 3) × (− 5) = 15
2. Sum of two negative numbers is equal to the sum of numbers with negative sign.
e.g.,
(− 12) + (− 7) = − 19
3. Difference of two equal negative numbers is zero.
e.g.,
(− 7) − (− 7) = 0
4. Division of two negative numbers is positive.
e.g.,
(− 28) ÷ (− 4) = 7
5. If a positive number and a negative number are added, then they are actually subtracted i.e., they
are added with their signs.
e.g.,
5 + (− 3) = 5 − 3 = 2 ,
− 5 + (3) = − 5 + 3 = − 2 ,
and
− 5 − (3) = − 5 − 3 = − 8
6. Multiplication of a positive and a negative numbers is always negative.
e.g.,
(− 3) × (9) = − 27
7. Negative numbers are commutative with respect to addition and multiplication.
e.g.,
(− 3) + (− 10) = − 13 = (− 10) + (− 3),
(− 7) × (− 2) = 14 = (− 2) × (− 7)
Note Remember that (−) × (−) = +
(−) × (+ ) = −
(+ ) × (−) = −
(+ ) × (+ ) = +
(+ ) + (+ ) = +
(−) + (−) = −
(+ ) − (−) = + or −
(−) − (+ ) = + or −
CTET Success Master 11
Some Important Rules for Addition of Natural Numbers
u
Sum of first n natural numbers ( Σn)
1+ 2 + 3 +…+ n =
u
Sum of squares of first n natural numbers ( Σn2 )
12 + 22 + 32 + … + n 2 =
u
u
n (n + 1) ( 2n + 1)
6
Sum of cubes of first n natural numbers ( Σn3 )
13 + 23 + 33 + … + n 3 =
u
n (n + 1)
2
n (n + 1) 


2
2
Sum of first n even natural numbers = n (n + 1)
Sum of first n odd natural numbers = n2
Note
¢
¢
Square of an even number is even and square of an odd number is odd.
2, 3 and 7 can not be the unit digit of a square number.
To Find Unit Digit
To Find the Unit Digit in the Multiplication of Numbers
If we want to the unit digit in the multiplication of some numbers, we can do so by multiplying only the unit
digits of the given numbers.
e.g., unit digit in 476 × 188 × 359 × 142
= unit digit in 6 × 8 × 9 × 2
= unit digit in 864 = 4
To Find the Unit Digit in a Cubic Manner
To find the unit digit of a number of mn , first divide n by 4. If the remainder is r, then last digit of the number
= (the last digit of m)r .
If remainder is 0 (zero), thus last digit of the number = (last digit of m) 4 .
e.g.,
(729)902 ⇒ (9)902 = (9)225 × 4 + 2
92 = 8 1 → unit digit
⇒
Point to Remember
4n
= Unit digit of ( 4)4n = Unit digit of ( 8)4n = 6
u
Unit digit of ( 2)
u
Unit digit of ( 3)4n = Unit digit of (7)4n = 1
u
Unit digit of ( 5)n = 5
u
Unit digit of ( 6)n = 6
u
u
Unit digit of (11)n [ or (1)n ] = 1
1, if n is even
Unit digit of ( 9)n = 
9, if n is odd
u
4, if n is even
Unit digit of 4n = 
6, if n is odd
12
Mathematics Number System
Example 1. Find the unit digit in 765 × 641 × 357 .
(1) 6
(2) 7
(3) 5
(4) 8
Sol. (1) Unit digit in (7)65
= Unit digit in 74 × 16 + 1
= Unit digit in 71 = 7
41
Unit digit in 6 = 6
Unit digit in 357
= Unit digit in 34 × 14 + 1
= Unit digit is 31 = 3
65
Thus, unit digit in 7 × 641 × 357
= unit digit in 7 × 6 × 3
= unit digit in 126 = 6
[Q unit digit in (6) n = 6]
Example 2. Find the unit digit is 896 − 416 .
(1) 0
(2) 6
Sol. (1) Unit digit in 896 = Unit digit in 84 × 24 = 6
and
Hence,
(3) 4
unit digit in 416 = unit digit in 4 4 × 4 = 6
unit digit in 896 − 416 = 6 − 6 = 0
(4) 2
(Q Unit digit in 84 n = 1)
(Q unit digit in 4 4n = 1)
Algebraic Formulae When we solve an expression which can be expressed as an algebraic formula, then
to solve such types of expressions, we use following algebaric formulae.
l
(a + b)2 = a2 + 2ab + b2
l
(a − b)2 = a2 − 2ab + b2
l
a2 − b2 = (a + b) (a − b)
l
(a + b)3 = a3 + b3 + 3ab (a + b)
l
(a − b)3 = a3 − b3 − 3ab (a − b)
l
a3 + b3 − (a + b) (a2 − ab + b2 )
l
a3 − b3 = (a − b) (a2 + ab + b2 )
l
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
l
a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 − ab − bc − ca) + 3abc
l
If a + b + c = 0, then a3 + b3 + c3 = 3abc
From the formulae 1 and 2, we can determine two more formulae
l
(a + b)2 + (a − b)2 = 2 (a2 + b2 )
l
(a + b2 ) − (a − b)2 = 4ab
Example 3. Find the value of
0.064 × 0.064 + 2 × 0.064 × 99.936 + 99936
.
× 99936
.
(1) 1000
(2) 10000
(3) 9999
(4) None of these
Sol. (2) If we solve the given expression by usual method, then calculation will be very complicated. If we
think slightly, we will come to know that the given expression is of the form a 2 + 2ab + b 2 which can be
written as ( a + b ) 2 .
If we put 0064
.
= a and 99936
.
= b, then given expression is
a 2 + 2ab + b 2 = (a + b)2 = (0064
.
+ 99936
.
)2 = (100) 2 = 10000
CTET Success Master 13
Example 4. Find the value of the expression
0096
.
× 0096
.
− 0004
.
× 0004
.
(1) 0.0092
(2) 0.092
(3) 0.00092
Sol. (1) Putting 0096
.
= a and 0004
.
= b, then given expression is
a2 − b2 = ( a + b) ( a − b)
= (0096
.
+ 0004
.
) (0096
.
− 0004
.
)
= (01
