Class-10-Mathematics-Chapter-1-CBSE-NCERT.ppsx

Prepared By:
AMIT KUMAR PANDEY
(SOFTCARE SOLUTION)

NUMBERS
Counting numbers
Include zero in natural numbers
WHOLE NUMBERS :
Include negative of natural numbers
INTEGERS :
NATURAL NUMBERS :

NUMBERS
INTEGERS :
Ratio of integers
RATIONAL NUMBERS :

IRRATIONAL NUMBERS
Which type of numbers are they?
These decimal numbers are
non-terminating and non-recurring

Lets find square root of numbers
that are not perfect squares
These decimal numbers are
non-terminating and non-recurring
IRRATIONAL NUMBERS

) (
8 ÷ 2
Example 1 :
8
2 4
8
-
0
) (
15 ÷ 2
Example 2 :
15
2 7
14
-
1
DIVIDEND
DIVISOR
QUOTIENT
REMAINDER
DIVIDEND = DIVISOR × QUOTIENT + REMAINDER
8 = 2 × 4 + 0 15 = 2 × 7 + 1
a a
= =
b b
× ×
q + q +
r r
; 0 < r < b
For two given positive integers a and b
a = bq + r
EUCLID’S DIVISION ALGORITHM
there exist unique integers q and r
satisfying

Sol.
Use Euclid’s division algorithm to find the HCF of :
Q.1 (i) 135 and 225
Since 225 > 135,
Applying Euclid’s Division Algorithm,
we get,
225 = 135 × 1 + 90
Now consider, divisor 90 and dividend 135
applying Euclid’s Division Algorithm,
we get,
135 = 90 × 1 + 45
we get,
90 = 45 × 2 + 0
since remainder = 0
HCF (135, 225) = 45
Dividend =
Divisor × Question + Reminder
) 225
135
135
-
90
1
) (
135 1
90
45
-
) (
90 2
90
-
0
Now, the divisor in this
division is required
HCF of 225 & 135
Exercise 1.1

Sol.
Use Euclid’s division algorithm to find the HCF of :
Q.1 (iii) 867 and 255
Since 867 > 255,
we get,
867 = 255 × 3 + 102
we get,
255 = 102 × 2 + 51
we get,
102 = 51 × 2 + 0
since remainder = 0
HCF (867, 255) = 51
Divide, 867 by 255
Dividend =
) 867
255
765
-
102
3
) (
255 2
204
51
-
) (
102 2
102
-
0
Now, the divisor in this
division is required
HCF of 867 & 255
Exercise 1.1

Use Euclid’s division Algorithm to find the HCF of :
(i) 184, 230, 276
230
184
184
-
46 184
184
-
4
1
0
230 = 184 × 1 + 46
Q.
184 = 46 × 4 + 0
276
46
276
-
0
6
276 = 46 × 6 + 0
HCF (184, 230, 276) = 46

Use Euclid’s division Algorithm to find the HCF of :
(ii) 136, 170, 255
170
136
136
-
34 136
136
-
4
1
0
170 = 136 × 1 + 34
Q.
136 = 34 × 4 + 0
255
34
238
-
17 34
34
-
2
7
0
HCF (136, 176, 255) = 17
255 = 34 × 7 + 17
34 = 17 × 2 + 0

Sol. Let be any positive odd integer
a and b = 6,
we get
a 6q
= + r where 0 < r < 6
The possible remainders are 0, 1, 2, 3, 4, 5
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
since a is odd and q is some integer, we neglect
6q, 6q + 2 and 6q + 4, as they are even
a = 6q + 1 or 6q + 3 or 6q + 5
+ve odd integer
denoted as ‘a’
Here, divisor b
is equal to 6
a = bq + r
0  r b
But, b = 6
Possible values
Of ‘r’ are
0, 1, 2, 3, 4, 5
q is some integer
even
odd
even
odd
even
odd
Any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5
Q.2. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5
where q is some integer.
(+ve odd integers) q + 1
6
q + 3
6
q + 5
6
=
=
=
Exercise 1.1

Sol.
Q.3
32 band
members
616 contingent
members
2 columns
16 16
308 308
4 columns
8 8 8 8
154 154 154 154
That means 2 is a common factor
For 32 & 616
We get the answer by dividing
numbers by 2
That means 4 is a common factor
For 32 & 616
We get the answer by dividing
numbers by 4
If we arrange them
in 2 columns.
Now, for 4 columns
We want to arrange in
maximum column
That means find maximum
i.e highest common factor
for 32 & 616
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?
Exercise 1.1

Sol.
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?
Q.3
Since 616 > 32,
we get,
616 = 32 × 19 + 8
we get,
32 = 8 × 4 + 0
since remainder = 0
HCF (616, 32) = 8
Dividend =
The number of columns must be selected in such a way that it must be
maximum and must divide both the numbers i.e. 616 and 32.
Maximum number of columns in which they can march is 8.
)616
32
32
-
296
288
-
8) (4
32
32
-
0
9
1
Exercise 1.1

