In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 PDF, MP Board Class 10 Maths Solutions, These solutions are solved subject experts from the latest edition books.

## MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

**Class 10 Maths Chapter 1 MP Board Question 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) HCF of 135 and 225

Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get

225 = (135 × 1) + 90, since 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and

90, we get 135 = (90 × 1) + 45

But 45 ≠ 0

∴ Applying Euclid’s lemma to 90 and 45, we get 90 = (45 × 2) + 0

Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45,

∴ HCF of 225 and 135 is 45.

(ii) HCF of 196 and 38220

We start dividing the larger number 38220 by 196, we get

38220 = (196 × 195) + 0

Here, r = 0

∴ HCF of 38220 and 196 is 196.

(iii) HCF of 867 and 255 Here, 867 > 255

∴ Applying Euclid’s Lemma to 867 and 255, we get

867 = (255 × 3) + 102, 102 ≠ 0

∴ Applying Euclid’s Lemma to 255 and 102, we get

255 = (102 × 2) + 51, 51 ≠ 0

∴ Applying Euclid’s Lemma to 102 and 51, we get

102 = (51 × 2) + 0, r = 0

∴ HCF of 867 and 255 is 51.

**MP Board Class 10 Maths Chapter 1 Question 2.**

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

Solution:

Let us consider a positive odd integer as ‘a’.

On dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.

∴ Using Euclid’s lemma, we get a = 6q + r

where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,

a = 6q + 0 = 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 are even values of ‘a’.

[∵ 6q = 2(3q) = 2m_{1} 6q + 2 = 2(3q + 1) = 2m_{2},

6q + 4 = 2(3 q + 2) = 2m_{3}]

But ‘a’ being an odd integer, we have :

a = 6q + 1, or a = 6q + 3, or a = 6q + 5

**Class 10 Maths Chapter 1 MP Board In English Medium Question 3.**

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:

Total number of members = 616

∴ The total number of members are to march behind an army band of 32 members is HCF of 616 and 32.

i. e., HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns.

∴ Applying Euclid’s lemma to 616 and 32, we get

616 = (32 × 19) + 8, since, 8 ≠ 0

Again, applying Euclid’s lemma to 32 and 8, we get

32 = (8 × 4) + 0, r = 0

∴ HCF of 616 and 32 is 8

Hence, the required number of maximum columns = 8.

**MP Board Solutions Class 10 Maths Chapter 1 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 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]

Solution:

Let us consider an arbitrary positive integer as ‘x’ such that it is of the form

3q, (3q + 1) or (3q + 2)

For x = 3q, we have x^{2} = (3q)^{2}

⇒ x^{2} = 9q^{2} = 3(3q^{2}) = 3m ………. (1)

Putting 3q^{2} = m, where m is an integer.

For x = 3q + 1,

x^{2} = (3q + 1)^{2} = 9q^{2} + 6q + 1

= 3(3q^{2} + 2q) + 1 = 3m + 1 ………… (2)

Putting 3q^{2} + 2q = m, where m is an integer.

For x = 3q + 2,

x^{2} = (3q + 2)^{2}

= 9q^{2} + 12q + 4 = (9q^{2} + 12q + 3) + 1

= 3(3q^{2} + 4q + 1) + 1 = 3m + 1 ……….. (3)

Putting 3q^{2} + 4q +1 = m, where m is an integer.

From (1), (2) and (3),

x^{2} = 3m or 3m + 1

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

**Class 10 Maths Chapter 1 Exercise 1.1 Solutions MP Board Question 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 us consider an arbitrary positive integer x such that it is in the form of 3q, (3q +1) or (3q + 2).

For x = 3q

x^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3}) = 9m ……… (1)

Putting 3q^{3} = m, where m is an integer.

For x = 3q + 1

x^{3} = (3 q + 1)^{3} = 27q^{3} + 27q^{2} + 9q + 1

= 9(3q^{3} + 3q^{2} + q) + 1 = 9m + 1 ………… (2)

Putting 3q^{3} + 3q^{2} + q = m, where m is an integer.

For x = 3q + 2,

x^{3} = (3q + 2)^{3} = 27q^{3} + 54q^{2} + 36q + 8

= 9(3q^{3} + 6q^{2} + 4q) + 8 = 9m + 8 ……………. (3)

Putting 3q^{3} + 6q^{2} + 4q = m, where m is an integer.

From (1), (2) and (3), we have

x^{3} = 9m, (9m + 1) or (9m + 8)

Thus, cube of any positive integer can be in the form 9m, (9m + 1) or (9m + 8) for some integer m.

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