. ) (0092
. ) = 00092
.
(4) None of these
Approximation Values
Sometimes, we face the conditions, where we have to depend on approximation values. For example, if we
are celebrating a party in our home, then we arrange the party for approximate number of guests.
Suppose, in a particular month, a person’s expenditure is ` 1976, he can assume it ` 2000.
Approximate Value of Numbers
Sometimes, we have the situations in our daily life where we have to depend on the estimation value of the
numbers. This estimation is called the approximate value of the number.
e.g., in a newspaper, it was written that about 5000 cricket lovers watched the cricket match in the
stadium. Here, it is clear that cricket watchers are approximate equal to 5000 but not exactly.
For an another example, we estimate the number of guests coming in a party of our home.
To Approximate till Tens
While rounding off a number to nearest ten, first replace the unit digit by 0. Now, if tens digit is 5 or more
than 5, then increase it by 1 otherwise let it unchange. Also, leave the other digits unchanged.
e.g., Approximation of 542 is 540.
Approximation of 7869 is 7870.
To Approximate till
Hundred While approximating a number to the nearest hundred, first replace ones and tens digits by 0. If
hundreds digit is equal or greater than 5, then increase it by 1 otherwise leave it unchanged. Also, leave the
other digits unchanged e.g., approximation of 5486 is 5500.
Approximation of 89174 is 89200.
Similarly, we can find the estimation of thousand, ten thousand etc. Also, we can estimate sum, difference,
multiplication and quotient also.
Example 1.
Find the approximate value of 5290 + 17986.
(1) 23000
(2) 2300
(3) 24000
Sol. (1) Approximate value of 5290 = 5000
Approximate value of 17986 = 18000
Approximate sum = 5000 + 18000
∴
= 23000
Alternate Method 5290 + 17986
= 23276 = 23000
(4) 24500
Example 2. Find the approximate value of 18 × 76.
Sol.
(1) 1300
(2) 1400
(2) 18 × 76 = 1368 = 1400 (approximate)
(3) 1500
(4) 1600
14
Mathematics Number System
Example 3. Find the approximation tens of 56937.
Sol.
(1) 56930
(2) 569 3 7 = 56940
(2) 56940
(3) 56950
(4) 56900
Example 4. Find the approximate thousand of 8198936.
(1) 8199000
Sol.
(2) 8199900
(3) 8198900
(4) 8190000
(1) 819 8 936 = 8199000
Fractions
When a unit is divided into some number of equal parts, then each part is called the fraction of the whole
part.
x
A number of the type , y ≠ 0, which represents x number of parts out of y number of equal parts of a thing
y
is called a fraction.
Here, x is called numerator and y is called denominator.
Some figures are given below to understand the fraction in better way.
1.
=
=1
2
=
2.
+
1
4
3.
=2
7
+
+
1
4
+
+
1
4
+
=
1
4
=
1
4.
=3=
4
5.
=5=1
10 2
Types of Fractions
1. Proper Fraction A fraction in which numerator is less than denominator, is called proper
fraction.
2 3 4
etc
e.g.,
, ,
3 7 11
2. Improper Fraction A fraction in which numerator is less than denominator, is called improper
fraction.
17 19 5
e.g.,
,
, etc.
13 11 2
3. Mixed Fraction If an improper fraction is written as a whole number with a proper fraction, then
the fraction is called mixed fraction.
3 5 2
e.g.,
2 , 3 , 1 etc
4 7 9
CTET Success Master 15
p
q
is a fraction, then is called its inverse function
q
p
8
3
e.g.,
is inverse function of .
3
8
5. Complex Fraction In complex fractions, numerator or demominator or both are complex fractions.
9
1 1
+
8
2
5 etc
,
e.g.,
2 4 7
1+
+
3 5 8
6. Continued Fraction It contains some additional fractions in numerator or in denominator.
1
e.g.,
etc
2+
5
2+
3
2+
7
7. Decimal Fraction A decimal fraction is a fraction whose denominator is 10 or a power of 10.
7 9
3
,
,
etc
e.g.,
10 100 10000
4. Inverse Fraction If
Addition of Fractions
When denominators are equal Here, we simply add the numerators and keep the denominators
same as all the denominators of the fraction.
a b
c
If fractions are , and , then
d d
d
a b c a+b+ c
+ + =
d d d
d
2 3 5 2 + 3 + 5 10
e.g.,
+ + =
=
7 7 7
7
7
When fractions are mixed with equal denominators
b e
g
If fractions be a , c and f , then
d d
d
b
e
g
(b + e + g)
a + c + f = (a + c + f ) +
d
d
d
d
2
7
7
4
1
2+ 4+1
e.g.,
1 + 5 + 3 = (1 + 5 + 3) +
=9 + =9
9
9
9
9
9
9
When denominators are unequal If denominators are unequal, then we take the LCM of
denominators and then add the fractions.
5 2
3 5 × 10 + 2 × 6 + 3 × 3 50 + 12 + 9 71
e.g.,
=
=
=
+ +
30
30
30
3 5 10
When fractions are mixed with unequal denominators Understand it with the following
example
5
2
1
1
 2 1 1
+ 4 + 3 = (5 + 4 + 3) +  + + 
 3 2 6
3
2
6
 2 × 2 + 3 × 1 + 1 × 1
= 12 + 