Sol.
Show that every positive even integer is of the form 2q, and that every
positive odd integer is of the form 2q + 1, where q is some integer.
Q
Let a be any positive integer
and b = 2.
By Euclid’s algorithm,
Then,
a = 2q + r,
for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2.
So, a = 2q or 2q + 1
If a is of the form 2q, then a is an even integer.
Also, a positive integer can be either even or odd.
Therefore, any positive odd integer is of the form 2q + 1

Q.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.
Sol. Let x be any positive integer and b = 3
Applying Euclid’s Division Algorithm

we get x 3q
= + r where 0 < r < 3
The possible remainders are 0, 1, 2
x 3q
= or 3q + 1 or 3q + 2
i) If x 3q
=
x2
 = (3q)2
= 9q2
= 3 (3q2)
3m
=
for some integer m, where m 3q²
=
ii) If x = 3q + 1
 x2
= (3q + 1)
2
9q2
= 6q
+ 1
+
3
= (3q2
+ 2q) 1
+
3m
= + 1
for some integer m, where m = (3q2 + 2q)
3q + 2
iii) If x =
 x2
= (3q + 2)2
9q2
= 12q
+ 4
+
9q2
= 12q
+ 3
+ 1
+
= 3 (3q2 4q
+ 1)
+ 1
+
3m
= + 1
for some integer m, where m = 3q2 4q
+ 1
+
+ve integer be
denoted as ‘x’
Here, divisor b
is equal to 3
a = bq + r
0  r b
But, b = 3
Possible values
Of ‘r’ are
0, 1, 2
We want square
of ‘+ve’ integer
Replace 3q2 by m
Now, for x = 3q + 1
Apply,
(a + b)2 = a2 + 2ab + b2
Replace (3q2 + 2q) by m
Now, for x = 3q + 2
Apply,
(a + b)2 = a2 + 2ab + b2
Replace
(3q2 + 4q + 1) by m
(+ve integer)2
= m
= m + 1
3
3
Square of any positive odd integer is
either of the form 3m or 3m+ 1
for some integer m.
Exercise 1.1

Soln.
Q.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 for some integer m.
Let x be any positive integer and b = 3
Here, divisor is 9
(x)3
= m
= m + 1
= m + 8
9
9
9
Possible remainders are
0, 1, 2, 3, 4, 5, 6, 7, 8
i.e. 9q
9q + 1
9q + 2
9q + 3
9q + 4
9q + 5
9q + 6
9q + 7
9q + 8
Lets consider divisor as 3
Possible remainders are
0, 1, 2
i.e. 3q
3q + 1
3q + 2
Applying Euclid’s Division Algorithm, we get

x = 3q, x = 3q + 1 or x = 3q + 2
i) If x 3q
=
x3
 = (3q)3
= 27q3
= 9
9m
=
for some integer m, where m 3q3
=
(3q3)
ii) If x = 3q + 1
 x3
= (3q + 1)3
(3q)3
= + 3(3q)2 (1) + 3(3q)(1)2 + (1)3
9
= (3q3 + 3q2 + q) + 1
Apply,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
27q3
= + 27q2 + 9q + 1
9m
= + 1
for some integer m, where m 3q3 + 3q2 + q
=
iii) If x = 3q + 2
 x3
= (3q + 2)3
(3q)3
= + 3(3q)2 (2) + 3(3q)(2)2 + (2)3
9
= (3q3 + 6q2 + 4q) + 8
Apply,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
27q3
= + 54q2 + 36q + 8
9m
= + 8
for some integer m, where m 3q3 + 6q2 + 4q
=
Cube of any positive integer is of the form
9m , 9m + 1 or 9m + 8 for some integer m.
Exercise 1.1

Let three consecutive positive integer’s be
n – 1, n and n + 1.
By Euclid’s Division Algorithm,
Prove that product of three consecutive positive integers is divisible by 6.
a 6q
= + r where 0 < r < 6
The possible remainders are 0, 1, 2, 3, 4, 5
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
6q + 4 or 6q + 5 , where q is some integer
Case 1 : If, n = 6q
n – 1 = 6q – 1
n + 1 = 6q + 1
So, (n – 1) (n + 1)
(n) = (6q – 1) (6q)(6q + 1)
= 6 [(6q – 1)(q)(6q + 1)]
= 6 m
[ where = (6q – 1) (q) (6q + 1) ]
Here, the above result is multiple of 6.
Hence, it is divisible by 6.
m
Q.

Case 2 : If, n = 6q + 1
n – 1 = 6q + 1 – 1
n + 1 = 6q + 1 + 1
=
=
6q
6q + 2
So, (n – 1) (n + 1)
(n) = (6q + 1)
= 6[q(6q + 1)(6q + 2)]
= 6
m
where =
(6q) (6q + 2)
m
[q(6q + 1)(6q + 2)]
n – 1, n and n + 1.
a 6q
= + r where 0 < r < 6
The possible remainders are
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
Q.
0, 1, 2, 3, 4, 5

Case 3 : If, n = 6q + 2
n – 1 = 6q + 2 – 1
n + 1 = 6q + 2 + 1
=
=
6q + 1
6q + 3
So, (n – 1) (n + 1)
(n) = (6q + 2)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6
m
where =
Here, the above result is multiple of 6. Hence, it is divisible by 6.
(6q + 1) (6q + 3)
m
[(6q + 1)(3q + 1)(2q + 1)]
= (3q + 1)
(6q + 1) (2q + 1)
(2) (3)
n – 1, n and n + 1.
a 6q
= + r where 0 < r < 6
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
6q + 4 or , where q is some integer
Q.
6q + 5
0, 1, 2, 3, 4, 5

