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 Written as per the syllabus prescribed by the Central Board of Secondary Education.
CBSE
CLASS X
MATHEMATICS
Salient Features
•
Extensive coverage of the syllabus in an effortless and easy to grasp format.
•
In alignment with the latest paper pattern of Central Board of Secondary Education.
•
Neat and labelled diagrams.
•
‘Things to Remember’ highlights important facts.
•
Variety of additional problems for practice.
•
Questions from previous years board papers have been solved.
•
Memory Maps at the end of each chapter to facilitate quick revision.
•
Sample Test Paper at the end of each chapter designed for student’s Self Assessment.
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P.O. No. 50860
10680_11520_JUP
Contents
Chapter No.
Chapter Name
Page No.
01
Real Numbers
1
02
Polynomials
24
03
Pair of Linear Equations in Two Variables
62
04
Quadratic Equations
136
05
Arithmetic Progressions
185
06
Triangles
240
07
Coordinate Geometry
315
08
Introduction to Trigonometry
363
09
Some Applications of Trigonometry
427
10
Circles
458
11
Constructions
490
12
Areas Related to Circles
536
13
Surface Areas and Volumes
594
14
Statistics
643
15
Probability
754
Board Question Paper : March 2017
779
Note:


* marked questions are not from examination point of view.
The pattern of Board Question Paper will be revised from the academic year 2017-2018. However, the
Board Question Paper of March 2017 has been included to give a fair idea to the students about the kind
of questions asked in the Board Examination.
01. Real Numbers
Euclid’s Division Lemma
Given positive integers a and b, there exist unique
integers q and r satisfying
a = bq + r ; 0  r < b
where ‘a’ is dividend, ‘b’ is divisor, ‘q’ is quotient
and r is the remainder.
Note: q and r can also be zero.
Examples:
Consider the following pair of integers:
i.
29, 8
Here, a = 29 and b = 8
By using Euclid’s Division lemma,
a = bq + r ; 0  r < b
i.e., 29 = 8  3 + 5 ; 0  5 < 8
…. divisor (b)
ii.
quotient (q)
dividend (a)
remainder (r)
77, 7
Here, a = 77 and b = 7
By using Euclid’s Division lemma,
a = bq + r ; 0  r < b
i.e., 77 = 7  11 + 0 ; 0  0 < 7
….
iii.
3
8 29
 24
5
divisor (b)
11
7 77
77
0
quotient (q)
dividend (a)
remainder (r)
9, 12
Here, a = 9 and b = 12
By using Euclid’s Division lemma,
a = bq + r ; 0  r < b
i.e., 9 = 12  0 + 9 ; 0  9 < 12
….
divisor (b)
0
12 9
0
9
quotient (q)
dividend (a)
remainder (r)
Euclid’s Division Algorithm:
Euclid’s division algorithm is a technique to compute
the Highest Common Factor (HCF) of two given
positive integers.
Euclid’s Division Algorithm to find HCF of two
positive integers ‘a’ and ‘b’ (a > b):
Step I: By Euclid’s division lemma, find whole
numbers ‘q’ and ‘r’
where a = bq + r ; 0  r < b
Step II: If r = 0, the HCF is b. If r  0, apply the
division lemma to b and r.
Chapter 01: Real Numbers
Step III: Continue the process till the remainder is
zero. When the remainder is zero the
divisor at that stage is the required HCF.
For the above algorithm HCF(a, b) = HCF(b, r)
Example:
Use Euclid’s division algorithm to find the HCF of
1467 and 453.
Solution:
Step I:Apply Euclid’s division lemma to 1467 and 453,
1467 = 453  3 + 108
3 q
…. b 453 1467 a
1359
108 r
Step II: Since r  0

apply Euclid’s division lemma to 453 and 108,
453 = 108  4 + 21
4 q
a
108
453
…. b
 432
21 r
Step III: Again, r  0

apply Euclid’s division lemma to 108 and 21,
108 = 21  5 + 3
5
…. 21 108
 105
3
Step IV: Apply Euclid’s division lemma to 21 and 3,
21 = 3  7 + 0
7
…. 3 21
 21
0
Since, r = 0

HCF(1467, 453) = 3

3 = HCF(21, 3) = HCF(108, 21)
= HCF(453, 108) = HCF(1467, 453)
Things to Remember

Euclid’s division algorithm can be extended
for all integers except zero i.e., b  0.
NCERT Exercise 1.1
1.
Use Euclid’s division algorithm to find the
HCF of:
i.
135 and 225
ii.
196 and 38220
iii. 867 and 255
Solution:
i.
Since 225 > 135, we apply the division lemma
to 225 and 135, to get
225 = 135  1 + 90
1
Class X: Mathematics
ii.
iii.
Since the remainder 90  0, we apply the
division lemma to 135 and 90, to get
135 = 90  1 + 45
We consider the new divisor 90 and the new
remainder 45, and apply the division lemma to get
90 = 45  2 + 0
As the remainder is zero, we stop.
Since, the divisor at this stage is 45, the HCF
of 135 and 225 is 45.
Since 38220 > 196, we apply the division
lemma to 38220 and 196, to get
38220 = 196  195 + 0
As the remainder is zero, we stop.
Since the divisor at this stage is 196, the HCF
of 196 and 38220 is 196.
Since 867 > 255, we apply the division lemma
to 867 and 255, to get
867 = 255  3 + 102
Since the remainder 102  0, we apply the
division lemma to 255 and 102, to get
255 = 102  2 + 51
We consider the new divisor 102 and the new
remainder 51, and apply the division lemma to
get
102 = 51  2 + 0
As the remainder is zero, we stop.
Since the divisor at this stage is 51, the HCF
of 867 and 255 is 51.
2.
Show that any positive odd integer is of the
form 6q + 1, or 6q + 3, or 6q + 5, where q is
some integer.
[CBSE 2014]
Solution:
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for
some integer q  0 and r = 0,1,2,3,4,5 because
0  r  6.

a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
or 6q + 5
Here, a cannot be 6q or 6q + 2 or 6q + 4, as
they are divisible by 2.
6q + 1
6 is divisible by 2 but 1 is not divisible by 2.
6q + 3
6 is divisible by 2 but 3 is not divisible by 2.
6q + 5
6 is divisible by 2 but 5 is not divisible by 2.
Since, 6q + 1, 6q + 3, 6q + 5 are not divisible
by 2, they are odd numbers.
Therefore, any odd integer is of the form
6q + 1, or 6q + 3, or 6q + 5.
3.
2 2
An army contingent of 616 members is to
march behind an army band of 32 members
in a parade. The two groups are to march
in the same number of columns. What is the
maximum number of columns in which
they can march?
Solution:
HCF(616, 32) will give the maximum number
of columns in which they can march.
Let us use Euclid’s algorithm, to find the
HCF.
616 = 32  19 + 8
32 = 8  4 + 0

the HCF of 616 and 32 is 8.

the maximum number of columns in which
they can march is 8.
4.
Use Euclid’s division lemma to show that
the square of any positive integer is either
of the form 3m or 3m + 1 for some integer
m.
[Hint: Let x be any positive integer then it is
of the form 3q, 3q + 1 or 3q + 2. Now square
each of these and show that they can be
rewritten in the form 3m or 3m + 1.]
[CBSE 2015]
Solution:
Let x be any positive integer and b = 3.
Then, by Euclid’s division lemma, x = 3q + r
for some integer q  0 and
r = 0,1,2 because 0  r < 3

x = 3q or 3q + 1 or 3q + 2
When x = 3q,
(x)2 = (3q)2 = 9q2
= 3(3q2)
= 3m, where m is a integer
When x = 3q + 1
(x)2 = (3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1, where m is a integer
When x = 3q + 2,
(x)2 = (3q + 2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m is a integer

the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m.
5.
Use Euclid’s division lemma to show that
the cube of any positive integer is of the
form 9m, 9m + 1 or 9m + 8.
Solution:
Let a be any positive integer and b = 3.
Then, by Euclid’s division lemma, a = 3q + r
for some integer q  0 and
r = 0,1,2 because 0  r < 3

a = 3q or 3q + 1 or 3q + 2
When a = 3q,
a3 = (3q)3 = 27q3
= 9(3q3)
= 9 m, where m is a integer
When a = 3q + 1,
a3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1
= 9m + 1, where m is a integer
Chapter 01: Real Numbers

When a = 3q + 2,
a3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8
= 9m + 8, where m is a integer
the cube of any positive integer is of the form
9m or 9m + 1 or 9m + 8.
Problems based on Exercise 1.1
1.
Using Euclid’s division algorithm, find the
[CBSE 2012]
HCF of 240 and 228.
Solution:
By Euclid’s division algorithm,
240 = 228  1 + 12
228 = 12  19 + 0

HCF (240, 228) = 12
2.
Find the HCF by Euclid’s division
algorithm of the numbers 92690, 7378 and
7161.
[CBSE 2013]
Solution:
By Euclid’s division algorithm,
92690 = 7378  12 + 4154
7378 = 4154  1+ 3224
4154 = 3224  1 + 930
3224 = 930  3 + 434
930 = 434  2 + 62
434 = 62  7 + 0