6
= 12 +
8
1
1
1
4
= 12 + = 12 + 1 = 12 + 1 + = 13
6
3
3
3
3
16
Mathematics Number System
Subtraction of Fractions
To solve the subtraction of fractions, we use the method which we used in addition. The difference is only
that we use − ve sign inplace of + ve.
2 1 2 ×2 −1×3 1
e.g., (i) − =
=
3 2
6
6
1
4 −3
1
1
1
 1 1
(ii) 9 − 8 = 9 − 8 +  −  = 1 +
=1+
=1
 3 4
3
4
12
12
12
3 4 6 3−4+6 9−4 5
(iii) − + =
=
=
7 7 7
7
7
7
Multiplication of Fraction
To multiply two or more fractions all the numerators are multiplied together to get the numerator of the
resultant and all the denominators are multiplied together to get the denominator of the resultant.
a c ac
(i) × =
b d bd
c
c

(ii) a × b = (a × b) +  a × 

d
d
b
e (ac + b) (df + e)
(iii) a × d =
×
c
f
c
f
(ac + d) (df + e)
=
cf
Division of Fraction
In division of fractions, we interchange the positions of numerator and denominator of the fractions which
are dividing and then multiply all the fractions.
a c a d ad
(i) ÷ = × =
b d b c bc
a c e a d f adf
(ii) ÷ ÷ = × × =
b d f b c e bce
1 1 1 2 1
e.g.,
(i) ÷ = × =
4 2 4 1 2
3
1 19 5 19 4 19
(ii) 2 ÷ 1 =
÷ =
× =
8
4 8 4 8 5 10
To Arrange the Fractions in Ascending or Descending Order
Method I. In this method, we compare the given fractions after converting them in their decimal form.
7 5
11
can be determined as the following way.
e.g., decending order of , and
12 8
15
7
5
11
= 059
. , = 063
. ,
= 074
.
12
8
15
Clearly,
074
. > 063
. > 059
.
11 5 7
∴
> >
15 8 12
Method II. In this method, we compare the fractions by making their denominators same and for it, we
take the LCM of the denominators. Remember that the fraction with greater numerator is greater.
CTET Success Master 17
e.g., if we are to find the decending order of
7 5
11
again, then first of all, we find the LCM of 12,
, and
12 8
15
8 and 15.
LCM of 12, 8 and 15 = 120
7
7 10 70
=
×
=
12 12 10 120
5 5 15 75
= ×
=
8 8 15 120
11 11 8 88
=
× =
15 15 8 120
88
75
70
>
>
120 120 120
11 5 7
> >
15 8 12
Now,
Therefore,
Hence,
Note
¢
¢
¢
n two or more fractions, if denominators are same, then fraction with greater numerator is greater and fraction
with lesser numerator is lesser.
In two or more fractions, if numerators are same, then fraction with greater denominator is lesser and fraction
with lesser denominator is greater.
If difference between numerator and denominator of given fractions, then the fraction having the greatest
numerator is greatest and the fraction having the lowest numerator is lowest.
Example 1. Which of the following fractions is greatest?
7 11 12 13 31
, , , ,
12 16 17 18 36
13
12
31
7
(2)
(3)
(4)
(1)
18
17
36
12
Sol. (3) Here, difference between numerator and denominator of all the fractions is 5. Therefore, the
fraction with greatest numerator is the greatest.
31
Hence,
is the greatest amongst the given fractions.
36
Example 2. If one-fifth of one-fourth of a number is
4
5
(2)
5
4
Sol. (2) Let the required number be x.
1 1
5
Then,
× ×x =
5 4
80
5
5
⇒
x=
×5×4=
80
4
5
Hence, the required number is .
4
(1)
3
4
Example 3. What is the th of
(1)
1
30
5
, find the number.
80
2
(3)
3
(4)
3
2
(4)
5
24
1
of given figure?
5
(2)
3
40
(3)
3
20
18
Mathematics Number System
Sol. (2)
=
3 1
=
6 2
3 1 1 3
× × =
4 5 2 40
3
2
Example 4. A wire of length m is cut into two parts. If the length of one of the parts is m, find the length
2
3
of other part.
5
6
2
1
(1) m
(2) m
(3) m
(4) m
6
5
3
6
3 2 3×3−2×2 9−4 5
Sol. (1) Length of other part of the wire = − =
=
= m
2 3
6
6
6
3 4 3
Example 5. Which fraction should be added to the sum of 5 , 4 , 7 to make the result a whole number?
4 5 8
1
3
7
(1)
(2)
(3)
(4) None of these
40
40
40
3
3 4 3
4
77
3
Sol. (2) Sum = 5 + 4 + 7 = (5 + 4 + 7) +  + +  = 16 +
 4 5 8
4
5
40
8
3
Here, we add
to make the sum as a whole number.
40
∴
Required value =
Example 6. Which of the following numbers are in ascending order?
12 12 12 12
12 12 12 12
(3)
,
,
,
,
,
,
29 37 19 25
37 29 19 25
12 12 12
12
Sol. (4) Given numbers are , , and .
19 25 37
29
Since, numerator are same.
12 12 12 12
Ascending order is
<
<
< .
∴
37 29 25 19
(1)
12 12 12 12
,
,
,
19 25 29 37
(2)
(4)
12 12 12 12
,
,
,
37 29 25 19
Recurring Decimals
A decimal number in which one or more digits are repeated infinite number of times, is called recurring
decimals.
To represent a recurring decimals, we put a bar (−) on the digits which repeat after the decimal point.
e.g., 146232323
(Mixed recurring decimal number)
.
… = 14623
.
166666
.
… = 16
. (Pure recurring decimal number).
Conversion of Pure Recurring Decimal into a Vulgar Fraction Write down the repeated
figures only once in the numerator and place as many nines in the denominator as the number of
figures repeating.
4
753
e.g.,
0.4 = , 0753
.
=
9
999
Conversion of Mixed Recurring Decimal into a Vulgar Fraction To convert a mixed
recurring decimal into a vulgar fraction, in the numerator take the difference between the number
formed by all the digits after decimal point (repeated digits will be taken only once) and the number
formed by non-repeating digits. In the denominator, place as many nines as there are repeating digits
and after nine put as many zeros as is the number of non-repeating digits.
467 − 4 463
e.g.,
0.467 =
=
990
990
Central Teacher Eligibility Test (CTET)
19
Factor
A number which divides a given number exactly, is called a factor of the given number.
e.g., Factors of 24 = 1 × 24 = 2 × 24 = 3 × 8 = 4 × 6
And prime factor of 24 = 2 × 2 × 2 × 3
Some Important Facts Related to Factors
1.
2.
3.
4.
5.
1 is the factor of every number.
Every number is a factor of itself.
A number is exactly divisible by its factors.
Factors of a number are less than or equal to that number.
Number of factors of a number are finite.
Multiple
A multiple of a number is obtained by multiplying it with a natural number.
e.g.,
Multiples of 5 = 5 × 1, 5 × 2, 5 × 3, 5 × 4,…
= 5, 10, 15, 20, …
Some Important Facts Related to Multiples
1.
2.
3.
4.
Multiple of a number is greater than or equal to that number.
Every number is a multiple of itself.
Every multiple of a number is exactly divisible by the number.
Number of multiples of a number are infinite.
Perfect Number If sum of the factors of a number (excluding the number itself) is equal to that