Case 4 : If, n = 6q + 3
n – 1 = 6q + 3 – 1
n + 1 = 6q + 3 + 1
=
=
6q + 2
6q + 4
So, (n – 1) (n + 1)
(n) = (6q + 3)
= 6[(3q + 1)(2q + 1)(6q + 4)]
= 6
m
where =
(6q + 2) (6q + 4)
m
[(3q + 1)(2q + 1)(6q + 4)]
= (3q + 1) (6q + 4)
(2q + 1)
(2) (3)
Q.
n – 1, n and n + 1.
a 6q
= + r where 0 < r < 6
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
0, 1, 2, 3, 4, 5

Case 5 : If, n = 6q + 4
n – 1 = 6q + 4 – 1
n + 1 = 6q + 4 + 1
=
=
6q + 3
6q + 5
So, (n – 1) (n + 1)
(n) = (6q + 4)
= 6[(2q + 1)(3q + 2)(6q + 5)]
= 6
m
where =
(6q + 3) (6q + 5)
m
[(2q + 1)(3q + 2)(6q + 5)]
= (2q + 1) (6q + 5)
(3q + 2)
(3) (2)
Q.
n – 1, n and n + 1.
a 6q
= + r where 0 < r < 6
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
0, 1, 2, 3, 4, 5

Case 6 : If, n = 6q + 5
n – 1 = 6q + 5 – 1
n + 1 = 6q + 5 + 1
=
=
6q + 4
6q + 6
So, (n – 1) (n + 1)
(n) = (6q + 5)
= 6[(6q + 4)(6q + 5)(q + 1)]
= 6
m
where =
(6q + 4) (6q + 6)
m
[(6q + 4)(6q + 5)(q + 1)]
= (6q + 4) (q + 1)
(6q + 5) (6)
Q.
n – 1, n and n + 1.
a 6q
= + r where 0 < r < 6
a 6q
= or 6q + 1 or 6q + 2 or 6q + 3 or
0, 1, 2, 3, 4, 5

The above Question can also be solved in the
following way
For any positive integer n,
prove that n3 – n is divisible by 6.
Hint : n3 – n = n (n2 – 1)
(n – 1)
= n
= (n + 1)
(n)
……..Since, (a2 – b2) = (a – b)(a + b)
•
Q.
(n + 1)
(n – 1)
Q.

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is
some integer.
Q.
Let us start with taking a, where a is a
positive odd integer.
We apply the division algorithm with a
and b = 4.
Since 0 ≤ r < 4,
That is, a can be or 4q + 1, or 4q + 2, or 4q + 3,
where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.
…….(since they are both divisible by 2).
a 4q
= + r where 0 < r < 4
the possible remainders are 0, 1, 2, 3,
4q,

and b = 4
Show that n2 – 1 is divisible by 8, if n is an odd positive integer.
Q.
Any odd positive integer n can be written in form
of 4q + 1 or 4q + 3.
a 4q
= + r where 0 < r < 4

The possible remainders are 0, 1, 2, 3

a 4q
= or 4q + 1 or 4q + 2 or 4q + 3,
where q is some integer

Case 1 : If, n = 4q + 1
n2
– 1 = (4q + 1)2 – 1
= 16q2 + 8q + 1 – 1
= 16q2 + 8q
= (8q) (2q + 1)
We know, (a + b)2 = (a2 + 2ab + b2)

Show that n2 – 1 is divisible by 8, if n is an odd positive integer.
Q.
Case 2 : If, n = 4q + 3
n2 – 1 = (4q + 3)2 – 1
= 16q2 + 24q + 9 – 1
= 16q2 + 24q
= (8)(2q2 + 3q + 1)
+ 8
and b = 4
Any odd positive integer n can be written in form
of 4q + 1 or 4q + 3.
a 4q
= + r where 0 < r < 4

The possible remainders are 0, 1, 2, 3

a 4q
= or 4q + 1 or 4q + 2 or 4q + 3,
where q is some integer

We know, (a + b)2 = (a2 + 2ab + b2)

Let x = 2m + 1
y 2n + 1
=
x2 + y2 = (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 + 4n + 1
= 4m2 + 4n2 + 4m + 4n + 2
= (m2 + n2 + m + n)
= 4 + 2 q
[ where = m2 + n2 + m + n ]
Here, in the above result 4q is even,
therefore 4q + 2 is also even .
And comparing with a = bq + r, here when
divisor is 4, remainder is 2.
Hence it is not divisible by 4.
Prove that if x and y are odd positive integers, then x2 + y2 is even but not divisible
by 4.
Q.
4 + 2
Therefore, x2 + y2 is even but not divisible by 4.
[ where m and n are positive integers ]
q
We know, (a + b)2 = (a2 + 2ab + b2)

