# NCERT solutions for class 10 Mathematics-Chapter-1

## Class 10 NCERT solutions chapter – 1

Hello students , In this page we provide you NCERT solutions for class 10 Science . In below  you will see class 10 science , it is available on written images content and later it will be appears on typing text content. You can scroll it and find your solution which do you want to get solutions of any questions in your chapter. Also we provide you some extra questions for your practice for regarding to your CBSE Board exam. You can share our link with your friends or classmates and to help them to understand better about their subjects and clear their doubts.

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

#### Some other extra questions :

Que1. Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1 for some integer q.
Ans. Let a be a given positive integer.
On dividing a by 2, let q be the quotient and r be the remainder.
Then by Euclid’s algorithm, we have
a = 2q + r , where 0 ≤ r < 2
⇒ a= 2q+ r , where r = 0 or r = 1.
⇒ a = 2q or a = 2q + 1.
When a = 2q for some integer q, then clearly a is even.
Also, an integer can be either even or odd.
Hence, an odd integer is of the form 2q + 1 for some integer q.

Que2. Show that every positive odd integer is of the form (6q+1) or (6q+3) or (6q+5) for some integer q.
Ans. Let a be a given positive integer.
On dividing a by 6, let q be the quotient and r be the remainder.
Then, by Euclid’s algorithm , we have
a = 6q+r , where 0 ≤ r < 6
⇒ a= 6q+r, where r = 0,1,2,3,4,5
⇒ a=6q or a= 6q + 1 or a= 6q+2 or a= 6q+3
or a= 6q + 4 or a= 6q + 5.
But, a = 6q , a = 6q + 2 , a = 6q + 4 give even values of a.
Thus, when a is odd , it is of the form
6q + 1 or 6q + 3 or 6q + 5 for some integer q.

Que3. Show that 2√3 is irrational.
Ans. If possible, let 2 √3be rational.
Let its simplest form be 2√3=, where a and b are positive integers having no common factor other than 1. Then,
2√3= a/b⇒ √3= a/
Since a and 2b are non-zero integers , so a/2b is rational.
Thus, from (i), it follows that √3is rational.
This contradicts the fact that √3is rational.
The contradiction arises by assuming that 2√3 is rational.
Hence, 2√3is irrational.

Que4. Show that (4 – √3) is irrational.
Ans. If possible, let (4 – √3) be rational. Then,
4 is rational, (4 – √3) is rational
⇒ {4 – ( 4 – √3)} is rational.
⇒ √3is rational.
This contradicts the fact that √3is irrational.
This contradiction arises by assuming that ( 4 – √3) is rational.
Hence, ( 4 – √3 )is irrational.

Que5. Prove that √11 is irrational.
Ans. If possible, let √11be rational and let its simplest form be a/b
Then, a and b are integers having no common factor other than 1, and b ≠ 0.
Now, √11 = a/b ⇒ 11 = a²/b²
⇒ 11b² = a²
⇒ 11 divides a²
⇒ 11 divides a
Let a = 11c for some positive integer c.
Putting a = 11c in (i), we get
11b² = 121c² ⇒ b² = 11c²
⇒ 11 divides b²
⇒ 11 divides b
Thus, 11 is a common factor of a and b.
But, this contradicts the fact that a and b have no common factor other than 1.
The contradiction arises by assuming that √11 is rational.
Hence, √11 is irrational.