**NCERT Solutions Class 10 Maths Chapter 1 Real Numbers** – Here are all the NCERT solutions for Class 10 Maths Chapter 1. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Real Numbers of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across chapter 1 Real Numbers. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers in one place.

## NCERT Solutions Class 10 Maths Chapter 1 Real Numbers

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 10 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 1 Real Numbers , Maths, Class 10.

Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 1 |

Chapter Name |
Real Numbers |

### NCERT Solutions Class 10 Maths chapter 1 Real Numbers

Class 10, Maths chapter 1, Real Numbers solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.1:Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Ans :(i) 135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 x 1 + 90 Since remainder \( 90 \neq 0 \), we apply the division lemma to 135 and 90 to obtain 135 = 90 x 1 + 45 We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 x 45 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 45, Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 x 195 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196. (iii) 867 and 255 Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 867 = 255 x 3 + 102 Since remainder \(102 \neq 0\) we apply the division lemma to 2S5 and 102 to obtain 255 = 102 x 2+ 51 We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 102 = 51 x 2+0 Since the remainder is zero, the process stops. Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

Q.2:Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer?

Ans :Let a be any positive integer and b = 6. Then, by Euclid's algorithm, a = 6q + r for some integer q 0, and r = 0, 1, 2, 3, 4, 5 because 0 \(\leq\) r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 x 3q +1 = \(2 k_{1}+1, \text { where } k_{1}\) is a positive integer \(6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_{2}+1, \text { where } k_{2}\) is an integer \(6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_{3}+1, \text { where } k_{3}\) is an integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers. And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5

Q.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?

Ans :HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid's algorithm to find the HCF. 616 = 32 x 19 + 8 32 = 8 x 19 +8 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.

Q.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 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Ans :Let a be any positive integer and b = 3. Then a = 3q + r for some integer \(q \geq 0\) And r = 0, 1, 2 because \(0 \leq r<3\) Therefore, a = 3q or 3q + 1 or 3q + 2 Or, \(\begin{array}{l}{a^{2}=(3 q)^{2} \text { or }(3 q+1)^{2} \text { or }(3 q+2)^{2}} \\ {a^{2}=\left(9 q^{2}\right) \text { or } 9 q^{2}+6 q+1 \text { or } 9 q^{2}+12 q+4} \\ {=3 \times\left(3 q^{2}\right) \text { or } 3\left(3 q^{2}+2 q\right)+1 \text { or } 3\left(3 q^{2}+4 q+1\right)+1} \\ {=3 k_{1} \text { or } 3 k_{2}+1 \text { or } 3 k_{3}+1}\end{array}\) Where \(k_{1}, k_{2}, \text{and } k_{3} \) are some positive integers Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1

Q.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.

Ans :Let a be any positive integer and b = 3 \(\begin{array}{l}{a=3 q+r, \text { where } q \geq 0 \text { and } 0 \leq r<3} \\ {\therefore a=3 q \text { or } 3 q+1 \text { or } 3 q+2}\end{array}\) Therefore, every number can be represented as these three forms. There are three cases. Case 1: When a = 3q, \(a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m\), Where m \( \text { is an integer such that } m=3 q^{3}\) \(\begin{array}{l}{\text {Case } 2 : \text { when } a=3 q+1} \\ {a^{3}=(3 q+1)^{3}} \\ {a^{3}=27 q^{3}+27 q^{2}+9 q+1} \\ {a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1} \\ {a^{3}=9 m+1} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+3 q^{2}+q\right)}\end{array}\) \(\begin{array}{l}{\text {Case } 3 : \text { when } a=3 q+2} \\ {a^{3}=(3 q+2)^{3}} \\ {a^{3}=27 q^{3}+54 q^{2}+36 q+8} \\ {a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8} \\ {a^{3}=9 m+8} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+6 q^{2}+4 q\right)}\end{array}\) Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

## NCERT / CBSE Book for Class 10 Maths

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