By Euclid’s Division Lemma,
a = 3q + r ; where =
r 0, 1, 2 ;
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where
n is any positive integer.
Q.
Let a be any positive integer for divisor 3
a 3q
= or 3q + 1 or 3q + 2 ;
Case 1 : When =
n 3q,
=
n + 2 3q + 2
=
and n + 4 3q + 4
= 3q + 3 + 1
= 3
Comparing with 3q + r, here only n is divisible by 3.
+ 1
+ 1)
(q

r 0, 1, 2 ;
Q.
a 3q
= or 3q + 1 or 3q + 2 ;
Case 2 : When =
n 3q + 1,
=
n + 2 3q + 1
Comparing with 3q + r, here only n + 2 is divisible by 3.
= 3q + 3
= 3 + 1)
= 3m …… where m
=
and n + 4 3q + 1
= 3q + 5
= 3 + 1)
= 3
+ 4
= 3q + 3 + 2
= (q + 1)
…… where m = (q + 1)
+ 2
(q
(q + 2
+ 2
m

r 0, 1, 2 ;
Q.
a 3q
= or 3q + 1 or 3q + 2 ;
Case 3 : When =
n 3q + 2,
=
n + 2 3q + 2
Comparing with 3q + r, here only n + 2 is divisible by 3.
= 3q + 4
= 3 + 1)
= 3m ……where m
=
and n + 4 3q + 2
= 3q + 6
= 3 + 2)
= 3
+ 4
= (q + 1)
= 3q + 3 + 1
+ 1
+ 1
……where m = (q + 2)
+ 2
(q
(q
m

If a and b are two odd positive integers such that a > b, then prove
+
a b
2
–
a b
2
that one of the two numbers and is odd and the other is even.

+
a b
2
=
+
(2m + 3) (2m + 1)
2
=
+
2m 2m
2
+ 4
Q.
[ where m and n are + ve integers ]
=
4m
2
+ 4
=
2
2
+ 2)
= 2m + 2 [ which is even ]
(2m
Let a = 2m + 3 and = 2m + 1

+
a b
2

If a and b are two odd positive integers such that a > b, then prove
+
a b
2
–
a b
2
that one of the two numbers and is odd and the other is even.
Now, –
a b
2
=
–
(2m + 3) (2m + 1)
2
=
+
2m 3
2
– 2m – 1
Q.
=
3
2
– 1
=
2
2
= 1 [ which is odd ]

+
a b
2
= +
2m 2 and 1
–
a b
2
=
One of them is odd and the other is even
Hence Proved
1
[ where m and n are + ve integers ]
Let a = 2m + 3 and = 2m + 1
= 2m + 2 [ which is even ]

+
a b
2

468
222
444
-
24 222
216
-
9
2
6 24
24
-
4
0
HCF (468, 222) = 6
Q. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two
different ways.
468 = 222 × 2 + 24
222 = 24 × 9 + 6
24 = 6 × 4 + 0
….(i)
….(ii)
….(iii)

222 = (24 × 9) + 6
9) =
222 – (24 × 6
= 222 – 9 × 24

= 222 – 9 × [468 –
= 222 – 9 × 468 + (9 × 2 × 222)
= (–9)468
= 468 + 222 (y)
where x = –9 and y = 19.
different ways.
To show :- 6 = 468x + 222y
HCF (468, 222) = 6
468 = 222 × 2 + 24
222 = 24 × 9 + 6
24 = 6 × 4 + 0
….(i)
….(ii)
….(iii)
= 222 – 9 × 468 + 18 × 222
= 222 – 9 × 468
+ 18 × 222
= 222 + 468
(1 + 18) (–9)
222 (19)
+
468(–9)
6

6

6

6

6

6

6
222(19) +
.…from (ii)
(x)
9)
= 222 – (24 ×
6
9
24 ×
.…from (i)
(222 × 2) ]
(2 × 222)
468(–9)
=
6
222(19) + 468(–9)

different ways.
6 = 222 × 19 + 468 × (–9)
= 222 × 19 + 468 × (–9) + [(468 × 222) – (468 × 222)]
= 222 × 19 + 468 × (–9) + (468 × 222) – (468 × 222)
= 222 [19 + 468 [(–9)
= 222(–449) + 468(213)
=
468(213) + 222(–449)
468 x + 222 y where x = 213 and y = 449.
To show :- 6 = 468x + 222y
HCF (468, 222) = 6
468 = 222 × 2 + 24
222 = 24 × 9 + 6
24 = 6 × 4 + 0
….(i)
….(ii)
….(iii)
where x = –9 and y = 19.
222 (19)
+
468(–9)
=
6
….(iv)
….from (iv)
0
= (222 × 19) – (468 × 222) + [468 × (–9)] + (468 × 222)
– 468] 222]
+

Sol.
If d is the HCF of 56 & 72, find x, y satisfying d = 56x + 72y. Also, show that x and y
are not unique.
Q.
Applying Euclid’s division lemma to 56 to 72,
72 = 56  …… (i)
1 + 16
Remainder is 16
So, we consider the divisor 56 and remainder 16
56 = 16  …… (ii)
3 + 8
16 = 8  2 + 0
The remainder at this stage is 0.
Last divisor 8 is HCF of 56 and 72
From (ii), we get
Q