HCF (92690, 7378) = 62
7161 = 62  115 + 31
62 = 31  2 + 0

HCF(7161, 62) = 31

HCF (92690, 7378, 7161) = 31
3.
Using Euclid’s division algorithm, find
whether the pair of numbers 231, 396 are
coprime or not.
Solution:
By Euclid’s division algorithm,
396 = 231  1 + 165
231 = 165  1 + 66
165 = 66  2 + 33
66 = 33  2 + 0

HCF(231, 396) = 33

the numbers are not coprime.
4.
Two tankers contain 850 litres and
680 litres of petrol. Find the maximum
capacity of a container which can measure
the petrol of each tanker in exact number of
times.
[CBSE 2012]
Solution:
HCF(850, 680) will give the maximum
capacity of container.
850 = 680  1 + 170
680 = 170  4 + 0

HCF (850, 680) = 170

the maximum capacity of a container which
can measure the petrol of each tanker in exact
number of times is 170 litres.
5.
A sweetseller has 420 kaju barfis and 130
badam barfis. She wants to stack them in
such a way that each stack has the same
number, and they take up the least area of
the tray. What is the maximum number of
barfis that can be placed in each stack for
this purpose?
Solution:
HCF(420, 130) will give the maximum
number of barfis that can be placed in each
stack.
By Euclid’s division algorithm,
420 = 130  3 + 30
130 = 30  4 + 10
30 = 10  3 + 0

HCF(420, 130) = 10

the sweetseller can make stacks of 10 for both
kinds of barfi.
6.
The length, breadth and height of a room
are 8m 25 cm, 6m 75 cm and 4 m 50 cm
respectively. Find the length of the longest
rod that can measure the three dimensions
of the room exactly.
[CBSE 2012]
Solution:
Since, 1m = 100 cm

8 m 25 cm = 825 cm
6 m 75 cm = 675 cm
4 m 50 cm = 450 cm
HCF(825, 675, 450) will give the length of the
longest rod.
825 = 675  1 + 150
675 = 150  4 + 75
150 = 75  2 + 0

HCF(825, 675) = 75
450 = 75  6 + 0

HCF(450, 75) = 75

HCF (825, 675, 450) = 75

the length of the longest rod is 75 cm.
NCERT Exemplar
1.
Write whether every positive integer can be
of the form 4q + 2, where q is an integer.
Justify your answer.
Solution:
No, every positive integer cannot be only of
the form 4q + 2.
Justification:
Let a be any positive integer. Then by Euclid’s
division lemma, we have
a = bq + r, where 0 ≤ r < b
3
Class X: Mathematics
Putting b = 4, we get
a = 4q + r, where 0 ≤ r < 4
Hence, a positive integer can be of the form,
4q, 4q + 1, 4q + 2 and 4q + 3.
“The product of two consecutive positive
integers is divisible by 2”. Is this statement
true or false? Give reasons.
Solution:
True.
Justification:
Let a, a + 1 be two consecutive positive
integers.
By Euclid’s division lemma, we have
a = bq + r, where 0 ≤ r < b
For b = 2 , we have
a = 2q + r, where 0 ≤ r < 2
...(i)
Putting r = 0 in (i), we get
a = 2q, which is divisible by 2.
a + 1 = 2q + 1, which is not divisible by 2.
Putting r = 1 in (i), we get
a = 2q + 1, which is not divisible by 2.
a + 1 = 2q + 2, which is divisible by 2.
Thus for 0 ≤ r < 2, one out of every two
consecutive integers is divisible by 2.

The product of two consecutive positive
integers is divisible by 2.


2.
“The product of three consecutive positive
integers is divisible by 6”. Is this statement
true or false? Justify your answer.
Solution:
True.
Justification:
At least one out of every three consecutive
positive integers is divisible by 2.

The product of three consecutive positive
integers is divisible by 2.
At least one out of every three consecutive
positive integers is divisible by 3.

The product of three consecutive positive
integers is divisible by 3.
Since the product of three consecutive positive
integers is divisible by 2 and 3.

It is divisible by 6 also.

The numbers are of the form 3q, 3q + 1 and
3q + 2.
(3q)2 = 9q2 = 3(3q2)
= 3m, where m is a integer.
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1
= 3m + 1,
where m is a integer.
(3q + 2)2 = 9q2 + 12q + 4,
which cannot be expressed in the
form 3m + 2.
Square of any positive integer cannot be
expressed in the form 3m + 2.
5.
A positive integer is of the form 3q + 1, q
being a natural number. Can you write its
square in any form other than 3m + 1, i.e.,
3m or 3m + 2 for some integer m? Justify
your answer.
Solution:
No.
Justification:
Consider the positive integer 3q + 1, where q
is a natural number.

(3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1, where m is an integer.
Thus (3q + 1)2 cannot be expressed in any
other form apart from 3m + 1.
3.
4.
Write whether the square of any positive
integer can be of the form 3m + 2, where m
is a natural number. Justify your answer.
Solution:
No.
Justification:
Let a be any positive integer. Then by Euclid’s
division lemma, we have
a = bq + r, where 0 ≤ r < b
For b = 3, we have
a = 3q + r, where 0 ≤ r < 3
...(i)
4 4
6.
The numbers 525 and 3000 are both
divisible only by 3, 5, 15, 25 and 75. What is
HCF (525, 3000)? Justify your answer.
Solution:
HCF(525, 3000) = 75
Justification:
3, 5, 15, 25 and 75 are the only common
factors of 525 and 3000.

75 is highest among the common factors.

HCF(525, 3000) = 75
7.
Show that the square of any positive integer
is either of the form 4q or 4q + 1 for some
integer q.
Solution:
Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4m + r for
some integer m ≥ 0 and r = 0, 1, 2, 3 because
0 ≤ r < 4.

a = 4m or 4m + 1 or 4m + 2 or 4m + 3.

(4m)2 = 16m2 = 4(4m2)
= 4q, where q is some integer.
(4m + 1)2 = 16m2 + 8m + 1
= 4(4m2 + 2m) + 1
= 4q + 1, where q is some integer.
(4m + 2)2 = 16m2 + 16m + 4
= 4(4m2 + 4m + 1)
= 4q, where q is some integer.
Chapter 01: Real Numbers

(4m + 3)2 = 16m2 + 24m + 9
= 4(4m2 + 6m + 2) + 1
= 4q + 1, where q is some integer.
The square of any positive integer is either of
the form 4q or 4q + 1, where q is some integer.
8.
Show that cube of any positive integer is of
the form 4m, 4m + 1 or 4m + 3, for some
integer m.
Solution:
Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4q + r for
some integer q ≥ 0 and r = 0, 1, 2, 3 because
0 ≤ r < 4.

a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
(4q)3 = 64q3 = 4(16q3)
= 4m, where m is some integer.
(4q + 1)3 = 64q3 + 48q2 + 12q + 1
= 4(16q3 + 12q2 + 3) + 1
= 4m + 1, where m is some integer.
(4q + 2)3 = 64q3 + 96q2 + 48q + 8
= 4(16q3 + 24q2 + 12q + 2)
= 4m, where m is some integer.
(4q + 3)3 = 64q3 + 144q2 + 108q + 27
= 4(16q3 + 36q2 + 27q + 6) + 3
= 4m + 3, where m is some integer.

The cube of any positive integer is of the form
4m, 4m + 1 or 4m + 3 for some integer m.
9.
Show that the square of any positive integer
cannot be of the form 5q + 2 or 5q + 3 for
any integer q.
Solution:
Let a be the positive integer and b = 5.
Then, by Euclid’s algorithm, a = 5m + r for
some integer m ≥ 0 and r = 0, 1, 2, 3, 4
because 0 ≤ r < 5.

a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or
5m + 4.

(5m)2 = 25m2 = 5(5m2)
= 5q, where q is any integer.
(5m + 1)2 = 25m2 + 10m + 1
= 5(5m2 + 2m) + 1
= 5q + 1, where q is any integer.
2
(5m + 2) = 25m2 + 20m + 4
= 5(5m2 + 4m) + 4
= 5q + 4, where q is any integer.
(5m + 3)2 = 25m2 + 30m + 9
= 5(5m2 + 6m + 1) + 4
= 5q + 4, where q is any integer.
(5m + 4)2 = 25m2 + 40m + 16
= 5(5m2 + 8m + 3) + 1
= 5q + 1, where q is any integer.

The square of any positive integer is of the
form 5q, 5q + 1, 5q + 4 and cannot be of the
form 5q + 2 or 5q + 3 for any integer q.
10.
Show that the square of any positive integer
cannot be of the form 6m + 2 or 6m + 5 for
any integer m.
Solution:
Let a be the positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for
some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5
because 0 ≤ r < 5.

a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
or 6q + 5.
(6q)2 = 36q2 = 6(6q2)
= 6m, where m is any integer.
(6q + 1)2 = 36q2 + 12q + 1
= 6(6q2 + 2q) + 1
= 6m + 1, where m is any integer.
(6q + 2)2 = 36q2 + 24q + 4
= 6(6q2 + 4q) + 4
= 6m + 4, where m is any integer.
(6q + 3)2 = 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3, where m is any integer.
(6q + 4)2 = 36q2 + 48q + 16
= 6(6q2 + 7q + 2) + 4
= 6m + 4, where m is any integer.
(6q + 5)2 = 36q2 + 60q + 25
= 6(6q2 + 10q + 4) + 1
= 6m + 1, where m is any integer.

