# NCERT Solutions for Class 10 Maths Chapter 1 Free PDF Download

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

NCERT Solutions for class 10 maths chapter 1 –Â  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 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.

Our team of expert teachers has created these NCERT Solutions according to the curriculum and pattern of board exams. Our app will help you to get complete NCERT Solutions for Class 10 Maths and other subjects. We are providing you the free pdf download links of the class 10 maths chapter 1.

### 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 Euclidâ€™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 further higher education also, our NCERT solutions for realÂ numbers class 10 Maths will definitely help.

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

This chapter helps the students to understand the fundamental theorem of arithmetic that has many 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: Euclidâ€™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 Euclidâ€™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. A 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.1Â 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

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

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

Substitute b = 3 in equation (1)

a = 3q + r where o srs3, r= 0, 1, 2

If r= 0, a = 39Â

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

If r = 1, a = 3q+1 Cube the value, we get 23 = (39+13 a3 = (2773 + 2772 +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 = (27q2 + 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 9m+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.Â

SolutionÂ

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

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

Substitute b = 6 in equation (1)Â

a = 6q+r where o srs 6 r= 0,1,2,3,4,5Â

If r = 0, a = 69, 6q is divisible by 6Â

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

If r= 2, a = 6q+2, 69+2 is divisible by 2Â

If r= 3, a = 69+3,6q+3 is not divisible by 2.Â

If r = 4, a = 69 +4, 69+4 is divisible by 2Â

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

So, the numbers 64, 69+ 2,69 + 4 are divisible by 2 and even numbers.

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

Q. 3Â 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 maximumÂ number of columns in which they can march?

SolutionÂ

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

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 the 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.Â

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

So, the H.C.F. of 616 and 32 is 8.Â

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 3m + 1 for some integer m.

SolutionÂ

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

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

Substitute b = 3 in equation (1)Â

a = 3q+r where o srs 6 r= 0,1,2 If r= 0, a = 39 On squaring we get, a2 = 3(37), where m= 392â€”(2)Â

If r= 1, a = 39+1Â

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

On squaring we get a2 = 3(3q2 +49 + 1) + 1, where m=372 +4q+1-â€” (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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Solved Questions For You:

Question 1. There is a circular path around a sports field. Sonia takesÂ Â minutes to drive one round of the field, while Ravi takesÂ Â minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Answer: Sonia and Ravi drive one round of the field at the same point, same time and same direction.

So, they will meet again after LCM of both values at the starting point.
To get the LCM we have factorize the number.
Factor ofÂ
Factor ofÂ
LCMÂ
Hence, they will meet together at the starting point afterÂ Â minutes.
Question 2. Find the LCM and HCF of the following pairs of integers and verify thatÂ LCMÂ HCF = product of the two numbers.
(i)Â Â andÂ
(ii)Â
(iii)Â Â andÂ

(i)Â Â andÂ
Factor ofÂ
Factor ofÂ
HCF ofÂ Â andÂ
LCM ofÂ Â andÂ
LCMÂ Â HCFÂ
So, LCM.HCFÂ Â product of the two numbersÂ
Hence proved.
(ii)Â
Factor ofÂ
Factor ofÂ
HCF ofÂ Â andÂ
LCM ofÂ Â andÂ
LCMÂ Â HCFÂ
So, LCM.HCFÂ Â product of the two numbersÂ
Hence proved.
(iii)Â Â andÂ
Factor ofÂ
Factor ofÂ
HCF ofÂ Â andÂ
LCM ofÂ Â andÂ
LCMÂ Â HCFÂ
So, LCM.HCF = product of the two numbersÂ
Hence proved.
Question 3. Use Euclid’s division algorithm to find the HCF of:Â Â andÂ
Step 1:Â First find which integer is larger.

Step 2:Â Then apply Euclid’s division algorithm toÂ Â andÂ Â to obtain

Repeat the above step until you will get remainder as zero.
Step 3:Â Now consider the divisorÂ Â and the remainderÂ Â and apply the division lemma to get

Since the remainder is zero, we cannot proceed further.
Step 4:Â Hence the divisor at the last process is 51.
So, the H.C.F. ofÂ Â andÂ Â isÂ

Question 4. Use Euclid’s division algorithm to find the HCF of:

(i)Â Â andÂ
(ii)Â Â andÂ
(iii)Â Â andÂ

(i)Â Â andÂ

Â

Step 1:Â First find which integer is larger.

Step 2:Â Then apply Euclid’s division algorithm toÂ Â andÂ Â to obtain

Repeat the above step until you will get remainder as zero.
Step 3:Â Now consider the divisorÂ Â and the remainderÂ Â and apply the division lemma to get

Since the remainder is zero, we cannot proceed further.
Step 4:Â Hence, the divisor at the last process isÂ
So, the H.C.F. ofÂ Â andÂ Â isÂ
(ii)Â Â andÂ
Â
Step 1:Â First find which integer is larger.

Step 2:Â Then apply the Euclid’s division algorithm to 38220 and 196 to obtain

Since the remainder is zero, we cannot proceed further.
Step 3:Â Hence, the divisor at the last process is 196.
So, the H.C.F. ofÂ Â andÂ Â isÂ
(iii)Â Â andÂ
Â
Step 1:Â First find which integer is larger.

Step 2:Â Then apply Euclid’s division algorithm toÂ Â andÂ Â to obtain

Repeat the above step until you will get remainder as zero.
Step 3:Â Now consider the divisorÂ Â and the remainderÂ Â and apply the division lemma to get

Since the remainder is zero, we cannot proceed further.
Step 4:Â Hence the divisor at the last process is 51.
So, the H.C.F. ofÂ Â andÂ Â isÂ

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