number, then the number is called perfect number.
e.g., 6 is a perfect number as because factors of 6 = 1, 2, 3 (leaving 6)
and
1+ 2 + 3 = 6
Common Multiple A common multiple of two or more numbers is a number which is exactly divisible
by each of them.
e.g., we can obtain common multiple of 3 and 5.
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, …
Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, …
Common Multiples of 3 and 5 = 15, 30, 45, …
Least Common Multiple (LCM)
The LCM of two or more numbers is the smallest number that is a common multiple of each of the numbers.
In above example,LCM of 3 and 5 is 15.
Finding LCM of Numbers
1. Find all the prime factors of both numbers.
2. Multiply all the prime factors of the larger number by those prime factors of the smaller number
that are not already included.
Example 1. Find the Lowest Common Multiple (LCM) of 240 and 924.
(1) 16
(2) 24
(3) 18480
Sol. (3) Prime factors of the given numbers are as follow
924 = 2 × 2 × 3 × 7 × 11
(4) 20402
20
Mathematics Number System
240 = 2 × 2 × 2 × 2 × 3 × 5
The LCM is 2 × 2 × 3 × 7 × 11 × 2 × 2 × 5 = 924 × 20 = 18480
Common Factor A common factor of two or more numbers is a number which divides each of the
numbers exactly.
e.g., we can obtain common factor of 9 and 12.
Prime factors of 9 = 1 × 3 × 3
Prime factors of 18 = 2 × 3 × 3
∴ Common factor of 9 and 12 = 3 and 9.
Highest Common Factor (HCF)
The HCF of two or more numbers is the greatest number which divides each of them exactly. In above
example, HCF of 9 and 18 is 9.
Division Method to Calculate HCF In this method, we first divide the larger number by the smaller
one. Then divide the divisor by the remainder. Repeat this process until the remainder becomes zero.
The least divisor is the required HCF.
Example 1. Find the HCF of 24 and 30.
(1) 6
(2) 8
(3) 9
(4) 4
Sol. (1) By prime factorization method,
24 = 2 × 2 × 2 × 3
30 = 2 × 3 × 5
HCF = Highest common factor of 24 and 30.
=2×3=6
By Division method,
24 ) 30 (1
24
6) 24 (4
24
×
∴ HCF = The last divisor = 6
Example 2. Find the HCF of 91, 112 and 49.
(1) 7
(2) 9
(3) 49
Sol. (3) First we find the HCF of 91 and 112.
91 ) 112 (1
91
21) 91 (4
84
7) 21 (3
21
×
Now, we find the HCF of 7 and the remaining number 49.
7) 49 (7
49
×
Hence, required HCF = 49
(4) 11
CTET Success Master 21
Example 3. Find the HCF of two consecutive number.
(1) 0
(2) 1
(3) 2
(4) 3
Sol. (2) Two consecutive numbers are always co-prime and the common factor in co-primes is 1 only. Hence,
required HCF = 1
Square
If a number is multiplied itself, then the result of this multiplication is called the square of the number.
e.g.,
Square of 8 = 8 × 8 = 64
square of 25 = 25 × 25 = 625
Note The square of an even number is always even and the square of an odd number is always odd.
Square Root
The square root of a number is the number, the square of which is equal to the given number. It is denoted
by the sign ‘ ’.
How to Calculate the Square Root?
Prime Factorization Method This method has the following steps
Step I Express the given number as the product of prime factors.
Step II Keep these factors in pairs.
Step III Take the product of these prime factors taking one out of every pair of the same primes. This
product gives us the square root of the given number.
Example 1. Find the square root of 1089.
(1) 32
(2) 33
(3) 36
Sol. (2) Prime factors of 1089 = 11 × 11 × 3 × 3
1089 = 11 × 11 × 3 × 3
⇒
Now, taking one number from each pair and multiplying them, we get
1089 = 11 × 3 = 33
(4) 39
11 1089
11 99
3
9
3
3
1
Note By Prime factorization method, we can find the square root of only perfect square number. If we don’t get
complete pairs of factors, then given number is not a perfect square number. This number may to converted
to a perfect number by multiplying or dividing it by uncomplete pair of factor.
Example 2. What least number should be multiplied to 245 to make it a perfect square?
(1) 2
(2) 3
(3) 5
(4) 7
Sol. (3) 245 = 5 × 7 × 7
5 245
7
49
7
7
1
Here, pair of factor 5 is not complete. If we multiply it by 5, we get a perfect square number.
Hence, required number is 5.
Division Method The steps of this method can be easily understood with the help of following
examples
22
Mathematics Number System
Example 3. Find the square root of 18769.
(1) 135
(2) 149
(3) 137
(4) 151
Sol. (3)
Step I In the given number, mark off the digits in pairs starting from the unit digit.
137
Each pair and the remaining one digit (if any), is called a period.
1 1 87 69
2
Step II Now, 1 = 1 on subtracting, we get 0 (zero) as remainder.
1
23
87
Step III Bring down the next period, i.e., 87. Now, the trial divisor is 1 × 2 = 2 and trial
69
dividend is 87. So, we take 23 as divisor and put 3 as quotient. The remainder
267
1869
is 18 now.
1869
Step IV Bring down the next period, which is 69. Now, the trial divisor is 13 × 2 = 26
×
and trial dividend is 1869. So, we take 267 is dividend and 7 is quotient. The
remainder is 0 now.
Step V The process (processes like III and IV) goes on till all the periods (pairs) come to an end and
we get remainder as 0 (zero).
Hence, the required square root = 137.
Example 4. What is the square root of 151321?
(1) 389
(2) 411
(3) 361
(4) 344
Sol. (1)
389
15 13 21
9
68
6 13
5 44
769 6921
6921
×
3
∴
Required square root = 389
Note We should apply division method when the number is so large that it is very difficult to find its square root by
prime factorization method.
Indices
If we multiply too or more numbers having same base, then their powers are added.
e.g.,
3 ×3×3 ×3×3
= 31 × 31 × 31 × 31 × 31
= 31 + 1 + 1 + 1 + 1 = 35
In general, when a number ‘P’ is multiplied by itself ‘n’ times, then the product is called nth power of ‘P’ and
is written as Pn . Here, P is called the base and ‘n’ is known as the index of the power.
Rules of Indices
1. Pm + Pn = Pm + n
e.g.,
2.
Pm
Pn
73 × 72 = 75
= Pm − n
e.g.,
64
62
= 64 − 2 = 62
CTET Success Master 23
3. (Pm )n = Pmn
(83 )6 = 83 × 6 = 818
e.g.,
4. (PQ)n = Pn × Qn
(3 × 7)4 = 34 × 74
e.g.,
n
Pn
 P
5.   = n
Q
Q
5
85
 8
  = 5
 7
7
e.g.,
6. P0 = 1
(1)0 = 1 = (2)0 = (3)0 = (− 7)0 = ( 2)0
e.g.,
7. P −n =
1
Pn
e.g., (i) 3− 7 =
 4
(ii)  
 5
1
37
−2
=
1
 4
 