The square of any positive integer is of the
form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot
be of the form 6m + 2 or 6m + 5 for any
integer m.
11.
Show that the square of any odd integer is
of the form 4q + 1, for some integer q.
Solution:
Let a be any odd integer and b = 4.
Then, by Euclid’s algorithm, a = 4m + r for
some integer m  0 and r = 0,1,2,3 because
0  r  4.

a = 4m or 4m + 1 or 4m + 2 or 4m + 3

a = 4m + 1 or 4m + 3
Here, a cannot be 4m or 4m + 2, as they are
divisible by 2.

(4m + 1)2 = 16m2 + 8m + 1
= 4(4m2 + 2m) + 1
= 4q + 1, where q is some integer.
(4m + 3)2 = 16m2 + 24m + 9
= 4(4m2 + 6m + 2) + 1
= 4q + 1, where q is some integer.

The square of any odd integer is of the form
4q + 1, for some integer q.
If n is an odd integer, then show that n2 – 1
is divisible by 8.
Solution:
Any odd integer n is of the form 4m + 1 or
4m + 3.
12.
5
Class X: Mathematics


n2 – 1 = (4m + 1)2 – 1
= 16m2 + 8m
= 8(2m2 + m),
which is divisible by 8.
Also, n2 – 1 = (4m + 3)2 – 1
= 16m2 + 24m + 8
= 8(2m2 + 3m + 1),
which is divisible by 8.
n2 – 1 is divisible by 8 for any odd integer n.
13.
Prove that if x and y are both odd positive
integers, then x2 + y2 is even but not
divisible by 4.
Solution:
Since x and y are odd positive integers, we
have
x = 2m + 1 and y = 2n + 1

x2 + y2 = (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 + 4n + 1
= 4(m2 + n2) + 4(m + n) + 2
2
2

x + y is an even number but not divisible by 4.
14.
Use Euclid’s division algorithm to find HCF
of 441, 567, 693.
Solution:
By Euclid’s division algorithm,
693 = 567  1 + 126
567 = 126  4 + 63
126 = 63  2 + 0

HCF (693, 567) = 63
441 = 63  7 + 0

HCF(441, 63) = 63

HCF (693, 567, 441) = 63
15.
Using Euclid’s division algorithm, find the
largest number that divides 1251, 9377 and
15628 leaving remainders 1, 2 and 3,
respectively.
Solution:
Since, 1, 2 and 3 are the remainders of 1251,
9377 and 15628 respectively.

1251 – 1 = 1250 is exactly divisible by the
required number,
9377 – 2 = 9375 is exactly divisible by the
required number,
15628 – 3 = 15625 is exactly divisible by the
required number.

required number = HCF of 1250, 9375 and
15625.
By Euclid’s division algorithm,
15625 = 9375  1 + 6250
9375 = 6250  1 + 3125
6250 = 3125  2 + 0

HCF (15625, 9375) = 3125
3125 = 1250  2 + 625
1250 = 625  2 + 0
6 6



HCF(3125, 1250) = 625
HCF (1250, 9375, 15625) = 625
the largest number is 625.
16.
Show that the cube of a positive integer of
the form 6q + r, q is an integer and r = 0, 1,
2, 3, 4, 5 is also of the form 6m+r.
Solution:
6q + r is a positive integer, where q is an
integer and r = 0, 1, 2, 3, 4, 5
Then, the positive integers are of the form 6q,
6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.
Taking cube of each term, we have,
(6q)3 = 216 q3 = 6(36q3) + 0
= 6m + 0, where m is an integer
(6q+1)3 = 216q3 + 108q2 + 18q + 1
= 6(36q3 + 18q2 + 3q) + 1
= 6m + 1, where m is an integer
(6q+2)3 = 216q3 + 216q2 + 72q + 8
= 6(36q3 + 36q2 + 12q + 1) +2
= 6m + 2, where m is an integer
(6q+3)3 = 216q3 + 324q2 + 162q + 27
= 6(36q3 + 54q2 + 27q + 4) + 3
= 6m + 3, where m is an integer
(6q+4)3 = 216q3 + 432q2 + 288q + 64
= 6(36q3 + 72q2 + 48q + 10) + 4
= 6m + 4, where m is an integer
(6q+5)3 = 216q3 + 540q2 + 450q + 125
= 6(36q3 + 90q2 + 75q + 20) + 5
= 6m + 5, where m is an integer

the cube of a positive integer of the form
6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is
also of the form 6m + r.
17.
Prove that one and only one out of n, n + 2
and n + 4 is divisible by 3, where n is any
positive integer.
Solution:
By Euclid’s division lemma, we have
a = bq + r; 0  r < b
For a = n and b = 3, we have
n = 3q + r,
…(i)
where q is an integer
and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
n = 3q

n is divisible by 3.
n + 2 = 3q + 2

n + 2 is not divisible by 3.
n + 4 = 3q + 4

n + 4 is not divisible by 3.
Putting r = 1 in (i), we get
n = 3q + 1

n is not divisible by 3.
n + 2 = 3q + 3 = 3(q + 1)

n + 2 is divisible by 3.
n + 4 = 3q + 5
Chapter 01: Real Numbers




n + 4 is not divisible by 3.
Putting r = 2 in (i), we get
n = 3q + 2
n is not divisible by 3.
n + 2 = 3q + 4
n + 2 is not divisible by 3.
n + 4 = 3q + 6 = 3(q + 2)
n + 4 is divisible by 3
Thus for each value of r such that 0 ≤ r < 3
only one out of n, n + 2 and n + 4 is divisible
by 3.
18.
Prove that one of any three consecutive
positive integers must be divisible by 3.
Solution:
Let the three consecutive positive integers be
n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0  r < b
For a = n and b = 3, we have
n = 3q + r
…(i)
Where q is an integer
and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
n = 3q

n is divisible by 3.
n + 1 = 3q + 1

n + 1 is not divisible by 3.
n + 2 = 3q + 2

n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
n = 3q + 1

n is not divisible by 3.
n + 1 = 3q + 2

n + 1 is not divisible by 3.
n + 2 = 3q + 3 = 3(q + 1)

n + 2 is divisible by 3.
Putting r = 2 in (i), we get
n = 3q + 2

n is not divisible by 3.
n + 1 = 3q + 3 = 3(q + 1)

n + 1 is divisible by 3.
n + 2 = 3q + 4

n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3
only one out of n, n + 1 and n + 2 is divisible
by 3.
For any positive integer n, prove that n3 – n
is divisible by 6.
[CBSE 2012]
Solution:
n3 – n = n (n2 – 1) = n (n – 1) (n + 1)

n3 – n is product of three consecutive positive
integers, where n is any positive integer.
Since one out of every two consecutive
integers is divisible by 2.
19.



The product n3 – n is divisible by 2.
Since one out of every three consecutive
integers is divisible by 3.
The product n3 – n is divisible by 3.
Any number which is divisible by 2 and 3 is
also divisible by 6.
The product n3 – n is divisible by 6.
20.
Show that one and only one out of n, n + 4,
n + 8, n + 12 and n + 16 is divisible by 5,
where n is any positive integer.
Solution:
By Euclid’s division lemma, we have
a = bq + r; 0  r < b
For a = n and b = 5, we have
n = 5q + r
…(i)
Where q is an integer
and 0 ≤ r < 5, i.e. r = 0, 1, 2, 3, 4.
Putting r = 0 in (i), we get
n = 5q

n is divisible by 5.
n + 4 = 5q + 4

n + 4 is not divisible by 5.
n + 8 = 5q + 8

n + 8 is not divisible by 5.
n + 12 = 5q + 12

n + 12 is not divisible by 5.
n + 16 = 5q + 16

n + 16 is not divisible by 5.
Putting r = 1 in (i), we get
n = 5q + 1

n is not divisible by 5.
n + 4 = 5q + 5 = 5(q + 1)

n + 4 is divisible by 5.
n + 8 = 5q + 9

n + 8 is not divisible by 5.
n + 12 = 5q + 13

n + 12 is not divisible by 5.
n + 16 = 5q + 17

n + 16 is not divisible by 5.
Putting r = 2 in (i), we get
n = 5q + 2

n is not divisible by 5.
n + 4 = 5q + 6

n + 4 is not divisible by 5.
n + 8 = 5q + 10 = 5(q + 2)