Remainder is 8
Q
8 = 56 –
… from (i)
16  3
8 = 56 – (72 – 56  1)  3
8 = 56 – 72  3 + 56  3
and apply division lemma.
72
56 1
56
–
16
56
16 3
48
–
8
16
8 2
16
–
0
From (i) we get
16 = 72 – 56 × 1

Sol.
If d is the HCF of 56 & 72, find x, y satisfying d = 56x + 72y. Also, show that x and y
are not unique.
Q.
Applying Euclid’s division lemma to 56 to 72,
72 = 56  …… (i)
1 + 16
Remainder is 16
56 = 16  …… (ii)
3 + 8
16 = 8  2 + 0
The remainder at this stage is 0.
Last divisor 8 is HCF of 56 and 72
From (ii), we get
Q

Remainder is 8
Q
8 = 56 –
… from (i)
16  3
8 = 56 – (72 – 56  1)  3
8 = 56 – 72  3 + 56  3
8 = 56 4 72  3
 –
8 = 56 4 72 (– 3)
 + 

8 = 56 4
 + 72 (– 3)
 + 56 72 56 72
–
Now,
8 = 56 4
 + 56 72+ 72 (– 3)
 56 72
–
8 = 56(4 + 72) + 72(– 3 – 56)
8 = 56 76
 + 72 (– 59)

Hence, x and y are not unique
Comparing with
56x + 72y
x = 4 & y = – 3
Comparing with
56x + 72y
x = 76 & y = – 59

Fundamental Theorem Of Arithmetic (Prime Factorisation)
HCF
Product of the least
powers of common prime
factors of the numbers
LCM
Product of the highest
powers of all the prime
factors of the numbers
Every composite number can be expressed
as a product of primes which is unique
Find HCF and LCM of 180 and 54
180 = 2 × 2 × 3 × 3
= 22
× 32
54 = 2 × 3 × 3 × 3
= 2 × 33
HCF
× 5
× 5
= 2 × 32
= 2 × 9
= 18
LCM = 22
× 33
= 4 × 27 × 5
= 540
× 5
HCF × LCM = 18 × 540 = 9720
PRODUCT OF NUMBERS = 54 × 180 = 9720
HCF(a, b) × LCM (a, b) = a × b

Sol.
Express each number as a product of its prime factors:
Q.1
140
(i) 140
140
70
35
2
2
5
7
7
1
140 = 22 × 5 × 7
Sol.
156
(ii) 156
156
78
39
2
2
3
13
13
1
156 = 22 × 3 × 13
Sol.
3825
(iii) 3825
3825
1275
425
3
3
5
85
5
17
3825 = 32 × 52 × 17
17
1
Exercise 1.2

Sol.
5005
(iv) 5005
5005
1001
143
5
7
11
13
13
1
5005 = 5 × 7 × 11 × 13
Sol.
7429
(v) 7429
7429
437
23
17
19
23
1
7429 = 17 × 19 × 23
Express each number as a product of its prime factors:
Q.1
Exercise 1.2

(i) 1000
1000
500
250
2
2
2
125
5
25
Q. Express each number as a product of its prime factors:
5
1
5
5
1000 = 2 × 2 × 2 × 5 × 5 × 5
= 23 × 53

(ii) 72
Q. Express each number as a product of its prime factors:
72
36
18
2
2
2
9
3
3
1
3
72 = 2 × 2 × 2 × 3 × 3
= 23 × 32

Sol.
Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = Product of the two numbers.
Q.2
(ii) 510 and 12
510
510
255
85
2
3
5
17
17
1
= × 3 × 5 × 17
12
12
6
3
2
2
3
1
= 22 × 3
HCF (510, 12) = 2 × 3 = 6 (Product of common factors raised to least power)
LCM (510, 12) = 22 × 3 × 5 × 17 1020
= …(Product of all the prime factors
raised to highest power)
Verification :
1020
= × 6 = 6120
Product of the two numbers 510
= × 12 = 6120
HCF
×
LCM = Product of the two numbers
HCF
×
LCM
2
Exercise 1.2

Sol.
(iii) 336 and 54
336
336
168
84
2
2
2
42
2
21
= × × 7
54
54
27
9
2
3
3
3
= × 33
HCF (336, 54) = 2 × 3 = 6 (Product of common factors raised to least power)
LCM (336, 54) = 24 × 33 × 7 3024
= …(Product of all the prime factors
raised to highest power)
Verification :
6
= × 3024 = 18144
Product of the two numbers 336
= × 54 = 18144
HCF
×
LCM = Product of the two numbers
HCF
×
LCM
24 3
7
3
7
1
3
1
2
Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = Product of the two numbers.
Q.2
Exercise 1.2

Sol.
Find the LCM and HCF of the following integers by applying the
prime factorisation method.
Q.3
(ii) 17, 23 and 29
17 = ×
23 = ×
29 = ×
HCF (17, 23, 29) = 1
(Product of common factors raised to least powers)
LCM (17, 23, 29) =17 × 23 × 29 = 11339
(Product of all the prime factors raised to highest powers)
17
23
29
1
1
1
Exercise 1.2