 5
2
 5
= 
 4
2
Example 1. Find the value of ( − 3)2 × (7)2 .
(1) 144
(2) 21
(3) 441
(4) − 441
(3) − 125 m6
(4) 125 m6
Sol. (3) (3) ( − 3)2 × (7)2
= ( − 3) × ( − 3) × 7 × 7
= 9 × 49 = 441
Example 2. Find the value of ( − 4 m )3 ÷
(1) − 256 m5
Sol. (1) ( − 4 m )3 ÷
1
(2m ) 2
(2) 256 m5
.
1
(2 m )2
1


Q a ÷ = a × b


b
= ( − 4 m )3 × (2 m ) 2
= ( − 4 m ) × ( − 4 m ) × ( − 4 m ) × (2 m ) × (2 m )
= − 64 m3 × 4 m 2
= − 256 m5
Example 3. Find the value of (220 ÷ 215 ) × 23 .
(2) 215
(1) 28
20
15
3
Sol. (1) (2 ÷ 2 ) × 2
= (220 − 15 ) × 23
= 25 × 23 = 25 + 3 = 28
(3) 24
(4) 25
Practice Questions
1. A student follows the rule 3 × 3 + 3 × 3 = 12,
4 × 4 + 4 × 4 = 16, 5 × 5 + 5 × 5 = 20, find the
value of the expression 12 × 12 + 12 × 12
according to this rule.
(1) 288
(2) 48
(3) 144
(4) 44
2. Sum of six consecutive odd numbers is 168.
Find the largest of these numbers.
(1) 23
(2) 27
(3) 31
(4) 33
3. Find the difference between the place values
of both 4’s in 847342.
(1) 39960
(2) 39060
(3) 36990
(4) 30960
4. Find the sum of place and face values of 6 in
437699.
(1) 504
(2) 606
(3) 540
(4) 660
5. Which of the following is natural numbers?
(1) 0
(2) − 1
(3) 1
(4) All of these
6. Which of the following is a real number?
3
(1) − 3
(2)
2
(3) 7
(4) All of these
7. Which of the following numbers is divisible by
3?
(1) 97632
(2) 88844
(3) 98765
(4) All of these
8. By which of the following, 555555 is not
divisible?
(1) 11
(2) 7
(3) 13
(4) All of these
9. By which of the following, 787878 is not
divisible?
(1) 7
(2) 11
(3) 13
(4) None of these
8
2
10. From 2 and 8 , which one is larger?
(1) 28
(2) 82
(3) Both are equal
(4) Can’t be determined
11. Find the value of 9 + 10 + 11 + 12 + … + 21
(1) 190
(2) 231
(3) 195
(4) 215
12. Find the value of 13 + 23 + 33 + … + upto 10
terms
(1) 55
(2) 3025
(3) 2025
(4) 95
13. What should come in place of ‘?’ so that the
number 873 ? 24 is divisible by 4?
(1) 1
(2) 3
(3) 6
(4) All of these
14. Find the value of 200 × 439 + 200 × 321.
(1) 152000
(2) 76000
(3) 172000
(4) 136000
15. 944 − 307 + 119 = ?
(1) 886
(3) 556
(2) 756
(4) 686
16. 2234 + 84 − 1273 = ? + 123
(1) 922
(2) 832
(3) 932
(4) 822
17. Find the sum of all prime numbers less than
20.
(1) 92
(2) 73
(3) 19
(4) 77
18. Which of the following is an irrational
numbers?
(1) 9
(2) 11
2
2
(3)
(4)
5
3
19. The first prime number is
(1) 1
(2) 0
(3) 2
(4) None of these
20. What is the number of even prime numbers?
(1) 0
(2) 1
(3) 2
(4) None of these
21. An even number cannot end with
(1) 0
(2) 1
(3) 2
(4) None of these
22. Find the remainder when 445 is divisible by 5.
(1) 0
(2) 1
(3) 2
(4) 3
23. Find the unit digit in the product 268 × 732.
(1) 0
(2) 6
(3) 2
(4) None of these
24. Find the unit digit in the product of
(268 × 539 × 826 × 102)
(1) 2
(2) 3
(3) 4
(4) 5
CTET Success Master 25
25. Find the unit digit in (636 )84 .
(1) 6
(2) 4
(3) 7
(4) 1
36. Find the HCF of 14, 42 and 84
(1) 7
(2) 2
(3) 14
26. Find the value of
38. Two positive integers differ by 4 and sum of
10
their reciprocals is
. Then, one of the
21
number is
(1) 1
(2) 5
(3) 3
(4) 21
+
(1)
1
2
(2)
1
3
(3)
1
6
(4)
5
6
39. On dividing 397246 by a certain number, the
quotient is 865 and the remainder is 211. The
divisor is
(1) 435
(2) 432
(3) 459
(4) 482
27. Find the value of
(1)
1
2
(2)
1
3
(3)
1
4
(4)
40. A number 774958 N1 96 N2 to be divisible by 8
and 9, the values of N1 and N2 will be
(1) 7, 8
(2) 0, 8
(3) 5, 8
(4) 6, 7
4
15
28. If m = (3 × 4) × (3 × 4), find square root of m.
(1) 3 × 3
(2) 4 × 4
(3) 3 × 4
(4) 2 3
30. Find the value of 98375 × 9999
(1) 983656125
(2) 986351625
(3) 99573815
(4) None of these
31. If 2525 is divisible by 26, find the remainder
(1) 21
(2) 23
(3) 24
(4) 25
32. Find the sum of first 21 odd numbers
(1) 213
(2) 231
(3) 441
(4) 144
112
576
×
12
196
(2) 12
(3) 16
Find the value of
(1) 8
(4) 18
34. Find the LCM of 2 × 3 × 5 × 7 and 3 × 5 × 7 × 11
(1) 2 × 3 × 5 × 7 × 11
(2) 2 × 32 × 52 × 72 × 11
(3) 3 × 5 × 7
(4) None of the above
35. Find the HCF
3 × 52 × 7 × 112
(1) 3 × 5 × 7
(3) 3 × 52 × 7
of
3
2
2
2 ×3 ×5 ×7
41. The total number of digits required to make a
book of 200 pages is
(1) 200
(2) 600
(3) 492
(4) 372
42. The total number of prime factors which are
contained in (30)6 is
(1) 15
(2) 16
(3) 17
(4) 18
29. Write 94 in roman system.
(1) CXIV
(2) XCVI
(3) XCIV
(4) IICX
33.
(4) 42
37. Find the LCM of 8, 14, and 26
(1) 728
(2) 782
(3) 872
(4) 278
and
(2) 2 × 3 × 5 × 7 × 11
(4) 33 × 52 × 72
43. Divide ` 53 amongst X, Y and Z, respectively
so that X may receive 7 more than Y and Z may
receive 8 less than Y.
(1) ` 10, ` 15, ` 18
(2) ` 15, ` 10, ` 25
(3) ` 20, ` 12, ` 27
(4) None of these
44. Suppose, P, Q and R are three consecutive odd
numbers in ascending order. If the value of
three times P is three less than two times R,
then value of R is
(1) 5
(2) 7
(3) 9
(4) 11
45. 5 of 4 of a number is 8 more than 2 of 4 of the
7 15
5 9
same number. What is half of the number?
(1) 315
(2) 630
(3) 210
(4) 105
46. A man reads 3/8 of a book on a day and 4/5 of
the remainder on the second day. If the
number of pages still unread are 40, how many
pages did the book contain?
(1) 320
(2) 330
(3) 340
(4) 350
26
Mathematics Number System
47. The sum of 2.75 and 3.78 is
(2) 4.53
(1) 5.53
(3) 1.03
(4) None of these
48. If 4096 = 64, then the value of
4096 +
(1) 7.09
(3) 7.1104
0.4096 +
0004096
.
is
+ 000004096
.
(2) 7.1014
(4) 7.12
49. If we multiply a fraction by itself and divide the
product by its reciprocal, the fraction thus
26
obtained is 18 . The original fraction is
27
2
1
5
8
(1) 2
(2)
(4) 2
(3) 1
3
3
7
27
50. The unit’s place of the number (3127)173 is
(1) 1
(2) 7
(3) 9
(4) 3
51. A train starts full of passengers. At the first
station, it drops one-third of these and takes in
96 more. At the next, it drops one-half of the
new total and takes in 12 more. On reaching the
next station, there are found to be 248 left. With
how many passengers did the train start?
(1) 564
(2) 568
(3) 579
(4) 600
52. If x and y are different integers, both divisible
by 5, then which of the following is not
necessarily true?
(1) x2 + y2 is divisible by 5
(2) (x − y) is divisible by 5