n + 8 is divisible by 5.
n + 12 = 5q + 14

n + 12 is not divisible by 5.
n + 16 = 5q + 18

n + 16 is not divisible by 5.
Putting r = 3 in (i), we get
n = 5q + 3

n is not divisible by 5.
n + 4 = 5q + 7

n + 4 is not divisible by 5.
7
Class X: Mathematics








n + 8 = 5q + 11
n + 8 is not divisible by 5.
n + 12 = 5q + 15 = 5(q + 3)
n + 12 is divisible by 5.
n + 16 = 5q + 19
n + 16 is not divisible by 5.
Putting r = 4 in (i), we get
n = 5q + 4
n is not divisible by 5.
n + 4 = 5q + 8
n + 4 is not divisible by 5.
n + 8 = 5q + 12
n + 8 is not divisible by 5.
n + 12 = 5q + 16
n + 12 is not divisible by 5.
n + 16 = 5q + 20 = 5(q + 4)
n + 16 is divisible by 5.
Thus for each value of r such that 0 ≤ r < 5
only one out of n, n + 4, n + 8, n + 12 and
n + 16 is divisible by 5.
Practice problems based on Exercise 1.1
1.
Find the HCF of 1656 and 4025 by Euclid’s
[CBSE 2013]
division algorithm.
2.
Use Euclid’s division algorithm to find HCF
[CBSE 2015]
of 65 and 175.
3.
Find the HCF of 1620, 1725 and 255 by
[CBSE 2012]
Euclid’s division algorithm.
4.
Using Euclid’s division algorithm, find
whether the pair of numbers 847, 2160 are
[CBSE 2012]
coprime or not.
5.
Find the largest number that divides 2053 and
967 and leaves remainders 5 and 7
respectively.
6.
Find HCF of 81 and 237 and express it as a
linear combination of 81 and 237 i.e., HCF of
81, 237 = 81x + 237y for some x and y.
[CBSE 2012]
7.
Show that the square of an odd positive
integer is of the form 8m + 1, for some whole
number m.
8.
Show that the square of any positive integer is
of the form 4m or 4m + 1, where m is any
[CBSE 2012]
integer.
9.
Show that any positive odd integer is of the
form 4q + 1 or 4q + 3, where q is some
[CBSE 2012]
integer.
10.
Prove that n2  n is divisible by 2 for every
positive integer n.
[CBSE 2012]
8 8
11.
12.
13.
An army contingent of 104 members is to
march behind an army band of 96 members in
a parade. The two groups are to march in the
same number of columns. What is the
maximum number of columns in which they
can march?
[CBSE 2011]
144 cartons of Coke cans and 90 cartons of
Pepsi cans are to be stacked in a canteen. If
each stack is of the same height and if it
contain cartons of the same drink, what would
be the greatest number of cartons each stack
would have?
[CBSE 2011]
Pens are sold in pack of 8 and notepads are
sold in pack of 12. Find the least number of
pack of each type that one should buy so that
there are equal number of pen and notepads.
[CBSE 2014]
Answers
1.
4.
11.
23
Yes
8
2.
5.
12.
5
64
18
3.
6.
13.
5
3
2, 3
Multiple Choice Questions
1.
2.
3.
4.
5.
6.
Euclid’s division lemma states that for two
positive integers a and b, there exist unique
integers q and r such that a = bq + r, where r
must satisfy
[NCERT Exemplar]
(A) 1 < r < b
(B) 0 < r  b
(C) 0  r < b
(D) 0 < r < b
For some integer m, every even integer is of
[NCERT Exemplar]
the form
(A) m
(B) m + 1
(C) 2m
(D) 2m + 1
For some integer q, every odd integer is of the
form
[NCERT Exemplar]
(A) q
(B) q + 1
(D) 2q + 1
(C) 2q
n2  1 is divisible by 8, if n is
[NCERT Exemplar]
(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer
The largest number which divides 70 and 125,
leaving remainders 5 and 8, respectively, is
[NCERT Exemplar]
(A) 13
(B) 65
(C) 875
(D) 1750
For any positive integer a and 3, there exist
unique integers q and r such that a = 3q + r,
[CBSE 2012]
where r must satisfy
(A) 0  r < 3
(B) 1 < r < 3
(C) 0 < r < 3
(D) 0 < r  3
Chapter 01: Real Numbers
The Fundamental Theorem of Arithmetic
Every composite number can be expressed
(factorised) as a product of primes and this
factorisation is unique, apart from the order in which
the prime factors occur.
This fundamental theorem of arithmetic help us to
find HCF and LCM of the numbers.
This method is also called the prime factorisation
method.
Example:
Consider a composite number 2352.
2352
2


HCF of two numbers is always a factor of
their LCM.
LCM is always a multiple of HCF.
HCF (p, q, r)  LCM(p, q, r)  p  q  r,
where p, q, r are positive integers. However,
the following results hold good for three
numbers p, q and r:
LCM(p, q, r) =
p.q.r.HCF(p,q,r)
HCF(p,q).HCF(q,r).HCF(p,r)
HCF(p, q, r) =
p.q.r.LCM(p,q,r)
LCM(p,q).LCM(q,r).LCM(p,r)
NCERT Exercise 1.2
294
1.
147
2
3
49
7
7
Prime factorization of 2352 is
2352 = 2  2  2  2  3  7  7
= 24  3  72
HCF and LCM:
HCF = Product of the smallest power of each
common prime factor in the numbers.
LCM =Product of the greatest power of each prime
factor in the numbers.
Relation between HCF and LCM of any two
positive integers:
For any two positive integers a and b,
HCF(a, b)  LCM(a, b) = a  b
Example:
Find HCF and LCM of the following pairs of integers.
i.
60 and 72
ii.
12, 30 and 144


588
2
i.
Things to Remember
1176
2


30 = 2  3  5
144 = 2  2  2  2  3  3 = 24  32
HCF = 2  3 = 6 and
LCM = 24  32  5 = 720
The prime factorization of 60 and 72 gives:
60 = 2  2  3  5 = 22  3  5
72 = 2  2  2  3  3 = 23  32
HCF = 22  3 = 12 and
LCM = 23  32  51 = 360
OR
LCM =
Express each number as a product of its
prime factors:
i. 140
ii.
156
iii. 3825
iv.
5005
v. 7429
Solution:
i.
140
LCM = 360
ii.
The prime factorization of 12, 30 and 144
gives:
12 = 2  2  3 = 22  3
35
2
7
5

140 = 2  2  5  7 = 22  5  7
ii.
156
2
78
39
2
3

13
156 = 2  2  3  13 = 22  3  13
iii.
3825
1275
3
60  72 60  72
=
HCF
12

70
2
425
3
85
5
5

17
3825 = 3  3  5  5  17 = 32  52  17
9
Class X: Mathematics
iv.
5005
1001
5
143
7
11

v.
5005 = 5  7  11  13
7429
17
437
19

13
23
7429 = 17  19  23
2.
Find the LCM and HCF of the following
pairs of integers and verify that
LCM  HCF = product of the two numbers.
i.
26 and 91
ii.
510 and 92
[CBSE 2011]
iii. 336 and 54
Solution:
i.
26 = 2  13
91 = 7  13
LCM(26, 91) = 2  7  13 = 182
HCF (26, 91) = 13
Verification:
LCM  HCF = 182  13 = 2366
Product of two numbers = 26  91 = 2366

LCM  HCF = Product of two numbers
ii.

iii.

3.
10
10 510 = 2  3  5  17
92 = 2  2  23 = 22  23
LCM(510, 92) = 22  3  5  17  23 = 23460
HCF(510, 92) = 2
Verification:
LCM  HCF = 23460  2 = 46920
Product of two numbers = 510  92 = 46920
LCM  HCF = Product of two numbers
336 = 2  2  2  2  3  7 = 24  3  7
54 = 2  3  3  3 = 2  33
LCM(336, 54) = 24  33  7 = 3024
HCF(336, 54) = 2  3 = 6
Verification:
LCM  HCF = 3024  6 = 18144
Product of two numbers = 336  54 = 18144
LCM  HCF = Product of two numbers
Find the LCM and HCF of the following
integers by applying the prime factorization
method.
i.
12, 15 and 21
ii.
17, 23 and 29
iii. 8, 9 and 25
Solution:
i.
12 = 2  2  3 = 22  3
15 = 3  5
21 = 3  7
LCM(12, 15, 21) = 22  3  5  7 = 420
HCF(12, 15, 21) = 3
ii.
17 = 1  17
23 = 1  23
29 = 1  29
LCM(17, 23, 29) = 1  17  29  23 = 11339
HCF(17, 23, 29) = 1
iii.
8 = 2  2  2 = 23
9 = 3  3 = 32
25 = 5  5 = 52
LCM(8, 9, 25) = 23  32  52 = 1800
HCF(8, 9, 25) = 1
4.
Given that HCF (306, 657) = 9, find LCM
(306, 657).
Solution:
LCM  HCF = Product of two numbers
 LCM  9 = 306  657
 LCM =
306  657
= 22338
9
Check whether 6n can end with the digit 0
for any natural number n.
Solution:
If the number 6n, for any natural number n, ends
with the digit zero, then it is divisible by 5. That
is, the prime factorization of 6n contains the
prime 5. This is not possible because
prime factorisation of 6n = (2  3)n = 2n  3n;
so the only primes in the factorisation of 6n are
2 and 3 and the uniqueness of the fundamental
theorem of arithmetic guarantees that there are
no other primes in the factorization of 6n.
So, there is no natural number n for which 6n
ends with the digit zero.
5.
Explain why 7  11  13 + 13 and
7  6  5  4  3  2  1 + 5 are composite
numbers.
Solution:
7  11  13 + 13 = (7  11 + 1)  13
= (77 + 1)  13 = 78  13
= (2  3  13)  13
= 2  3  132
Since 7  11  13 + 13 can be expressed as a
product of primes, it is a composite number.
7654321+5
= (7  6  4  3  2  1 + 1)  5
= (1008 + 1)  5 = 1009  5
= 5  1009
Since 7  6  5  4  3  2  1 + 5 can be
expressed as a product of primes, it is a
composite number.
6.
Chapter 01: Real Numbers
7.
There is a circular path around a sports field.
Sonia takes 18 minutes to drive one round of
the field, while Ravi takes 12 minutes for the
same. Suppose they both start at the same
point and at the same time, and go in the
same direction. After how many minutes will
they meet again at the starting point?
Solution:
Sonia and Ravi start at the same time from the
same point and go in the same direction and at
the same time, so to find the time when they
will meet again at the starting point, we have
to find the LCM of 12 and 18.
12 = 2  2  3 = 22  3
18 = 2  3  3 = 2  32

LCM(12, 18) = 22  32 = 36

Sonia and Ravi will meet again at the starting
point after 36 minutes.
Solution:
Every composite number can be expressed
(factorised) as a product of primes and this
factorisation is unique, apart from the order in
which the prime factors occur.
Yes.
Justification:
HCF of two numbers is always a factor of
their LCM.
But here 18 is a factor of 378.

