Complete solutions of Ex 1.1 Class 10 Maths Chapter 1 with additional questions and answers from new NCERT syllabus textbook Class 10 Maths.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

- Real Numbers Class 10 Ex 1.1
- Real Numbers Class 10 Ex 1.2
- Real Numbers Class 10 Ex 1.3
- Real Numbers Class 10 Ex 1.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Real Numbers |

Exercise |
Ex 1.1 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.

**Ex 1.1 Class 10 Maths 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)** By Euclid’s Division Algorithm, we have

225 = 135 x 1 + 90 135

= 90 x 1 + 45 90

= 45 x 2 + 0

∴ HCF (135, 225) = 45.

**(ii)** By Euclid’s Division Algorithm, we have

38220 = 196 x 195 + 0

196 = 196 x 1 + 0

∴ HCF (38220, 196) = 196.

**(iii)** By Euclid’s Division Algorithm, we have

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102 = 51 x 2 + 0

∴ HCF (867, 255) = 51.

**Ex 1.1 Class 10 Maths 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 a be a positive odd integer. Also, let q be the quotient and r the remainder after dividing a by 6.

Then, a = 6q + r, where 0 ≤ r < 6.

Putting r = 0, 1, 2, 3, 4, and 5, we get:

a = 6q, a = 6q + 1, a = 6q + 2, a = 6q + 3, a = 6q + 4, a = 6q + 5

But a = 6q, a = 6q + 2 and a = 6q + 4 are even.

Hence, when a is odd, it is of the form 6q + 1, 6q + 3, and 6q + 5 for some integer q.

Hence proved.

**Ex 1.1 Class 10 Maths 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:**

Let n be the number of columns such that the value of n be maximum and it must divide both the numbers 616 and 32.

Then, n = HCF (616, 32)

By Euclid’s Division Algorithm, we have:

616 = 32 x 19 + 8 32 = 8 x 4 + 0

∴ HCF (616, 32) = 8

i. e., n = 8

Hence, the maximum number of columns is 8.

**Ex 1.1 Class 10 Maths Question 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.

**Solution:**

Let a be a positive integer, q be the quotient and r be the remainder.

Dividing a by 3 using the Euclid’s Division Lemma,

we have:

a = 3q + r, where 0 ≤ r < 3

Putting r = 0, 1 and 2, we get:

a = 3q

⇒ a

^{2}= 9q

^{2}

= 3 x 3q

^{2}

= 3m (Assuming m = q

^{2})

Then, a = 3q + 1

⇒ a

^{2}= (3q + l)

^{2}= 9q

^{2}+ 6q + 1

= 3(3q

^{2}+ 2q) + 1

= 3m + 1 (Assuming m = 3q

^{2}+ 2q)

Next, a = 3q + 2

⇒ a

^{2}= (3q + 2)

^{2}=9q

^{2}+ 12q + 4

= 9q

^{2}+ 12q + 3 + 1

= 3(3q

^{2}+ 4q + 1) + 1

= 3m + 1. (Assuming m = 3q

^{2}+ 4q+l)

Therefore, the square of any positive integer (say, a

^{2}) is always of the form 3m or 3m + 1.

Hence, proved.

**Ex 1.1 Class 10 Maths Question 5.
**Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.

**Solution:**

Let a be a positive integer, q be the quotient and r be the remainder.

Dividing a by 3 using the Euclid’s Division Algorithm, we have,

a = 3q + r, where 0 ≤ r < 3

Putting r = 0, 1 and 2, we get:

a = 3q, a = 3q + 1 and a = 3q + 2

If a = 3q, then a

^{3}= 27q

^{3}= 9(3q

^{3}) = 9m. (Assuming m = 3q

^{3}.)

If a = 3q + 1, then

a

^{3}= (3q + l)

^{3}= 27q

^{3}+ 9q(3q + 1) + 1 = 9(3q

^{3}+ 3q

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

^{3}+ 3q

^{2}+ q)

If a = 3q + 2, then a

^{3}= (3q + 2)

^{3 }= 27q

^{3}+ 18q(3q + 2) + (2)

^{3 }= 9(3q

^{3}+ 6q

^{2}+ 4q) + 8

= 9m + 8, (Assuming m – 3q

^{3}+ 6q

^{2}+ 4q)

Hence, a

^{3}is of the form 9m, 9m + 1 or 9m + 8.

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