(3) xy is divisible by 25
(4) (x + y) is divisible by 10
53. 165 + 215 is divisible by
(1) 31
(2) 13
(3) 27
(4) 33
54. The remainder, when 2851 × (2862)2 × (2873)3 is
divided by 23 is
(1) 5
(2) 10
(3) 11
(4) 18
55. The number of divisors of 540 is
(1) 20
(2) 22
(3) 23
(4) 24
56. Which of the following fractions is less than 7/8
1
and greater than ?
3
1
23
11
17
(1)
(2)
(3)
(4)
4
24
12
24
57. Descending order of
11 15 13
,
,
20 16 25
15 13 11
(3)
,
,
16 25 20
(1)
15 11 13
is
,
,
16 20 25
15 11 13
(2)
,
,
16 20 25
(4) None of these
58. If any two irrational numbers are added,
which of the following statements is always
true?
(1) It is always a rational number
(2) It is always an irrational number
(3) It may be rational or irrational
(4) It is always an integer
59. The sum of the series
12 + 22 + 32 + … + 502 is
(1) 42000
(2) 42925
(3) 42900
(4) 42800
60. If a number is divided by first 5 and then the
quotient by 3, then remainders are 4 and 1,
respectively. If it is divided by 15, find the
remainder.
(1) 8
(2) 7
(3) 9
(4) 6
61. The product of two numbers is 5/4. If one
5
number is , what is the other number?
6
1
(1) 2
(2)
2
3
2
(3)
(4)
2
3
2
62. th part of a certain sum was donated and
7
1
th was spent on education. The balance
4
amount will be
13
11
(1)
(2)
28
28
5
14
(3)
(4)
28
28
547527 547527
.
63. Find the value of
if
= x.
82
00082
.
[CTET June 2011]
(1) 10 x
x
(3)
100
(2) 100 x
x
(4)
10
64. The value of 2 + 3 + 2 − 3 is
[CTET June 2011]
(1) 6
(3) 2 3
(2) 2 2
(4) 6
CTET Success Master 27
65. Which is greatest among 33 1 %, 4 and 0-35?
2
15
(1)
(2)
(3)
(4)
(1) 10.3
(3) 8.5
[CTET Jan. 2012]
(1) 1600
(3) 3600
[CTET June 2011]
(2) 7
(4) 1
[CTET June 2011]
(2) 8
(4) 5
[CTET June 2011]
(2) 20%
(4) 33.33%
69. A student observed the following examples
(10)2 = (5 + 5)2 = 52 + 2(5)(5) + (5)2 = 100
= (6 + 4)2 = 62 + 2(6)(4) + (4)2 = 100
= (8 + 2)2 = 82 + 2(8)(2) + (2)2 = 100
= (1 + 9)2 = 12 + 2(1)(9) + (9)2 = 100
and concluded that
(a + b)2 = a2 + 2(a)(b) + b2
The above method of drawing conclusions is
[CTET Jan. 2012]
(2) activity
(4) inductive
70. How many times will I be writing 2 if wrote
down numbers from 11 to 199?
[CTET Jan. 2012]
(1) 38
(3) 36
74. 4 − (2 − 9)0 + 32 ÷ 1 + 3 is equal to
(1) 15
(3) 17
75. Given
(2) 12
(4) 16
the
numbers 375
. × 10−7 , 3
3
× 10−7 ,
4
3
× 10−7 . Which of these is not
8
equal to 0.000000375?
[CTET Jan. 2012]
3
−9
(2) × 10−7
(1) 375 × 10
8
3
−7
(4) 3 × 10−7
(3) 375
. × 10
4
1
76. Given n numbers, n > 1, of which, one is 1 −
n
and all others are 1’s. The mean of the n
numbers is
[CTET Jan. 2012]
1
1
1
(1) n −
(2) 1 − 2 (3) 1
(4) n − 2
n
n
n
3 5
77. If A = ÷ ,
4 6
375 × 10−9 and
68. ‘Buy three, get one free.’ What is the
percentage of discount being offered here?
(1) analytical
(3) deductive
(2) 900
(4) 2500
[CTET Jan. 2012]
67. The smallest number by which 68600 must be
multiplied to get a perfect cube is
(1) 25%
(3) 28.56%
(2) 10.5
(4) 8.8
73. The least number which is a perfect square
and is also divisible by 10, 12, 15 and 18 is
66. Unit’s digit in 132003 is
(1) 10
(3) 12
[CTET Jan. 2012]
[CTET June 2011]
4
15
0.35
Can’t be compared
1
33 %
2
(1) 3
(3) 9
72. The mean of the median, mode and range of
the observations 6, 6, 9, 14, 8, 9, 9, 8 is
(2) 39
(4) 37
71. The symbol a drawn to any size means a + 4
and the symbol b drawn to any size means b2 ,
where a and b are numbers. Then, the value of
B = 3 ÷ [(4 ÷ 5) ÷ 6],
C = [3 ÷ (4 ÷ 5)] ÷ 6 and
[CTET Jan. 2012]
D = 3 ÷ 4 (5 ÷ 6), then
(1) A and D are equal (2) All are equal
(3) A and B are equal (4) A and C are equal
78. Shown here are expressions given to Seema,
Anees, Asha and Tessy with their answers
Seema 4 × 1 + 8 ÷ 2 = 8
Anees 6 + 4 ÷ 2 − 1 = 4
Asha 9 + 3 × 2 − 4 ÷ 2 = 10
Tessy 27 ÷ 3 − 2 × 3 = 21
Who has got the correct answer?
[CTET Jan 2012]
3
+
2
–
4
(1) Asha
(3) Seema
is
[CTET Jan. 2012]
(1) 32
(3) 75
(2) 9
(4) 35
(2) Tessy
(4) Anees
79. When the number 398 is divided by 5, the
remainder is
[CTET Jan 2012]
(1) 3
(2) 4
(3) 1
(4) 2
Answers
1.
11.
21.
31.
41.
51.
61.
71.
(2)
(3)
(2)
(4)
(3)
(1)
(3)
(4)
2.
12.
22.
32.
42.
52.
62.
72.
(4)
(2)
(1)
(3)
(4)
(4)
(1)
(3)
3.
13.
23.
33.
43.
53.
63.
73.
(1)
(4)
(2)
(3)
(4)
(4)
(4)
(2)
4.
14.
24.
34.
44.
54.
64.
74.
(2)
(1)
(3)
(1)
(3)
(4)
(4)
(1)
5.
15.
25.
35.
45.
55.
65.
75.
(3)
(2)
(1)
(3)
(1)
(4)
(2)
(2)
6.
16.
26.
36.
46.
56.
66.
76.
(4)
(1)
(4)
(3)
(1)
(4)
(2)
(2)
7.
17.
27.
37.
47.
57.
67.
77.
(1)
(4)
(3)
(1)
(2)
(2)
(4)
(1)
8.
18.
28.
38.
48.
58.
68.
78.
(4)
(2)
(3)
(3)
(3)
(3)
(1)
(3)
9.
19.
29.
39.
49.
59.
69.
79.
(2)
(3)
(3)
(3)
(1)
(2)
(4)
(2)
10.
20.
30.
40.
50.
60.
70.
(1)
(3)
(4)
(2)
(2)
(3)
(2)
Hints and Solutions
1. Given pattern is
3 × 3 + 3 × 3 = 12
4 × 4 + 4 × 4 = 16
5 × 5 + 5 × 5 = 20
Here, the numbers of every expression are
added. Therefore,
12 × 12 + 12 × 12 = 12 + 12 + 12 + 12 = 48
2. Let six consecutive odd numbers be x, x + 2 ,
x + 4 ,x + 6 , x + 8 and x + 10.
According to the question,
x+ x+2+ x+ 4+ x+6
+ x + 8 + x + 10 = 168
⇒
6x + 30 = 168
⇒
6x = 168 − 30
⇒
6x = 138
138
⇒
x=
= 23
6
Hence, largest number = 23 + 10 = 33
3. Given number is 847342
Place value of first 4 = 40000
Place value of second 4 = 40
∴ Required difference = 4000 − 40 = 39960
4. Place value of 6 in 437699 = 600
7. A number is divisible by 3 if the sum of all its
digits is divisible by 3.
From the given options, only 97632 is divisible
by 3 because 9 + 7 + 6 + 3 + 2 = 27, which is
divisible by 3.
8. A number made by writing a digit 6 times is
divisible by each of 7, 11 and 13. So, 555555 is
divisible by each of 7, 11 and 13.
9. We know that a number which is made by
writing a 2-digit number 3-times, then the
number is divisible by 7 and 13.
Now we are to check the divisibility of 11 only.
Sum of digits at odd places in
787878 = 7 + 7 + 7 = 21
Sum of digits at even places in 787878
= 8 + 8 + 8 = 24
Difference = 24 − 21 = 3, which is not divisible
by 11. Therefore, 787878 is not divisible by 11.
10. Since, 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
Therefore, 28 is larger.
11. 9 + 10 + 11 + 12 + … + 21
= (1 + 2 + 3 + 4 + … + 21) − (1 + 2 + 3 + … 8)
21 × 22 8 × 9
=
−
2
2
n(n + 1) 