Sol.
(iii) 8, 9 and 25
8 = × 1
9 = × 1
25 = × 1
HCF (8, 9, 25) = 1
(Product of common factors raised to least powers)
LCM (8, 9, 25) = 23 × 32 × 52 = 1800
(Product of all the prime factors raised to highest powers)
23
32
52
Find the LCM and HCF of the following integers by applying the
prime factorisation method.
Q.3
Exercise 1.2

Sol.
Find the LCM and HCF of the following integers by applying the prime
factorisation method.
Q.3
(i) 12, 15 and 21
12
12
6
3
2
2
3
1
= ×
15
15
5
1
3
5
= × 5
22
21
21
7
1
3
7
= × 7
HCF (12, 15,21) = 3 (Product of common factors raised to least power)
LCM (12, 15,21) = 3 × 22 × 5 × 7 420
= (Product of all the prime factors
raised to highest powers)
3
3
3
Exercise 1.2

Sol.
Given that HCF (306, 657) = 9,
Q.4 find LCM (306, 657)
HCF (306, 657) = 9
Now, HCF (306, 657) × LCM (306, 657) 306
= × 657
LCM (306, 657) =
306 × 657
9
=
= 22338
We know the relation
HCF(a,b) × LCM(a,b) = a × b
34 × 657
Exercise 1.2
3
102
34

Sol.
Q.7
L.C.M of 18 and 12
18 = 2 × 32
12 = 22 × 3
L.C.M = 22 × 32 (Product of all the prime factors raised to
highest powers)
= 36
In 36 minutes Sonia arrived at the starting point after making 2 rounds,
while in 36 minutes, Ravi arrived at the same starting point after making 3 rounds.
Starting point
Time taken to drive 1 round
Sonia Ravi
18 min
36 min
54 min
12 min
24 min
36 min
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.
That means she will come
to starting point
after every 18 min
Therefore LCM of 18 & 12
will give us the time they
meet again at starting point
That means he will come
to starting point
after every 12 min
Sonia takes 18 min
to drive 1 round
Ravi takes 12 min
to drive 1 round
Hence they meet each other at the starting point after 36 minutes.
Exercise 1.2

Sol.
Check whether 6n can end with the digit 0 for any natural number n.
Q.5
For e.g. 10, 20, 30,…
These nos. are
divisible by 5
If the number 6n for any n  N ends with the digit ‘0’,
then it is divisible by 5.
That means the prime factorisation of 6n must contain the
prime number 5.
But,
6n = 6  6  6  ….
That can also be written as,
6n = 2  3  2  3  2  3  ….
It is not possible to
get prime number 5
But this is not possible, because the primes in the
prime factorisation of 6n are 2 and 3.
By Fundamental Theorem of Arithmetic
there are no other prime numbers except 2 and 3 in the
factorisation of 6n.
So, there is no natural number for which 6n ends with the digit 0.
Exercise 1.2

Sol.
Q.6
7 × 11 × +
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 ×
= 13 × 13 × 2 × 3
Also, 7 × 6 × × 4 × 3 × 2 × 1 +
= 5 (7 × 6 × 4 × 3 × 2 + 1)
= 5 (1008 + 1)
= 5 × 1009
Product of primes
Product of primes
Composite numbers are those
numbers which can be
expressed as product of primes
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
13 13
78
5 5
7 × 11 × 13 + 13 is a composite number
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
Exercise 1.2

Sol.
Q.
There are 156, 208 and 260 student in Groups A , B and C respectively. Buses
are to be hired to take them for a field trip. Find minimum number of buses
to be hired if the same number of students be accommodated in each bus.
The number of Buses will be minimum if each bus accommodated
maximum number of students.
The no. of students in each bus must be HCF of 156, 208 and 260 students.
The prime factorisation of 156 , 208 and 260 are :
156
156
78
39
2
2
3
13
13
1
= × 3 × 13
22
208
104
52
2
2
2
26
2
13
13
1
208 = × 13
24
260
130
65
2
2
5
13
13
1
260 = × 5 × 13
22
HCF of 156, 208 and 260 = × 13
22 = 52
In each bus 52 students can be accommodated
=
Minimum no. of buses required Total no. of students
52
=
156 + 208 + 260
52
= 624
52
= 18 buses.
18 no. of buses will be hired if the same number of students be accommodated in each bus.

96
Q.1 A mason has to fit a bathroom with square marble tiles of largest possible size. The size of the
bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut
and how many such tiles are required.
Length of
bathroom = 10 ft
= 10  12
= 120 inches
Breadth of
bathroom
= 8 ft
= 8  12
= 96 inches
[Q 1 ft = 12 inche]
[Q 1 ft = 12 inches]
Area of bathroom = 120  96

120
96
-
24 96
96
-
4
0
HCF (120, 96) = 24
120 = 96 × 1 + 24
96 = 24 × 4 + 0
….(i)
….(ii)
1
120
96

Q.1 A mason has to fit a bathroom with square marble tiles of largest possible size. The size of the
bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut
and how many such tiles are required.
The side of the tile

Number of tiles =
Total Area of bathroom
Area of Each Tile
=
120  96
24  24

HCF (120, 96) = 24
Area of bathroom = 120  96
= 24
Area of a tile
= (Side)2
= (24)2
24  24
=
Area of a tile

5 4
= 20
No. of tiles required is 20.