Two numbers can have 18 as their HCF and
380 as their LCM.
6.
Find the missing numbers a, b, c and d in
the given factor tree.
18018
9009
2
Given that HCF (306, 1314) = 18. Find
LCM (306, 1314).
[CBSE 2013]
Solution:
LCM  HCF = Product of two numbers
 LCM  18 = 306  1314
3003
a
Problems based on Exercise 1.2
 LCM =
306  1314
= 22338
18
Find the HCF and LCM of 90 and 144 by
the method of prime factorization.
[CBSE 2012]
Solution:
90 = 2  3  3  5 = 2  32  5
144 = 2  2  2  2  3  3 = 24  32
HCF (90, 144) = 2  32 = 18
LCM (90, 144) = 24  32  5 = 720
State Fundamental theorem of arithmetic.
Is it possible for the HCF and LCM of two
numbers to be 18 and 378 respectively.
[CBSE 2014]
Justify your answer.
[CBSE 2012]
Solution:
9009
=3
3003
1001
=7
1001 = b  143  b =
143
9009 = a  3003  a =

7.
Find the LCM of 96 and 360 by using
fundamental theorem of arithmetic.
[CBSE 2012]
Solution:
96 = 2  2  2  2  2  3 = 25  3
360 = 2  2  2  3  3  5 = 23  32  5

LCM (96, 360) = 25  32  5 = 1440
5.
d
c
3.
Determine the values of p and q so that the
prime factorization of 2520 is expressible as
[CBSE 2014]
23  3p  q  7.
Solution:
Since, 2520 = 2  2  2  3  3  5  7
= 23  32  5  7

p = 2 and q = 5
143
b
2.
4.
1001
3
1.
143 = c  d
Since, 143 = 11  13
c = 11, d = 13 or c = 13, d = 11
Complete the following factor tree and find
the composite number x.
x
3381
2
3
?
161
7
7
Solution:
6762
[CBSE 2012]
2  3381 = 6762
3381
2
3
1127
x = 6762
7  161 = 1127
161
7

?
7
23
161 = 7  23
11
Class X: Mathematics
Explain why 7  13  11 + 11 and
(7  6  5  4  3  2  1) + 3 are composite
[CBSE 2012]
numbers.
Solution:
7  13  11 + 11 = (7  13 + 1)  11
= (91 + 1)  11 = 92  11
= (2  2  23)  11
= 2  2  11  23
Since 7  13  11 + 11 can be expressed as a
product of primes, it is a composite number.
(7  6  5  4  3  2  1) + 3
= (7  6  5  4  2  1 + 1)  3
= (1680 + 1)  3 = 1681  3
= (41  41)  3 = 3  41  41
Since (7  6  5  4  3  2  1) + 3 can be
expressed as a product of primes, it is a
composite number.
8.
9.
Three bells toll at intervals of 9, 12, 15 minutes
respectively. If they start tolling together, after
what time they next toll together? [CBSE 2011]
Solution:
9 = 3  3 = 32
12 = 2  2  3 = 22  3
15 = 3  5

LCM (9, 12, 15) = 22  32  5 = 180

the bells will toll together after 180 minutes.
10.
The HCF of 65 and 117 is expressible in the
form 65m  117. Find the value of m. Also
find the LCM of 65 and 117 using prime
[CBSE 2011]
factorization method.
Solution:
117 = 65  1 + 52
65 = 52  1 + 13
65 = 13  5 + 0

HCF(65, 117) = 13
Since, HCF(65, 117) = 65m  117

13 = 65m  117
 65m = 117 + 13 = 130
m=

11.
130
=2
65
Now, 65 = 5  13
and 117 = 3  3  13 = 32  13
LCM(65, 117) = 32  5  13 = 585
If two positive integers x and y are
expressible in terms of primes as x = p2q3
and y = p3q, what can you say about their
LCM and HCF. Is LCM a multiple of
HCF? Explain.
[CBSE 2014]
Solution:
x = p2q3 and y = p3q
HCF(x, y) = p2  q
LCM(x, y) = p3  q3
LCM(x, y) = (p  q2)  HCF(x, y)

LCM is a multiple of HCF.
12
12 NCERT Exemplar
1
Explain why 3 × 5 × 7 + 7 is a composite
number.
Solution:
3 × 5 × 7 + 7 = (3  5 + 1) × 7
= (15 + 1)  7
= 16 × 7 = 7  16
Since (3 × 5 × 7 + 7) can be expressed as a
product of primes, it is a composite number.
2.
Can two numbers have 18 as their HCF and
380 as their LCM? Give reasons.
Solution:
No.
Justification:
HCF of two numbers is always a factor of
their LCM.
But here 18 is not a factor of 380.

Two numbers cannot have 18 as their HCF
and 380 as their LCM.
Show that 12n cannot end with the digit 0 or
5 for any natural number n.
Solution:
If the number 12n, for any natural number n,
ends with the digit 0 or 5, then it is divisible by
5. That is, the prime factorization of 12n contains
the prime 5. This is not possible because prime
factorisation of 12n = (22  3)n = 22n  3n; so
the only primes in the factorisation of 12n are
2 and 3 and the uniqueness of the fundamental
theorem of arithmetic guarantees that there are
no other primes in the factorization of 12n.
So, there is no natural number n for which 12n
ends with the digit zero.
3.
4.
On morning walk, three persons step off
together and their steps measure 40 cm,
42 cm and 45 cm, respectively. What is the
minimum distance each should walk so that
each can cover the same distance in
complete steps?
Solution:
Since, the three persons start walking together.

The minimum distance covered by each of
them in complete steps = LCM of the
measures of their steps
= LCM (40, 42, 45)
40 = 2  2  2  5 = 23  5
42 = 2  3  7
45 = 3  3  5 = 32  5

LCM (40, 42, 45) = 23  32  5  7
= 2520

Each person should walk a minimum distance
of 2520 cm in complete steps.
Chapter 01: Real Numbers
Practice problems based on Exercise 1.2
Multiple Choice Questions
1.
Find the HCF of 960 and 432.
1.
2.
Find the HCF and LCM of 404 and 96 and
verify HCF  LCM = Product of the two given
numbers.
[CBSE 2012]
If the HCF of 65 and 117 is expressible in the
form 65m  117, then the value of m is
[NCERT Exemplar]
(B) 2
(A) 4
(C) 1
(D) 3
3.
Express 2120 as the product of its prime
[CBSE 2012]
factors.
2.
4.
Find the HCF and LCM of 60, 120 and 288.
[CBSE 2012]
5.
The HCF and LCM of two numbers are 9 and
360 respectively. If one number is 45, write
the other number.
If two positive integers a and b are written as
a = x3y2 and b = xy3; x, y are prime numbers,
then HCF (a, b) is
[NCERT Exemplar]
(B) xy2
(A) xy
3 3
(D) x2y2
(C) x y
3.
6.
The HCF of 45 and 105 is 15. Write their
[CBSE 2010]
LCM.
7.
The LCM of 2 numbers is 14 times their HCF.
The sum of LCM and HCF is 600. If one
number is 280, then find the other number.
[CBSE 2012]
If two positive integers p and q can be
expressed as p = ab2 and q = a3b; a, b being
prime numbers, then LCM (p, q) is
[NCERT Exemplar]
(A) ab
(B) a2b2
(C) a3b2
(D) a3b3
4.
The least number that is divisible by all the
numbers from 1 to 10 (both inclusive) is
[NCERT Exemplar]
(A) 10
(B) 100
(D) 2520
(C) 504
5.
Given that LCM (91, 26) = 182, then HCF
(91, 26) is
[CBSE 2011]
(A) 13
(B) 26
(C) 7
(D) 9
6.
If the HCF of 55 and 99 is expressible in the
form 55m  99, then the value of m is
[CBSE 2011]
(A) 4
(B) 2
(C) 1
(D) 3
7.
The values of x and y in the given figure are
8.
Find the HCF of the numbers 520 and 468 by
prime factorization method.
[CBSE 2014]
9.
Find LCM of the numbers given below:
m, 2m, 3m, 4m and 5m, where m is any
[CBSE 2014]
positive integer.
10.
Show that the number 4, when n is a natural
number cannot end with the digit zero.
[CBSE 2012]
11.
Show that 7n cannot end with the digit zero,
[CBSE 2012]
for any natural number n.
12.
Can two numbers have 15 as their HCF and
175 as their LCM? Give reasons.
[CBSE 2012]
13.
14.
y
State Fundamental theorem of arithmetic.
Is it possible that HCF and LCM of two
numbers be 24 and 540 respectively. Justify
your answer.
[CBSE 2015]
Explain why the number 7  5  3  2 + 3 is
not a prime number?
[CBSE 2015]
(A)
(C)
8.
1.
2.
3.
4.
5.
7.
9.
13.
48
HCF = 4, LCM = 9696
23  5  53
HCF = 12, LCM = 1440
72
6.
80
8.
120 m
12.
No
x
3
7
x = 10; y = 14
x = 21; y = 25
(B)
(D)
[CBSE 2012]
x = 21; y = 84
x = 10; y = 40
LCM of 23  32 and 22  33 is
(A)
(C)
Answers
4
23
23  33
(B)
(D)
[CBSE 2012]
33
22  32
Revisiting Irrational Numbers
315
52
No
Rational Number:
A number r is called a rational number, if it can be
p
q
written in the form , where p and q are integers and
q  0.
13
Class X: Mathematics
Irrational Number:
A number s is called a irrational number, if it cannot
be written in the form
p
, where p and q are integers
q
and q  0.
Theorem 1.3:
Let p be a prime number. If p divides a2, then p
divides a, where a is a positive integer
Example:
Let prime number p be 5
and a = 15

