# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

**Exercise- 1.1 Class 10 Mathematics Solutions Chapter 1 Real Numbers **

**Question 1. **NCERT Solutions for Class 10 Maths

**Use Euclid’s division algorithm to find the H.C.F. of the following numbers:-**

(I) 135 and 225

(II) 196 and 38220

(III) 867 and 255

**Solution- **NCERT Solutions for Class 10 Maths Chapter 1

**(I) Numbers given in question = 135 and 225**

**Step I.** Using Euclid’s division lemma for the given numbers 225 and 135 According to the question,

225 = (135 x 1) + 90 (∴ remainder 90 ≠ 0)

**Step II.** Using Euclid’s division lemma for the numbers 135 and 90,

135 = (90 x 1) + 45 (∴ remainder 45 ≠ 0)

**Step III.** Using Euclid’s division lemma for the numbers 90 and 45,

90 = (45 x 2) + 0 (∴ remainder = 0)

Remainder is zero and divisor = 45,

Therefore, the H.C.F. of 135 and 225 is equal to 45.

**(II) Numbers given in the question = 196 and 38220**

**Step I.** Using Euclid’s division lemma for the given numbers 196 and 38220 According to the question,

38220 = (196 x 195) + 0 (∴ remainder = 0)

Remainder is zero and divisor = 196, so H.C.F. = 196.

**(III) Numbers given in the question = 867 and 255**

**Step I.** Using Euclid’s division lemma for the given numbers 867 and 255 According to the Question,

867 = (255 x 3) + 102 (∴ remainder 102 ≠ 0)

**Step II.** using Euclid’s division lemma for the numbers 255 and 102, 255 = (102 x 2) + 51 (∴ remainder 51 ≠ 0)

**Step III.** From Euclid’s division lemma for numbers 102 and 51, 102 = (51 x 2) + 0 (∴ remainder = 0)

Remainder is zero and divisor = 51, so H.C.F. = 51.

**Read Also **

**Exercise- 1.1 Class 10 Mathematics Solutions Chapter 1 in Hindi Medium**

**NCERT Class 10th Science Chapter 1**

**NCERT Class 10th Science Chapter 2**

**Question 2. **NCERT Solutions for Class 10 Maths

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

**Solution- **NCERT Solutions for Class 10 Maths Chapter 1

Let a be any odd positive integer. which is greater than 6 and b is a positive integer such that b = 6.

Then, from Euclid’s division lemma,

a = bq + r

a = 6q + r (∴ b = 6)

Then, the value of r must be less than 6.

Then, possible values of r = 0,1,2,3,4,5

Then, a = 6q + 0

a = 6q + 1

a = 6q + 2

a = 6q + 3

a = 6q + 4

a = 6q + 5

a is any odd number; So a = 6q + 0, 6q + 2 and 6q + 4 cannot be because these quantities are divisible by 2.

Then, the odd number will be a = 6q + 1 or 6q + 3 or 6q + 5.

So a positive odd integer will be of the form 6q + 1 or 6q + 3 or 6q + 5.

**Question 3. **NCERT Solutions for Class 10 Maths

**In a parade an army contingent of 616 members is to march behind an army band of 32 members. Both the groups have to march in the same number of columns. What is the maximum number of columns in which they can march?**

**Solution- **NCERT Solutions for Class 10 Maths Chapter 1

The maximum number of lines will be the maximum multiple of 616 for the number of troop soldiers and 32 for the band members.

Then,

**Step I.** Using Euclid’s division lemma for 616 and 32 according to the question,

616 = (32 x 19) + 8 (∴ remainder 8 ≠ 0)

Then,

**Step II.** From Euclid’s division lemma for 32 and 8,

32 = (8 x 4) + 0 (∴ remainder = 0)

The remainder is zero and the divisor is 8. Highest Factor (H.C.F.) = 8 so the army can march in 8 columns.

**Question 4. **NCERT Solutions for Class 10 Maths Chapter 1

**Using Euclid’s division lemma, show that the square of a positive integer is of the form 3m or 3m + 1 for some integer m.**

**Solution- **NCERT Solutions for Class 10 Maths Chapter 1

Let a and b be two positive integers such that a > b and b = 3

Then, from Euclid’s division lemma according to question,

a = 3b + r while 0 ≤ r < 3

Then, possible values of k = 0,1,2

Then, a = 3b + 0

⇒ a = 3b + 1

⇒ a = 3b + 2

Then, a^{2} = (3b + 0)^{2}

⇒ a^{2} = (3b + 1)^{2}

⇒ a^{2} = (3b + 2)^{2}

Then, a^{2} = (3b + 0)^{2} then a^{2} = 9b^{2} = 3(3b^{2})

If a^{2} = (3b + 1)^{2} then a^{2} = 9b^{2 }+ 6b + 1

= 3(3b^{2} + 2b) + 1

If a^{2} = (3b + 2)^{2} then a^{2} = 9b^{2} + 12b + 4

= (9b^{2} + 12b + 3) + 1

= 3(3b^{2} + 4b + 1) + 1

It is clear from all the expansions of a^{2} that a^{2} is divisible by 3 and the remainder is either zero or 1.

a^{2} = 3m + 0 ⇒ a^{2} = 3m +1

So the square of a positive integer is of the form 3m or 3m + 1 for any integer m.

**Question 5. **NCERT Solutions for Class 10 Maths

**Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.**

**Solution- **NCERT Solutions for Class 10 Maths Chapter 1

Let a and b be two positive integers such that a > b and b > 9

Then, using Euclid’s division lemma According to the question, a = 9b + r

Then, the value of r must be less than 9.

Then, possible values of r = 0,1,2,3,4,5,6,7,8

Then,

⇒ a = 9b + 0

⇒ a = 9b + 1

⇒ a = 9b + 2

⇒ a = 9b + 3

⇒ a = 9b + 4

⇒ a = 9b + 5

⇒ a = 9b + 6

⇒ a = 9b + 7

⇒ a = 9b + 8

When a = 9b + 0

then

⇒ a^{3} = (3b + 0)^{3} = 27b^{3}

⇒ a^{3} = 9(3b^{3}) ………….. (1)

When a = 9b + 1 then a^{3} = (3b + 1)^{3}

⇒ a^{3} = (3b)^{3} + 9b(3b + 1) + (1)^{3}

⇒ a^{3} = (27b^{3} + 27b^{2} + 9b) + 1

⇒ a^{3} = 9[3b^{3} + 3b^{2} + b] +1 ………..(2)

When a = 9b + 2 then a^{3} = (3b + 2)^{3}

⇒ a^{3} = (3b)^{3} + 3.3b.2(3b + 2) + (2)^{3}

⇒ a^{3} = [27b^{3} + 54b^{2} + 36b] + 8

⇒ a^{3} = [27b^{3} + 18b(3b +2)] + 8

⇒ a^{3} = 9[3b^{3} + 6b^{2} + 4b] + 8 ……….. (3)

Then, look at equations (1), (2) and (3) to see that it is divisible by 9.

Then, respectively a^{3} = 9m,

or a^{3} = 9m + 1,

Or a^{3} = 9m + 8 can be written.

Hence, the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.