Sol.
Q. Find the largest number that will divide 445 , 572 and 699 leaving remainders
4, 5 and 6 respectively.
The required number when divides 445 , 572 and 699 leaves
remainder 4 , 5 and 6.
The numbers completely divisible by required number are
445 – 4 = 441,
441
441
147
49
3
3
7
7
7
1
= × 72
32
567
189
63
3
3
3
21
3
7
7
1
567 = × 7
34
693
231
77
3
3
7
11
11
1
693 = × 7 × 11
32
HCF of 441, 567 and 693 = × 7
32 = 63
The largest number that divides 445 , 572 and 699 leaves remainder 4, 5 and 6 is 63
572 – 5 = 567 and
699 – 6 = 693
The required number will be HCF of 441, 567 and 693

Find the smallest number which when increased by 17 is exactly
divisible by both 520 and 468.
Q.
Therefore, 4663 is the required answer.
The least number divisible by both 520 and 468 is the
Least Common Multiple of 520 and 468(L.C.M)
520 5
= 2  2  2  13

468 3
= 2  2  3  13

Clearly, =
L.C.M. 13
23  32
 5  = 4680
So, least number divisible by both 520 and 468 is 4680.
Let the number which when increased by 17 gives the least number
divisible by 520 and 468 (i.e. LCM of the two i.e. 4680) be x
So, x + 17 = 4680
x = 4663
x – 17
= 4680
520
2
2
2
13
2
6
5
1
13
1
0
3
60
468
2
2
2
11
3
3
3
1
13
1
7
3
34
= 23 × 5 × 13
= 22 × 32 × 13
9
5

16
36
=
4
9
Co-prime numbers two numbers having no common factor
other than 1 are co-prime numbers.
If a and b are co-prime numbers then they have no common factor other than 1
Example:
12 & 17, 21 & 22, 33 & 40,
Let p be a prime number,
If p divides a2 , then p divides a
Example:
If 2 divides (8)2 then 2 divides 8
If 7 divides (35)2 then 7 divides 35

Prove that 5 is irrational.
Q.1
Let us assume that is a rational number.
There exist co-prime integers a and b, (b ≠ 0) such that,

=
a
b Rational number =
a
b
, (b 0)
& a, b are co-prime integers
We will prove it by
Contradiction method
5
5
= a
squaring both sides,
5b2 = a2
5 divides a2
 5 divides a
... (1)
Let a = 5c where c is some integer
... (2)
substituting this value of a in (1)
(5c)2
=
5b2 = 25c2
b2 = 5c2

5b2
5 divides b2  5 divides b ... (4)
From (3) and (5), we get,
a and b both have common factor 5.
Our assumption that is a rational number is wrong.

This contradicts the fact that a and b are co-prime.
If 5 divides 15
That means 5 × integer = 15
Dividing both side by 5
5 divides a & b both
b
5
5
is an irrational number.
5
Exercise 1.3
5 is a factor of a… 3
After equation 2
5 is a factor of b….5
After equation 4

Prove that 3 + 2 5 is irrational.
Q.2
Let us assume that 3 2
+ is a rational number.
There exist co-prime integers a and b ( b ≠ 0) such that,

3 2
+ =
a
b
2
=
a
b
3
–
2
=
a – 3b
b
=
a – 3b
2b
Since a and b are integers,
a – 3b
2b
is rational  is also rational
This contradicts the fact that is irrational.
Our assumption that
3 2
+ is a rational number is wrong.

Rational number =
a
b
, (b 0)
& a, b are co-prime integer
We will prove it by
This also implies
That is rational
5
5
5
5
5
5
5
5
5
3 2
+ is irrational.
5
Exercise 1.3

(i)
1
2
1
2
= × =
Let us assume that is rational.
There exist co-prime integers
a and b (b  0) such that,
=
a
b

= 2a
b

2a
b
is rational  is also rational,
but this contradicts the fact that is irrational.
Our assumption that
i.e. is rational is wrong.
2
We will first rationalise
the denominator
Multiply both numerator
& denominator by 2
Now we will prove by
2a is integer
b is integer & b  0
Arrange this equation in
such a way that we get
only in L.H.S
2
This also implies
That is rational
2
1
2 2
2 2
2
2
2
2
2
2
2
2
2 1
2
is an irrational number
1
2
Prove that the following are irrationals.
Q.3
Exercise 1.3

Q.3
(ii) 7 5
Let us assume that 7 5 is rational.
There exist co-prime integers a and b (b  0) such that,
=
a
b
5 =
7
a

7b
a
is rational it implies that 5 is also rational,
But this contradicts the fact that 5 is irrational.
Our assumption that 7 5 is rational is wrong.
Lets prove this by
contradiction
It should be in
form of a/b
it implies that 7b is also integer
Integer
Integer
7 5
7 5 is irrational.
b
Exercise 1.3

(iii) 6 + 2
Let us assume that 6 + 2 is rational.
There exist co-prime integers a and b (b  0) such that,
=
a
b
2 =
b
a – 6b

b
a – 6b
is rational which implies that 2 is also rational,
But this contradicts the fact that 2 is irrational.
Our assumption that 6 + 2 is rational is wrong.
Lets prove this by
contradiction
It should be in
form of a/b
it implies that a – 6b is also integer
Integer
Integer
6 + 2
6 + 2 is irrational.
2 =
a
b
– 6
Q.3
Exercise 1.3