a2 = 225
Here, p divides a2 i.e.,

p divides a i.e.,
2=
a 15
= =3
p 5
Things to Remember


14
14 such that



a
b
Suppose a and b have a common factor other than 1,
then we can divide by the common factor, and assume
that a and b are coprime.

2b = a
 2b2 = a2
…(i) [Squaring both the sides]
2
 a is divisible by 2
 a is divisible by 2
So, we can write a = 2c for some integer c.

a2 = (2c)2
…[Squaring both the sides]
 2b2 = 4c2 …[From (i)]
 b2 = 2c2
 b2 is divisible by 2
 b is divisible by 2

2 divides both a and b.

a and b have at least 2 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because of our incorrect
assumption that 2 is rational.

2 is an irrational number.

1.
Prove that 5 is irrational.
Solution:
Let us assume, to the contrary, that 5 is
rational.
So, we can find coprime integers a and b(b  0)
a 2 225
=
= 45
5
p
Theorem 1.4:
Prove that 2 is an irrational number:
Proof:
Let us assume, to the contrary, that 2 is a rational
number.
So, we can find coprime integers a and b(b  0) such
that
NCERT Exercise 1.3
Any two numbers whose highest common
factor is 1 are said to be co-prime numbers.
Eg. i.
15, 22
ii.
11, 39
The sum or difference of a rational and an
irrational number is irrational.
The product and quotient of a non-zero
rational and irrational number is irrational.
5=
a
b
 5b=a
 5b2 = a2
….(i) [Squaring both the sides]
 a2 is divisible by 5
 a is divisible by 5
So, we can write a = 5c for some integer c.
a2 = 25c2
….[Squaring both the sides]
 5b2 = 25c2
….[From (i)]
 b2 = 5c2
 b2 is divisible by 5
 b is divisible by 5
5 divides both a and b.
a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are
coprime.
This contradiction arises because we have
assumed that 5 is rational.

5 is irrational.
2.
Prove that 3 + 2 5 is irrational.
Solution:
Let us assume, to the contrary, that 3 + 2 5 is
rational.
So, we can find coprime integers a and
b(b  0) such that 3 + 2 5 =

2 5=

5=
a
b
a
3
b
1  a  3b 


2 b 
Since a and b are integers,
1  a  3b 

 is
2 b 

rational.
5 is rational

But this contradicts the fact that 5 is
irrational.
3 + 2 5 is irrational.
3.
Prove that the following are irrationals:
i.
iii.
1
ii.
2
6+
2
7 5
Chapter 01: Real Numbers
Solution:
i.
Let us assume, to the contrary, that
1
is
2
rational.
So, we can find coprime integers a and b(b  0)
such that





ii.
2=
1
a
=
b
2
b
a
 2a=b
 2a2 = b2
….(i) [Squaring both the sides]
2
 b is divisible by 2
 b is divisible by 2
So, we can write b = 2c for some integer c.
b2 = 4c2
….[Squaring both the sides]
 2a2 = 4c2
….[From (i)]
 a2 = 2c2
 a2 is divisible by 2
 a is divisible by 2
2 divides both a and b.
a and b have at least 2 as a common factor.
But this contradicts the fact that a and b are
coprime.
Let us assume, to the contrary, that 7 5 is
rational.
So, we can find coprime integers a and b(b  0)
5=
a
b
a
7b
Since 7, a and b are integers,


iii.
a
is rational.
7b
But this contradicts the fact that
irrational.
7 5 is irrational.
5

such that




a
b
a
a  6b
2 = 6=
b
b



2 is rational.
2
a
b
If p is a prime number, then prove that p
is irrational.
[CBSE 2014]
Solution:
Let us assume, to the contrary, that p is
rational.
So, we can find coprime integers a and b(b  0)
2=
But this contradicts the fact that
irrational.
6 + 2 is irrational.
3=
2.
is
Let us assume, to the contrary that 6 + 2 is
rational.
So, we can find coprime integers a and b(b  0)
3 is an irrational number.
 3b=a
 3b2 = a2
….(i) [Squaring both the sides]
 a2 is divisible by 3
 a is divisible by 3
So, we can write a = 3c for some integer c.
a2 = 9c2
….[Squaring both the sides]
 3b2 = 9c2
….[From (i)]
 b2 = 3c2
 b2 is divisible by 3
 b is divisible by 3
3 divides both a and b.
a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are
coprime.
This contradiction arises because we have
assumed that 3 is rational.
3 is irrational.
such that
a  6b
Since a and b are integers,
is rational.
b

Show that
[CBSE 2015]
Solution:
Let us assume, to the contrary, that 3 is
rational.
So, we can find coprime integers a and b(b  0)
5 is rational.
such that 6 +

1.
1
is irrational.
2
such that 7 5 =

Problems based on Exercise 1.3
is

p=
a
b
 pb=a
 pb2 = a2
….(i) [Squaring both the sides]
 a2 is divisible by p
 a is divisible by p
So, we can write a = pc for some integer c.
a2 = p2c2
….[Squaring both the sides]
 pb2 = p2c2
….[From (i)]
 b2 = pc2
 b2 is divisible by p
 b is divisible by p
p divides both a and b.
a and b have at least p as a common factor.
But this contradicts the fact that a and b are
coprime.
This contradiction arises because we have
assumed that p is rational.
p is irrational.
15
Class X: Mathematics
Show that 4  5 2 is an irrational number.
[CBSE 2010]
Solution:
Let us assume, to the contrary that 4 – 5 2 is
rational.
So, we can find coprime integers a and b(b  0)
3.
such that 4 – 5 2 =
a
b


4b  a
is rational.
5b
2 is rational.
But this contradicts the fact that
irrational.
4 – 5 2 is irrational.
2
is
Show that 5 + 3 2 is an irrational number.
[CBSE 2012]
Solution:
Let us assume, to the contrary that 5 + 3 2 is
rational.
So, we can find coprime integers a and b(b  0)
a
b

5.
a
5
b
a  5b
2=
3b



2
a2  2
, which is a contradiction as the
2a
right hand side is rational number, while
irrational.
3 + 5 is irrational.
5 is
2
p  q = a, where a is rational.
q=a–
p
Squaring on both sides, we get
q = a2 + p  2a p
2 is rational.
But this contradicts the fact that
irrational.
5 + 3 2 is irrational.
5=
Let
is
Is product of a rational number and an
irrational number, a rational number? Is
product of two irrational numbers a
rational number or irrational number?
[CBSE 2015]
Justify giving examples.
Solution:
The product of a rational number and an
irrational number can be a rational number or
an irrational number.
Examples:
i.
rational number = 0
irrational number = 2

0  2 = 0, which is a rational number.
ii.
rational number = 2
irrational number = 2
16
16 2
Prove that p  q is irrational, where p, q
are primes.
Solution:
Let us suppose that p  q is rational.
a  5b
is rational.
Since a and b are integers,
3b

 3  = a  5 
2.
3 2 =

1.
Prove that 3  5 is irrational.
Solution:
Let us suppose that 3 + 5 is rational.
Let 3 + 5 = a, where a is rational.

3 = a 5
Squaring on both sides, we get
 3 = a2 + 5 – 2a 5
 2a 5 = a2 + 2
4.
such that 5 + 3 2 =
2  2 = 2 2 , which is an irrational number.
The product of two irrational numbers can be
a rational number or an irrational number.
Examples:
i.
2  2 = 4 = 2, which is a rational number
2  3 = 6 , which is an irrational number
ii.
NCERT Exemplar
a
5 2 =4–
b
4b  a
 2 =
5b
Since a and b are integers,



p=
a2  p  q
, which is a contradiction as the
2a
right hand side is rational number, while
irrational.
p  q is irrational.
p is
Practice problems based on Exercise 1.3
1.
2.
Prove that 2 is irrational.
Show that 3 2 is irrational.
3.
Prove that
2 3
is an irrational number.
5
4.
Prove that
13 5
is an irrational number.
7
5.
[CBSE 2014]
Prove that 3 + 5 is an irrational number.
[CBSE 2011]
[CBSE 2011]
Chapter 01: Real Numbers
8.
Prove that 3  2 is an irrational number.
Show that 2 3  1 is an irrational number.
[CBSE 2010]
Prove that 2 3  5 is an irrational number.
9.
Prove that
6.
7.
10.
2.
The product of a non zero rational and an
[NCERT Exemplar]
irrational number is
(A) always irrational
(B) always rational
(C) rational or irrational
(D) one
(A)
(B)
(C)
(D)
3.
4.
x=
=
[CBSE 2012]
Prove that 5 is an irrational number and
hence show that 2  5 is also an irrational
[CBSE 2011]
number.


