Class 10 Maths Chapter 1 Exercise 1.1 Solutions

 Class 10 Maths Chapter 1 Exercise 1.1 Solutions

Here you will find Class 10 Maths Chapter 1 Exercise 1.1 Solutions that are prepared by our experienced teachers. In the previous classes, you must have heard about numbers. 

In this class 10th maths ch 1 ex 1.1, you can read all questions with a complete solution. This exercise 1.1 class 10 maths is important for CBSE, RBSE, and other board students.

So, you are advised to solve all these questions of class 10 maths chapter 1 real numbers exercise 1.1. you will be provided a free PDF of class 10 maths NCERT chapter 1 exercise 1.1.

Exercise 1.1
Question 1
Question 2
Question 3
Question 4
Question 5

Maths Class 10 Chapter 1 Exercise 1.1 Question 1

Q1. Use Euclid’s division algorithm to find the HCF of

(i).135 and 225 (ii). 196 and 38220  (iii). 867 and 255

 

Ex 1.1 class 10 maths question 1 solution

(i). 135 and 225

Using Euclid’s Division Algorithm

Step 1 ∵225>135

Therefore according to Euclid’s Division Lemma

225= 135 ×1 + 90 

Step 2 ∵ Remainder 90≠0, therefore applying Euclid’s Division Lemma on 135 and 90

135=90×1+45

Step 3 ∵ Remainder 45≠0 , therefore applying Euclid’s Division Lemma on 90 and 45

90=45×2 +0

The remainder has now become zero, so this procedure stops. The divisor in this step is 45. 

Therefore, the HCF of 135 and 225 is 45.

Maths Class 10 Chapter 1 Exercise 1.1 Question 2

Q2.Show that any positive odd integer is of the form 6q + 1,or 6q + 3,or 6q + 5, where q is some integer.

Ex 1.1 class 10 maths question 2 solution

let a be a positive odd integer, now for a and b = 6 by the application of Euclid’s Division Algorithm 0 ≤ r < 6,i.e. the values of a can be 6q or 6q +1or 6q +2 or 6q +3 or 6q +4 or 6q+5, where q is some quotient.

Now since a is an odd positive integer. Therefore it can’t be of the form 6q, 6q +2 or 6q +4 , since al these being divisible by 2 are even positive integers.

So any positive odd integer of form 6q+1,6q+3 or 6q+5, where q is some integer.

Maths Class 10 Chapter 1 Exercise 1.1 Question 3

Q3.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?

Ex 1.1 class 10 maths question 3 solution

Group of members parade = 616 and 32

according to the questions, we are to find out the maximum numbers of columns i.e. we have to find out the HCF of 616 and 32.

Step 1 ∵616 >32

therefore, according to Euclid’s Division Lemma

616= 32 × 19 + 8 

Step 2 ∵ Remainder of 8 ≠0, therefore now applying Euclid’s division Lemma on 32 and 8

32 = 84 + 0

Since the remainder obtained now is zero. So this procedure stops.

The divisor in this step is 8, so the HCF of  616 and 32 is 8.

The maximum number of columns in which they can parade is 8.

Maths Class 10 Chapter 1 Exercise 1.1 Question 4

Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m +1 for some integer m.

[Hint: Let x be any positive integer then it is of form 3q, 3q +1, or 3q +2. Now square each of these and show that they can be rewritten in form 3m or 3m +1]

Ex 1.1 class 10 maths question 4 solution

Let x be any positive integer. Then it can be of form 3q, 3q +1 pr 3q +2.

Therefore if x = 3q

Squaring both sides

`x^2=(3q^2)` 

`= 9q^2` 

`=3(3q^2)` 

`=3m` 

Where m=3q2 and m is also an integer.

Hence `x^2=3m`  …………..(i)

Again if  x=3q+1

Squaring both sides

`x^2=(3q+1)^2`   

⟹ `x^2=9q^2  +2×3q ×1+1` 

⟹ `x^2=3(3q^2+2q)+1` 

⟹ `x^2=3m +1`     ………..(ii)

Where `m= 3q^2+2q` and m is also an integer.

Now   if x=3q+2

Squaring both sides

`x^2=(3q+2)^2` 

`=9q^2  12q +4` 

`=3(3q^2+4q+1)+1` 

=3m+1 

Where `m=3q^2+4q+1`  and m is also an integer.

Hence from (i),(ii), and (iii)

`x^2= 3m` or 3m+1 

Therefore the square of any positive integer is of the form 3m or 3m +1 for some integer m.

Maths Class 10 Chapter 1 Exercise 1.1 Question 5

Q5. use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,9m +1, or 9m +8

Ex 1.1 class 10 maths question 5 solution

Let a be any positive integer and b=3

∴a=3q+r 

Where q is the quotient and r is the remainder.

Here 0 ≤  r <  3

Therefore 

 If r=0 then a=3q

If r=1 then a=3q+1 

If r=2 then a=3q+2

Now if a = 3q

Cubing both sides

`a^3= (3q)^3` 

`a^3=27q^3=9(3q)^3=9m` 

Here `m=3q^3` and m is also an integer 

`a^3=9m` …………(i)

Again  if a=3q+1

Cubing both sides

`a^3= (3q+1)^3` 

⟹`a^3= 27q^3+27q^2+9q+1` 

=`9(3q^3+3q^2+q)+1` 

=9m+1 

Here `m= 3q^3+ 3q^3+q` and m is also an integer.

Hence  `a^3=9m+1` ………….(ii)

Now if a=3q+2

Cubing both sides

`a^3= (3q+2)^3` 

`a^3= 27q^3+ 54q^2+36q+8` 

`a^3=9(3q^3+ 6q^2+4q)+8` 

`a^3=9m+8` 

Here   `m= 3q^3+ 6q^2+4q` and m is also an integer.

Hence  `a^3=9m+8`

Now from (i) ,(ii) and (iii) 

We find that the cube of any positive integer is of the  form 9m, 9m +1, or 9m + 8

We hope now You must have understood math 10th class chapter 1 exercise 1.1.

Topics for you

    1.  Exercise 1.2

    2.  Exercise 1.3