Q 1 + 2 + ...... + n =

2 
Face value of 6 is 437699 = 6
Required sum = 606
∴
5. We know that natural numbers are counting
numbers starting from 1.
Hence, 1 is a natural number.
6. Since, all rational and irrational numbers are
3
real numbers, therefore − 3, and 7, all are
2
real numbers.
82 = 8 × 8 = 64
and
= 231 − 36 = 195
3
3
12. 1 + 2 + 33 + … + 103
2
10 × (10 + 1) 
=

2


CTET Success Master 29
2
 3
 n (n + 1) 
3
3
3
 
Q 1 + 2 + 3 + … + n = 
2  

26.
+
=
2
 10 × 11
2
=
 = (55) = 3025
 2 
13. A number is divisible by 4 if the number made
by last two digits is divisible by 4. In the given
number, last two digits i.e., 24 is divisible by 4.
So, ‘?’ can occur any of the values.
=
1 1 3+2 5
+ =
=
2 3
6
6
27. Total number of smaller parts = 16
Number of shaded parts = 4
4 1
∴Required fraction =
=
16 4
14. 200 × 439 + 200 × 321
= 200 (439 + 321) (Taking 200 as common)
= 200 (760) = 152000
28. Given, m = (3 × 4) × (3 × 4)
= (3 × 4)2
15. 944 − 307 + 119
∴Square root of m
= (944 + 119) − 307
= 1063 − 307 = 756
m = (3 × 4)2
=3 × 4
16. 2234 + 84 − 1273 = ? + 123
⇒ 2234 + 84 − 1273 − 123 = ?
⇒ ? = (2234 + 84) − (1273 + 123)
= 2318 − 1396 = 922
29.
94 = 100 − 10 + 4
= XCIV
30. 98375 × 9999
17. Sum of all prime numbers less than 20
= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77
18. From the given options, 11 can’t be written in
p
the form of . So, it is an irrational number.
q
= 98375 × (10000 − 1)
= 983750000 − 98375
= 983651625
31. We know that if xn + 1is divisible by x + 1when
n is odd.
19. According to the definition of prime number,
the first prime number is 2.
In
⇒ (2525 + 1 − 26) is divisible by 26
21. Even numbers are those which are divisible by
2. So, number ending 1 can’t be even.
23. Unit digit in 268 × 732
= Unit digit in 8 × 2
= Unit digit in 16 = 6
24. Unit digit in 268 × 539 × 826 × 102
= unit digit in 8 × 9 × 6 × 2
= unit digit in 72 × 12
= unit digit in 2 × 2 = 4
25. We know that unit digit in 6n , for every n, is 6.
36 84
Unit digit in (6 )
=6
2525 , 25 is odd.
∴ (2525 + 1) is divisible by 25 + 1 or 26
20. There is only one even prime number which
is 2.
22. Since, 445 is exactly divisible by 5, so
remainder is 0.
4 2
+
8 6
⇒
(2525 − 25) is divisible by 26
∴When 2525 is divided by 26, the remainder is
25.
32. Q Sum of first n odd numbers = n2
∴Sum of first 21 odd numbers = 212 = 441
33.
112
576 112 24
×
=
×
= 16
12
14 12
196
34.
First number = 2 × 3 × 5 × 7
Second number = 3 × 5 × 7 × 11
LCM = 2 × 3 × 5 × 7 × 11
35. First number = 2 × 33 × 52 × 72
Second number = 3 × 52 × 7 × 112
HCF = 3 × 52 × 7
30
Mathematics Number System
Now, we will check the divisibility be 8. For it,
take last three digits 96N2 = 968 which is
divisible by 8.
36. First, we will find the HCF of 14 and 42.
14) 42 (3
42
×
Now, we will find the HCF of 14 and 84.
14) 84 (6
84
×
Hence, the number 7749580968 is divisible
by both 8 and 9.
41. Number of one digit pages from 1 to 9 = 9
Number of two digit pages from 10 to 99
= 90
Hence, required HCF = 14
37.
2
2
2
7
13
Number of three digit pages from 100 to 200
= 101
8, 14, 26
4, 7, 13
2, 7, 13
1, 7, 13
1, 1, 13
1, 1, 1
Total number of digit
= 9 × 1 + 90 × 2 + 101 × 3 = 492
∴Prime factor of (30)6 = 26 × 36 × 56
Hence, required LCM = 2 × 2 × 2 × 7 × 13 = 728
38. Let the two positive integers be x and x + 4.
Then,
⇒
⇒
⇒
1
1
10
+
=
x x + 4 21
x + 4 + x 10
=
x (x + 4) 21
43. Let Y receives ` u.
Also,
u + 7 + u + u − 8 = 53
⇒
3u − 1 = 53
⇒
u = 18
Y’s share = ` 18
⇒
∴
X’s share = 18 + 7 = ` 25
Z’s share = 18 − 8 = ` 10
(2x + 4) 21 = 10x (x + 4)
21x + 42 = 5x2 + 20x
5x2 − x − 42 = 0
⇒ 5x (x − 3) + 14 (x − 3) = 0
⇒
(x − 3) (5x + 14) = 0
x =3
∴Total number of prime factors
= 6 + 6 + 6 = 18
Then, X receives ` (u + 7) and Z receives
` (u − 8).
⇒ 5x2 − 15x + 14x − 42 = 0
⇒
Prime factor of 30 = 2 × 3 × 5
42.
14

Q x ≠ − 

5
Thus, two positive integers are 3 and 3 + 4 i.e.,
3 and 7.
Dividend − Remainder
39. Divisor =
Quotient
397246 − 211
=
= 459
865
40. First, we will check the divisibility by 9.
Sum of the digits of the given number
7 + 7 + 4 + 9 + 5 + 8 + N1 + 9 + 6 + N2
= 55 + N1 + N2
44. Let the values of three consecutive odd
numbers P, Q and R be 2n + 1, 2n + 3 and 2n + 5,
respectively.
According to the given condition,
3 × (2n + 1) + 3 = 2 × (2n + 5)
⇒
6n + 6 = 4n + 10
⇒
n=2
∴ R = 2n + 5 = 2(2) + 5 = 9
45. Let the number be x.
5 4
2 4
Then, ×
x − × x =8
7 15
5 9
8
4
−
 x =8

 21 45
 60 − 56 

 x =8
 7 × 3 × 15
By taking option one-by-one, we take
1. 55 + 7 + 8 = 70 is not divisible by 9.
2. 55 + 0 + 8 = 63 which is divisible by 9.
⇒
4x
=8
7 × 3 × 15
CTET Success Master 31
8 × 21 × 15
= 630
4
x 630
∴Required number = =
= 315
2
2
⇒
x=
46. Let total number of pages = n
Second day, number of pages read
4
5
n
= of n =
5
8
2
5n n n
Remaining pages =
− =
8 2 8
According to the question,
n
= 40
8
⇒
n = 320
47. 275
. + 378
. = ( − 2 + 075
. ) + ( − 3 + 078
. )
= − 5 + 153
.
= − 4 + 053
. = 4.53
4096
+
100
4096
+
10000
4096
1000000
+
4096
100000000
64
64
64
64
+
+
+
10 100 1000 10000
= 6.4 + 064
. + 0064
.
+ 00064
.
= 71104
.
a
49. Let the fraction be , then
b
26
 a a b
 ×  ÷ = 18
 b b a
27
a a a 512
⇒
× × =
b b b 27
=
⇒
⇒
⇒
a3
b3
3
=
512
27
 a
 8
  = 
 b
 3
a 8
=
b 3
2
=2
3
3
∴Unit’s place of (3127)173 = (1)43 × 7 = 7
After dropping one-third and taking in
96 passengers, the train has left
x
= x − + 96
3
2x + 288
passengers
=
3
Similarly, the second station, the number of
passengers left
2x + 288
=
+ 12
6
According to the question,
2x + 288
+ 12 = 248
6
⇒
2x + 288 = 1416
⇒
x = 564
52. Suppose, x = 0, y = 5
4096
. + 0.4096 + 0004096
.
+ 000004096
.
=
(3127)173 = (74 )43 × 71
∴
51. Let the train start with x passengers.
3
First day, number of pages read = n
8
3
5