5 + 2 is an irrational number.
Multiple Choice Questions
1.
Example:
Consider rational number x = 0.0945
22
is
7
[CBSE 2012]
a rational number
an irrational number
a prime number
an even number
x=
33  5  7 33  5  7
= 4 4
(5  2) 4
5 2
33  7
24  53
Here x is expressed in
p
form, where p and q are
q
co-prime numbers.
Also prime factorization of q is of the form
2n5m. i.e., 2453
Theorem 1.6:
Let x =
p
be a rational number, such that the prime
q
factorization of q is of the form 2n5m, where n, m are
non-negative integers. Then x has a decimal
expansion which terminates.
Example:
Consider a rational number
The reciprocal of an irrational number is
[CBSE 2012]
(A) an integer
(B) a rational
(C) a natural number
(D) an irrational
The product of two irrational numbers is
[CBSE 2012]
(A) always a rational
(B) always an irrational
(C) one
(D) always a non-zero number
Revisiting Rational Numbers and Their Decimal
Expansions
Rational number:
A number whose decimal expansion is terminating or
non-terminating recurring is rational.
Irrational number:
A number whose decimal expansion is non-terminating
non-recurring is irrational.
Theorem 1.5:
Let x be a rational number whose decimal expansion
terminates. Then x can be expressed in the form

945
945
=
10000 104
p
,
q
where p and q are co-prime and the prime
factorization of q is of the form 2n5m, where n, m are
non-negative integers.

x=
648
625
p 648 23  34 23  34  24
=
= 4 = 4 4
q 625
5
5 2
=
10368
= 1.0368
104
Theorem 1.7:
Let x =
p
be a rational number, such that the prime
q
factorisation of q is not of the form 2m5n, where m, n
are non-negative integers. Then, x has a decimal
expansion which is non-terminating repeating
(recurring).
Example:
Consider a rational number
133
= 12.0909…
11
Here, q = 11, which is not of the form 2m5n.
NCERT Exercise 1.4
1.
Without actually performing the long
division, state whether the following
rational numbers will have a terminating
decimal expansion or a non-terminating
repeating decimal expansion:
i.
iii.
v.
13
3125
64
455
29
343
ii.
iv.
vi.
17
8
15
1600
23
2352
17
Class X: Mathematics
vii.
ix.
129
2 2 57 7 5
35
50
viii.
x.
6
15
77
210
Solution:
i.
17
8
8 = 2  2  2 = 23 = 23  50
Since the denominator is of the form 2m  5n
the rational number has a terminating decimal
expansion.
iii.
64
455
455 = 5  7  13
Since the denominator is not of the form
2m  5n, the rational number has a nonterminating repeating decimal expansion.
iv.
vi.
23
23  52
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
vii.
129
22  57  75
Since the denominator is not of the form
2m  5n, the rational number has a nonterminating repeating decimal expansion.
viii.
6
3 2 2
=
=
15 3  5 5
5 = 20  51
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
ix.
35
5 7
7
=
=
50 5  10 10
10 = 2  5
18
18 2.
Write down the decimal expansions of those
rational numbers in Question 1 above
which
have
terminating
decimal
expansions.
Solution:
i.
13
13 13  25 13  32
416
= 5= 5 5=
= 5 = 0.00416
3125 5
10
5 2
(5  2)5
ii.
17 17 17  53 17  125 2125
= 3= 3 3=
=
= 2.125
8
2
103
2 5
(2  5)3
iv.
15
15
= 4
1600
2  102
15  625
15  54
=
= 4 4
(2  5) 4  102
2  5  102
=
vi.
29
343
343 = 7  7  7 = 73
Since the denominator is not of the form
2m  5n, the rational number has a nonterminating repeating decimal expansion.
77
7  11 11
=
=
210 7  30 30
30 = 2  3  5
Since the denominator is not of the form
2m  5n, the rational number has a nonterminating repeating decimal expansion.
15
5 3
3
=
=
1600
5  320
320
320 = 2  2  2  2  2  2  5 = 26  51
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
v.
x.
13
3125
3125 = 5  5  5  5  5 = 55 = 20  55
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
ii.
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
9375
9375
=
= 0.009375
104  102
106
23  53  22
23
23  125  4
= 3 2 3 2=
2
2 5
25  55
2 5 5 2
11500
11500
=
= 0.115
=
5
105
(2  5)
3
viii.
6
2 2 2
4
= =
=
= 0.4
15 5 5  2 10
ix.
35
7
=
= 0.7
50 10
3.
The following real numbers have decimal
expansions as given below. In each case,
decide whether they are rational or not. If
they are rational, and of the form
p
, what
q
can you say about the prime factors of q?
i.
43.123456789
ii.
0.120120012000120000…
iii.
43.123456789
Solution:
i.
Since the given number has a terminating
decimal expansion, it is a rational number of
the form
p
and q is of the form 2m  5n.
q
The prime factors of p and q will be either 2 or
5 or both.
Chapter 01: Real Numbers
ii.
iii.
Since the decimal expansion is neither
terminating nor recurring, the given number is
an irrational number.
Since the decimal expansion is nonterminating repeating, the given number is a
rational number of the form
p
and q is not of
q
4.
If
241
241
=
, find the values of m and n
4000 2m 5n
where m and n are non-negative integers.
Hence, write its decimal expansion without
actual division.
[CBSE 2012]
Solution:
241
241
= m n
4000 2 5
241
241
 5 3= m n
2 5
2 5
the form 2m  5n.
The prime factors of q will also have a factor
other than 2 or 5.
 m = 5 and n = 3
Problems based on Exercise 1.4
Now,
1.
Write the decimal expansion of
without actual division.
Solution:
27
1250
241
241
= 5 3
4000
2 5
241  52
25  53  52
241  25
= 5 5
2 5
6025
6025
=
=
105
(2  5)5
=
[CBSE 2015]
27
27
23  27
= 3
= 3 3
1250
5  10 2  5  10
8  27
=
(2  5)3  10
=
216
216
=
103  10 104
= 0.0216
 15
5 
Express    as a decimal without
 4 40 
actual division.
[CBSE 2011]
Solution:
2.
15 5
15  10 5
=
+

4 40
4  10 40
150 5
=

40 40
155
155
= 3
=
40
2 5
155  52
155  25
= 3 3
= 3
2 5
2  5  52
3875
3875
=
=
103
(2  5)3
= 3.875
3.
Write the denominator of the rational
number
257
in the form 2m  5n,
500
where m and n are non-negative integers.
Hence write its decimal expansion without
actual division.
[CBSE 2012]
Solution:
Denominator = 500 = 22  53
257
257
2  257
= 2 3=
500
2 5
2  22  53
514
514
514
= 3 3=
=
= 0.514
2  5 (2  5)3 103
= 0.06025
5.
What is the condition for the decimal
expansion of a rational number to
terminate? Explain with the help of an
example.
[CBSE 2015]
Solution:
Let x =
p
be a rational number, such that the
q
prime factorization of q is of the form 2m5n,
where n, m are non-negative integers. Then x
has a decimal expansion which terminates.
Example:
49
49