2 10
2 10
Let us assume that 2 + 5 is a rational number.
a
b
=
2 + 5 (where b  0)
Squaring both sides, we get
[
a
b
= [ ]
2
2 + 5 + 2 10 =
a2
b2
7 + =
a2
b2
=
a2
b2
– 7
2 10 =
a2
b2
– 7b2
10 =
a2
2
– 7b2
So, there exist co-prime integers a and b such that

Prove that 2 + 5 is an irrational number.
Q.
2 + 5]2
We know, (a + b)2 = (a2 + b2 + 2ab)
a
b
2
1
b2

Here,
a2
2b2
– 7b2
is an rational number
This implies, 10 is also rational number.
But, we know that 10 is an irrational number
2 + 5 is an irrational number.
Therefore, there is a contradiction and our
assumption is wrong
2 + 5 is not a rational number.
𝐇𝐞𝐧𝐜𝐞 𝐏𝐫𝐨𝐯𝐞𝐝.
Prove that 2 + 5 is an irrational number.
Q.
Let us assume that 2 + 5 is a rational number.
a
b
=
2 + 5 (where b  0)
So, there exist co-prime integers a and b such that
10 =
a2
2
– 7b2

b2

Let us assume
𝟑
5 is a rational number
So, there exist co-prime integers a and b where b  0
𝟑
5
a
b
=
Cubing on both sides, we get
5 =
a3
b3  5 =
a3
b3
5 divides a3  5 divides a
5 is a factor of a
…from (1)
Let a = 5c
b3 =
5c
5
b3 =
5
5
 5  5c3
b3
5
= 5c3
5 divides b3  5 divides 5.
5 is a factor of b … (3)
From (2) and (3)
5 is a common factor of a and b
a and b are not co-prime
our assumption is wrong
𝟑
5 is an irrational number.
Q. Prove that
𝟑
5 is an irrational number.
such that
[
a
b
= [ ]
3
𝟑
5 ]3
a
b
3
b3
5
( )3
… (1)
… (2)

Decimal expansion of rational numbers is terminating
or non – terminating but repeating.
Example:
3.5, 7.272727…, 3.142857142857…, 25,
3.5 =
35
10
=
35
2 × 5
47.1245 =
471245
10000
=
471245
24 × 54
0.222 =
222
1000
=
222
23 × 53
A decimal expansion that terminates when expressed in form
of
𝒑
𝒒
, then prime factorisation of q is of the form 2n5m
3.5 can be expressed
as i.e
P
Q
35
10
Terminating decimal expansion

Q.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)
13
3125
=
13
55
=
20 × 55
13
(ii) 17
8
=
17
23
=
23 × 50
17
has terminating decimal expansion.
13
3125

17
8

13
3125
17
8
Since denominator is in the form of 2n5m

(iii) 15
1600
=
3
320
=
26 × 51
3
(iv) 77
210
=
11
30
=
2 × 3
11
15
1600

has non - terminating decimal expansion.
77
210

15
1600
77
210
Since denominator is not in the form of 2n5m
× 5
3
320
11
30

(v) 64
455
=
5 × 7
64
(iv) 23
2352
has non-terminating repeating decimal
expansion.
64
455

23
2352

64
455
23
2352
Since denominator is not in the form of 2n5m
× 13
455
5
7
1
13
1
3
91

(i) 13
3125
13
3125
=
13
=
×
13 × 25
=
416
10
= 0.00416
55 55 25
(ii) 17
8
17
8
=
17
=
×
17 × 53
=
2125
( )3
= 2.125
23 23 53
Q.2 Write down the decimal expansions of the rational numbers which have
terminating decimal expansions.
× 25
× 25
13
3125
= 0.00416

× 53
× 53
17
8
= 2.125

10
2 5
3 3
5 2
5 5
( )5

(iii) 15
1600
15
1600
=
3
320
=
3
26 × 51
=
3 × 55
26 × 51 × 55
=
3 × 3125
26 × 56
=
9375
10
= 0.009375
To express it as 10m we
need to multiply it by 55
23
23 × 52
=
23 × 51
23 × 52 × 51
=
23 × 5
23 × 53
=
115
10
= 0.115
(iv)
23
23 × 52
3
320
15
1600
= 0.009375

23
23 × 52
2 5
6 6 6
= 0.115
3
2 5
3 3

(v) 6
15
6
15
=
2
5
=
2 × 2
5 × 2
=
4
10
= 0.4
(vi) 35
50
35
50
=
7
10
= 0.7
6
15
= 0.4

35
50
= 0.7

Q.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 ,
what can you say about the prime factors of q ?
p
q
(i) 43.123456789
43.123456789

Decimal number is
terminating
Hence prime factors of q
i.e. denominator will be of
the form 2n5m
43.123456789 is a rational number
and q will be in the form of 2n5m
(ii) 0.120120012000120000...
0.120120012000120000...
is a non-terminating and non-repeating decimal
0.120120012000120000... is not a rational number.

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