Remaining pages = 1 −  n = n

8
8
48.
50. We know, 71 = 7, 72 = 49, 73 = 343, 74 = 2401
Here, x andy both are divisible by 5.
But x + y is not divisible by 10.
53. 165 + 215 = (24 )5 + 215
= 220 + 215
= 215 (32 + 1)
= 33 × 215
Hence, it is divisible by 33.
54. When the numbers 2851, 2862 and 2873 are
divided by 23, the remainders are 22, 10 and
21 respectively.
Required remainder,
22 × (10)2 × (21)3
23
22 × 100 × 9261
=
23
22 × 8 × 15
=
23
R=
[When 100, 9261 are divisible
remainders are 8, 15]
2640 114 × 23 + 18
=
=
23
23
Hence, remainder is 18.
by
23
32
Mathematics Number System
55. Let N = 540 = 22 × 33 × 51
∴Number of divisor = (2 + 1) (3 + 1) (1 + 1)
= 3 × 4 × 2 = 24
1
7
56. Given fractions are and .
3
8
LCM (8, 3) = 24
The given fractions can be rewritten as
8
and
24
21
17
. From the given options,
lies between
24
24
the given two numbers.
57. LCM (16, 20, 25) = 400
So, given fractions can be written as
375 220 208
,
,
400 400 400
Hence, descending order is
15 11 13
,
,
16 20 25
58. Let us take an example of two irrational
numbers 2 + 3 and 3 + 7.
Their sum = 2 + 3 + 3 + 7
=5 + 3 + 7
= an irrational number
Now, taking an another example of two
irrational numbers 3 + 3 and 3 − 3.
Their sum = 3 + 3 + 3 − 3 = 6
61. Let the number is x.
5 5
Then, x × =
6 4
5 5
x= ÷
⇒
4 6
5 6 6 3
= × = =
4 5 4 2
3
x=
⇒
2
62. Let total amount is 1.

 2 1 
Then, balance amount = 1 −  +  
 7 4 

=
1  8 + 7
−

1  28 
28 − 15 13
=
=
28
28
.
547527
63. Given, x =
.
00082
 547527
10 × 
⇒

 82 
=
∴
∴ x2 = ( 2 + 3 + 2 − 3 )2
= ( 2 + 3 )2 + ( 2 − 3 )2
Hence, sum of two irrational numbers may be
rational or irrational.
n (n + 1) (2n + 1)
59. Q12 + 22 + … + n2 = Σn2 =
6
+2 2+ 3⋅ 2− 3
= 2 + 3 + 2 − 3 + 2 (2)2 − ( 3)2
= 4+ 2 4−3
12 + 22 + … + 502 = Σ502
50 (50 + 1) (2 × 50 + 1)
6
50 × 51 × 101
=
= 42925
6
=4+2 1
=
60. Let number be x
Then,
x = 5p + 4
Again, p = 3q + 1
∴
x = 5 (3q + 1) + 4
= 15q + 9
Hence, remainder will be 9.
547527 x
=
82
10
64. Let 2 + 3 + 2 − 3 = x
= a rational number
∴
1  4 × 2 + 1 × 7
−


1 
7×4
∴
ie,
= 4 + 2 =6
x= 6
2+ 3 + 2− 3 = 6
1
335
.
65. 33 % =
= 0335
.
2
100
4
= 026
.
15
035
. = 035
.
Clearly, 0.35 is greater.
CTET Success Master 33
85
. +9+8
3
255
.
=
= 85
.
3
66. The unit digit in 132003 is
Hence, required mean =
Unit digit in 13(500 × 4 + 3)
⇒ Unit digit in 133 = 7
67. 68600 = 23 × 73 × 52
Clearly, in order to make 68600 a complete
cube, it must be multiplied by 5.
68. On buying 3, one is free meant by that in total
4 objects, one is free.
1
∴ Discount % = × 100 = 25%
4
69. The method of drawing conclusions using the
examples is known as Inductive method.
70. Number of 2’s from 1 to 100 = 20
(2, 12, 20, 21, 22, ........, 29, 32, 42, 52, 62,
72, 82, 92)
Similarly, number of 2’s from 101 to 199 = 20
Total number of 2’s from 1 to 199
= 20 + 20 = 40
But we are to take 2’s from 11 to 199.
Therefore, required number of 2’s
= 40 − 1 = 39
71.
73. The number which is divisible by 10, 12, 15
and 18 is also divisible by their LCM.
2
3
5
3
LCM of 10, 12, 15, 18 = 2 × 2 × 3 × 3 × 5
To make it perfect square, we should multiply
it by 5.
Hence, required number
= 2 × 2 × 3 × 3 × 5 × 5 = 900
74. 4 − (2 − 9)0 + 32 ÷ 1 + 3
= 4 −1+
⇒ 3
3
× 10−7 = 375
. × 10−7
4
= 0000000375
.
⇒ 375 × 10−9 = 0000000375
.
b means b2 .
+
9
+ 3 = 4 − 1 + 9 + 3 = 15
1
75. 375
. × 10−7 = 0000000375
.
a means a + 4.
3
10, 12, 15, 18
5, 6, 15, 9
5, 2, 5, 3
1, 2, 1, 3
1, 2, 1, 1
2
–
4
= 3 + 4 + 2 + 4 − 42
= 7 + 6 − 16
= 72 + 6 − (16 + 4)
= 49 + 6 − 20
= 55 − 20
= 35
72. Median of the observations 6, 6, 9, 14, 8, 9, 9,
8 or 6, 6, 8, 8, 9, 9, 9, 14
4th term + 5th term
is
2
8+9
Median =
⇒
= 85
.
2
and
Mode = 9
and
Range = 14 − 6 = 8
3
× 10−7 = 0375
.
× 10−7
8
= 00000375
.
3
−7
Hence, × 10 is not equal to 0000000375
.
.
8
⇒
76. Sum of n numbers
1
n
1
1
= n −1+ 1− = n −
n
n
∴ Mean of these numbers
1
n−
1
n
=
=1− 2
n
n
3 5 3 6 9
77. A = ÷ = × =
4 6 4 5 10
B = 3 ÷ [(4 ÷ 5) ÷ 6]
4 
=3÷ ÷6
 5 
{1 + 1 + 1 + ......(n − 1) times } + 1 −
34
Mathematics Number System
4
30 
30 45
=3×
=
4
2
C = [3 ÷ (4 ÷ 5)] ÷ 6
 4
= 3÷
÷6
 5 
5

= 3 ×  ÷ 6

4
= 3÷
15
15 5
÷6 =
=
4
24 8
D = 3 ÷ 4 (5 ÷ 6)
5
20
= 3÷ 4 × = 3÷
6
6
6 18 9
=3×
=
=
20 20 10
Hence, A and D are equal.
=
78. Seema 4 × 1 + 8 ÷ 2
= 4+ 4=8
Anees 6 + 4 ÷ 2 − 1
= 6 + 2 −1
= 8 −1 = 7 ≠ 4
Asha 9 + 3 × 2 − 4 ÷ 2 = 9 × 6 − 2
= 54 − 2 = 52 ≠ 10
Tessy 27 ÷ 3 − 2 × 3
9 − 6 = 3 ≠ 21
Hence, answer of Seema is correct.
79. Unit digit in 398 = Unit digit in (34 )24 ⋅ 32 = 9
∴ Dividing 398 by 5 or dividing 9 by 5, we get
the remainder 4.