**NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers**

NCERT Solutions for class 10 maths chapter 1 – real numbers lets the students solve and revise the whole syllabus very effectively. After going through the stepwise solutions given by our subject expert teachers, the student will be able to score better marks.

Class 10 is the board exam so preparation is very important for marks as well as for a career. Our NCERT for class 10 Maths chapter 1 is very much complete in covering all topics, contents, problems and their self-explanatory solutions. NCERT solutions for class 10 Maths chapter 1 will help you to solve all types of problems of class 10 Maths. The Toppr app provides the solution to your problems of any kind of real numbers class 10 Maths.

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**CBSE Class 10 Maths Chapter 1 NCERT Solutions**

NCERT solutions for class 10 Maths Chapter 1 introduces the very fundamental but essential topic of mathematics. In the beginning, the chapter real numbers class 10 gives the introduction of real numbers and then two very important topics Euclid’s Division Algorithm and The Fundamental Theorem of Arithmetic. This fundamental theorem of arithmetic has many real life and scientific applications. Other related fields also have use of these. So for a strong base in Maths which will support the further higher education also, our NCERT solutions for real numbers class 10 Maths will definitely help.

**Topics Covered under CBSE Chapter 10 Maths Chapter 1 – Real Numbers**

- Ex. 1.1: Introduction to Real Numbers
- Ex. 1.2: Euclid’s Division Lemma
- Ex. 1.3: The Fundamental Theorem of Arithmetic
- Ex. 1.4: Revisiting Irrational Numbers
- Ex. 1.5: Revisiting Rational Numbers and Their Decimal Expansions.

**NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers**

This chapter helps the students to understand the fundamental theorem of arithmetic that has much real life and scientific applications. Other related fields also have use of the Real numbers which are explained in this chapter. So for a strong base in Maths which will support the further higher education also, our NCERT solutions for class 10 Maths will definitely help. Rational and irrational numbers are major types of real numbers. Theorems will explain these with proper examples and applications.

**Let us discuss the sub-topics in detail –**

**1.1: Introduction**:

In this chapter, the student will explore the world of real numbers and their related applications. This chapter contains some very important properties of positive numbers.

**1.2: Euclid’s Division Lemma:**

In this chapter, the student will learn the technique to compute the Highest Common Factor (HCF) of two given positive integers by Euclid’s algorithm.

**1.3: The Fundamental Theorem of Arithmetic:**

The student will learn that every composite number can be expressed as a product of primes uniquely. This property is a fundamental theorem of arithmetic.

**1.4: Revisiting Irrational Numbers:**

This chapter redefines the irrational numbers. Some relevant examples will help to understand the concept easily. Method of contradiction will help to prove them.

**1.5: Revisiting Rational Numbers and Their Decimal Expansions:**

The student will revisit the concept of rational numbers using fraction expression as well as using decimal expansions. This is because that decimal expansion of every rational number is either terminating or repeatedly non-terminating.

**Some Questions from NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers**

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

Solution

Using Euclid division algorithm, we know that a = *bq *+ r*, *Osrs*b*—- (1)

Let a be any positive integer, and b = 3.

Substitute b = 3 in equation (1)

*a *= 3*q *+ r where o s*r*s3, r= 0, 1, 2

If r= 0, a = 3*9 *

Cube the value, we get 23 *= 2703 **2*3 = 9(393), where m=393 —(2)

If r = 1, a = 3q+1 Cube the value, we get 23 = (39+13 a3 = (27*7*3 *+ 277*2 +9q+1) 23 = 9/393 +382 + 1) +1, where m= 393 +372 +*9*—(3)

If r= 2, a = 39+2

Cube the value, we get an2 = 139+213 *43 *= (27*q*2 + 5302 + 36q+8) 23 = 9393 +692 + 49) + 8, where m=393 +692 +49—-(4)

From equation 2, 3 and 4,

The cube of any positive integer is of the form 9m, 9m + 1 or 9*m*+8.

**Q. 2. Show that any positive odd integer is of form 6q+1, or 69 +3, or 6q+5, where q is some integer. **

**Solution **

Using Euclid division algorithm, we know that a = *bg*+r*, *Osrs b—-(1)

Let a be any odd positive integer and b = 6.

Substitute b = 6 in equation (1)

*a *= 6*q*+r where o s*r*s 6 *r*= 0,1,2,3,4,5

If r = 0, a = 6*9, *6q is divisible by 6

If r = 1, a = 69+1,6q+1 is not divisible by 2.

If r= 2, a = 6*q*+2, 6*9*+2 is divisible by 2

If r= 3, a = 6*9*+3,6*q*+3 is not divisible by 2.

If r = 4, *a *= 6*9 *+4, 6*9*+4 is divisible by 2

If r= 5, a = 6q+5, 69+5 is not divisible by 2.

So, the numbers 6*4*, 6*9*+ 2,6*9 *+ 4 are divisible by 2 and even numbers.

The remaining numbers 6q+1, 69+ 3 and 69+5 are odd.

**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?**

**S****olution **

HCF (616, 32) is the maximum number of columns in which they can march.

First, find which integer is larger.

616 > 32

Step 2: Then apply Euclid’s division algorithm to 616 and 32 to obtain

616 = 32 x 19+8

Repeat the above step until you will get remainder as zero.

Step 3: Now consider the divisor 32 and the remainder 8, and apply the division lemma to get

32 = 8 + 4 +0

Since the remainder is zero, we cannot proceed further.

Step 4: Hence the divisor at the last process is 8

So, the H.C.F. of 616 and 32 is 8.

Therefore, 8 is the maximum number of columns in which they can march,

**Q. 4 Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3 m + 1 for some integer m.**

**Solution **

Using Euclid division algorithm, we know that a = *bq*+r*, *Osrs…… (1)

Let a be any positive integer, and b = 3.

Substitute b = 3 in equation (1)

*a = *3q+r where o s*r*s 6 r= 0,1,2 If r= 0, a = 39 On squaring we get, a2 = 3(37), where *m*= 392—(2)

If r= 1, *a *= 3*9*+1

On squaring we get a2 = 33q2 + 2q +1, where *m*= 392 +24—(3) If r= 2, *a *= 39+2

On squaring we get *a*2 = 3(3q2 +49 + 1) + 1, where *m*=372 +4q+1-— (4)

From equation 2,3 and 4,

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

**You can download complete NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers by clicking on the button below.**

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