500 22  53
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
6.
Express the number 0.3178 in the form of
rational number
a
.
b
[CBSE 2011]
Solution:
Let x = 0.3178

x = 0.3178178178…
….(i)
Multiplying both sides of (i) by 10, we get
….(ii)
10x = 3.178178…
Multiplying both sides of (i) by 10000, we get
10000x = 3178.178… ….(iii)
Subtracting (ii) from (iii), we get
9990x = 3175
x=
3175
635
=
9990 1998
19
Class X: Mathematics
2.
NCERT Exemplar
1.
i.
Without actually performing the long
division, find if
987
will have terminating
10500
or non-terminating (repeating) decimal
expansion. Give reasons for your answer.
Solution:
47
987
3  7  47
47
=
=
=
10500 3  7  500 500 2 2  53
Since the denominator is of the form 2m  5n,
the rational number has a terminating decimal
expansion.
A rational number in its decimal expansion
is 327.7081. What can you say about the
prime factors of q, when this number is
p
expressed in the form ? Give reasons.
q
Solution:
3277081 3277081 p
327.7081 =
= 4 4 =
q
10000
2 5
m
n
Here, q is of the form 2 × 5 , where m and n
are natural numbers.
The prime factors of p and q will be either 2 or
5 or both.
ii.
3.
1.
2.
3.
1.
1.
iii.
v.
vii.
20
20 7
80
49
2
2  53  7 2
175
4000
13
875
ii.
iv.
vi.
27
40
7294
625
22
7
expansion
of
[CBSE 2012]
i.
terminating
ii.
terminating
iii. non-terminating repeating
iv.
terminating
v.
terminating
vi.
non-terminating repeating
vii. non-terminating repeating
i.
1.512
ii.
0.375
iii. 0.0266
0.00512
The decimal expansion of the rational number
(A)
(B)
(C)
(D)
2.
one decimal place
two decimal places
three decimal places
more than 3 decimal places
The decimal expansion of the rational number
14587
will terminate after
1250
(A)
(B)
(C)
(D)
3.
[NCERT Exemplar]
one decimal place
two decimal places
three decimal places
four decimal places
The decimal expansion of
17
will terminate
8
after how many places of decimals?
[CBSE 2011]
(A) 1
(B) 2
(C) 3
(D) will not terminate
Without actually performing the long division,
state whether the following rational numbers
will have a terminating decimal expansion or a
non-terminating repeating decimal expansion:
i.
decimal
33
will terminate after [NCERT Exemplar]
22.5
257
257
2  257
= 3 4=
5000 2  5
2  23  54
514
514
514
=
= 0.0514
= 4 4=
2 5
(2  5) 4 104
Practice problems based on Exercise 1.4
the
Multiple Choice Questions
257
in the form 2m  5n, where m,
5000
n are non-negative integers. Hence, write its
decimal expansion, without actual division.
Solution:
Denominator = 5000 = 23  54
down
75
23  5 2
ii.
Answers
Write the denominator of the rational
number
Write
189
125
133
5000
16
without actual division.
3125
2.
3.
Write down the decimal expansions of the
following rational numbers:
4.
From the following, the rational number
whose decimal expansion is terminating is
[CBSE 2011]
(A)
(C)
2
15
17
60
(B)
(D)
11
160
6
35
Chapter 01: Real Numbers
5.
The decimal expansion of  is
[CBSE 2011]
(A)
(B)
(C)
(D)
6.
(C)
125
441
15
1600
(B)
(D)
Write
2 45  3 20
whether
2 5
2 45  3 20
2 95  3 45
=
2 5
2 5
=
6 5  6 5 12 5
=
2 5
2 5
= 6, which is a rational number
77
210
129
22  52  72
6.
non-negative integers m and n, prime factors
[CBSE 2012]
of q are of the form
m
n
(A) 2  3
(B) 3m  5n
(C) 3n  7n
(D) 2m  5n
p
is a rational number (q  0), what is
q
If
the condition of q so that the decimal
The decimal representation of a rational
p
number
is a terminating decimal only if for
q
representation of
p
is terminating?
q
Solution:
For a rational number
p
to have terminating
q
decimal representation, the prime factorisation
of q should be of the form 2m  5n, where m
and n are non-negative integers.
HOTS Questions
One Mark Questions
1.
State Fundamental theorem of arithmetic.
Solution:
Every composite number can be expressed
(factorised) as a product of primes and this
factorisation is unique, apart from the order in
which the prime factors occur.
1.
Show that there is no positive integer n, for
which n  1  n  1 is rational.
[CBSE 2012]
Solution:
Let us assume that there is a positive integer n
for which n  1 + n  1 is rational and equal
A
, where A and B are positive integers
B
to
2.
Express 960 as the product of its prime
factors.
Solution:
960 = 2  2  2  2  2  2  3  5
= 26  3  5
(B  0). Then,

B
=
A
3.

B
=
A

n 1  n 1

n 1  n 1
B
=
=
A (n  1)  (n  1)
n 1  n 1
2
2B
A
....(ii)
Write the HCF of smallest composite
number and smallest prime number.
Solution:
The smallest composite number is 22 = 4
and the smallest prime number is 2.

HCF (4, 2) = 2
4.
Has the rational number
on
simplification gives a rational or an
irrational number.
Solution:
Which of the following rational numbers have
a terminating decimal expansion?
[CBSE 2011]
(A)
7.
terminating
non-terminating and non-recurring
non-terminating and recurring
doesn’t exist
5.
441
,
22.57 .7 2
n 1 +

A
B
n 1 =
....(i)
1
n 1  n 1

n 1  n 1
n 1  n 1
n 1 –
n 1 =

Adding (i) and (ii), we get
a
terminating or a non-terminating decimal
representation?
Solution:
Since the denominator is not of the form
2m  5n, the rational number has a nonterminating repeating decimal expansion.
A 2B A 2  2B2
=

AB
B A
2
2
A  2B
 n 1 =
2AB
2 n 1 =
Subtracting (ii) from (i), we get
n 1 =
A 2  2B2
2AB
21
Class X: Mathematics


Since, A and B are positive integers.
n  1 and n  1 are rationals.
But it is possible only when n + 1 and n  1
both are perfect squares. But they differ by 2
and any two perfect squares differ at least by
3.
n + 1 and n  1 cannot be perfect squares.
Hence, there is no positive integer n for which
n  1 + n  1 is rational.
Value Based Questions
1.
For a morning walk, three persons steps off
together. Their steps measure 80 cm, 85 cm
and 90 cm respectively. What is the
minimum distance each should walk to
show that they can cover the distance in
complete steps? Which value is preferred in
this situation?
[CBSE 2013]
Solution:
Since, the three persons start walking together.

The minimum distance covered by each of
them in complete steps = LCM of the
measures of their steps
= LCM (80, 85, 90)
80 = 2  2  2  2  5= 24  5
85 = 5  17
90 = 2  3  3  5 = 2  32  5

LCM (80, 85, 90) = 24  32  5  17
= 12240

Each person should walk a minimum distance
of 12240 cm in complete steps.
Morning walk is good for healthy life.
2.
A school library has 280 science journals
and 300 maths journals. Students were told
to stack these journals in such a way that
each stack contains equal number of
journals. Determine the number of stacks of
science and maths journals. What is the
benefit of library in student life?
Solution:
Since, each stack contains equal number of
journals.

Number of journals in each stack
= HCF (280, 300)
3
280 = 2  2  2  5  7 = 2  5  7
300 = 2  2  5  5  3 = 22  52  3

HCF (280, 300) = 22  5 = 20

Number of stacks of science journals
=
280
= 14
20
Number of stacks of maths journals
=
22
22 300
= 15
20

There are 14 stacks of science journals and 15
stacks of maths journals.
Library helps students to develop life-long
learning skills.
Chapter 01: Real Numbers
Memory Map
Real Numbers
a = bq + r; 0  r < b
where, ‘a’ is dividend,
‘b’ is divisor ‘q’ is
quotient and ‘r’ is the
remainder.
The Fundamental
Theorem of Arithmetic
Euclid’s Division
Algorithm
Euclid’s Division
Lemma
HCF of two positive integers ‘a’
and ‘b’ (a > b):
I: By
Euclid’s
division
lemma,
find
whole
numbers ‘q’ and ‘r’ where a
= bq + r; 0  r < b
II: If r = 0, the HCF is b.
If r  0, apply the division
lemma to b and r.
III: Continue the process till
the remainder is zero.
When the remainder is
zero the divisor at that
stage is the required HCF.
Every
composite
number
can
be
expressed (factorised)
as a product of primes
and this factorisation
is unique, apart from
the order in which the
prime factors occur.
Example:
60 = 2  2  3  5
Relation between LCM
and HCF of two numbers
For any two positive
integers a and b
HCF (a, b)  LCM(a, b)
=ab
HCF(a, b) = HCF(b, r)
Rational numbers and their
decimal expansions
If rational number x has terminating
p
decimal expansion, then x = and
q
m n
q=2 5 ,
where p and q are co-prime;
m and n are non-negative integers.
If x =
p
m n
and q = 2 5
q
p
and
q
are
(where,
co-prime; m and n are nonnegative integers), then x has
terminating decimal expansion.
If x =
p
m n
and q  2 5
q
(where, p and q are co-prime; m
and n are non-negative integers),
then x has decimal expansion
non-terminating repeating.
Sample Test Paper
1.
2.
3.
Total Marks: 20
[1]
Find HCF of 24  32 and 2  33
Express 140 as a product of its prime factors.
[1]
State whether the following rational number has a terminating decimal expansion or a nonterminating repeating expansion:
4.
425
x
8.
[2]
Complete the following factor tree and find the numbers x and y.
?
3
5.
6.
7.
3
3125
85
y
5
Find the HCF of the numbers 4052 and 12576 by Euclid’s division algorithm.
Explain why 7  12  11 + 11 and 7  6  4  3 + 6 are composite numbers.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m
or 3m + 1 for some integer m.
Show that 5 is an irrational number.
[2]
[3]
[3]
[4]
[4]
23