Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM ×HCF = product of the two numbers. (i) 26 and 91 (ii) 510×92 (iii) 336 and 54

Answer: (i) 26 and 91 Factor of 26=2×13×1 Factor of 91=7×13×1 HCF of 26 and 91=13×1=13 LCM of 26 and 91=2×7×13=182 LCM × HCF =182×13=2366 26×91=2366 So, LCM.HCF = product of the two numbers =26×91. Hence proved. (ii) 510×92 Factor of 510=2×3×5×17×1 Factor of 92=2×2×23 HCF of 510 and 92=2×2=2 LCM of 510 and 92=2×2×3×5×17×23=23,460 LCM × HCF =23,460×2=46,920 510×92=46,920 So, LCM.HCF = product of the two numbers =510×92. Hence proved. (iii) 336 and 54 Factor of 336=2×2×2×2×7×3×1 Factor of 54=2×3×3×3 HCF of 336 and 54=2×3=6 LCM of 336 and 54=2×3×2×2×2×7×3×3=3,024 LCM × HCF =3,024×6=18,144 366×54=18,144 So, LCM.HCF = product of the two numbers =336×54. Hence proved.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Q: Question 1. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 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 of 18=2×3×3 Factor of 12=2×2×3 LCM =2×2×3×3=36 Hence, they will meet together at the starting point after 36 minutes.

Q: Question 3. Use Euclid's division algorithm to find the HCF of: 867 and 225

Answer: Step 1: First find which integer is larger. 867>255 Step 2: Then apply Euclid's division algorithm to 867 and 255 to obtain 867=255×3+102 Repeat the above step until you will get remainder as zero. Step 3: Now consider the divisor 225 and the remainder 102, and apply the division lemma to get 255=102×2+51 102=51×2=0 Since the remainder is zero, we cannot proceed further. Step 4: Hence the divisor at the last process is 51. So, the H.C.F. of 867 and 255 is 51.

Q: Question 4. Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 225

Answer: (i) 135 and 225 Step 1: First find which integer is larger. 225>135 Step 2: Then apply Euclid's division algorithm to 225 and 135 to obtain 225=135×1+90 Repeat the above step until you will get remainder as zero. Step 3: Now consider the divisor 135 and the remainder 90, and apply the division lemma to get 135=90×1+45 90=2×45+0 Since the remainder is zero, we cannot proceed further. Step 4: Hence, the divisor at the last process is 45 So, the H.C.F. of 135 and 225 is 45. (ii) 196 and 38220 Step 1: First find which integer is larger. 38220>196 Step 2: Then apply the Euclid's division algorithm to 38220 and 196 to obtain 38220=196×195+0 Since the remainder is zero, we cannot proceed further. Step 3: Hence, the divisor at the last process is 196. So, the H.C.F. of 196 and 38220 is 196. (iii) 867 and 225 Step 1: First find which integer is larger. 867>255 Step 2: Then apply Euclid's division algorithm to 867 and 255 to obtain 867=255×3+102 Repeat the above step until you will get remainder as zero. Step 3: Now consider the divisor 225 and the remainder 102, and apply the division lemma to get 255=102×2+51 102=51×2=0 Since the remainder is zero, we cannot proceed further. Step 4: Hence the divisor at the last process is 51. So, the H.C.F. of 867 and 255 is 51.

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 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 further higher education also, our NCERT solutions for real numbers class 10 Maths will definitely help.

ncert solutions for class 10 maths chapter 15

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 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: 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. 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.1 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 = 39

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.

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

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

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

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

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

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

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

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.

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?

Solution

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 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.

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 3m + 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 srs 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 = 39+1

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

On squaring we get a2 = 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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Solved Questions For You:

Question 1. There is a circular path around a sports field. Sonia takes $18$ minutes to drive one round of the field, while Ravi takes $12$ 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 of

18 = 2 \times 3 \times 3

Factor of

12 = 2 \times 2 \times 3

LCM

= 2 \times 2 \times 3 \times 3 = 36

Hence, they will meet together at the starting point after

36

minutes.

Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers.

(i) $26$ and $91$

(ii) $510 \times 92$

(iii) $336$ and $54$

Answer:

(i)

26

and

91

Factor of

26 = 2 \times 13 \times 1

Factor of

91 = 7 \times 13 \times 1

HCF of

26

and

91 = 13 \times 1 = 13

LCM of

26

and

91 = 2 \times 7 \times 13 = 182

LCM

\times

HCF

= 182 \times 13 = 2366

26 \times 91 = 2366

So, LCM.HCF

=

product of the two numbers

= 26 \times 91 .

Hence proved.

(ii)

510 \times 92

Factor of

510 = 2 \times 3 \times 5 \times 17 \times 1

Factor of

92 = 2 \times 2 \times 23

HCF of

510

and

92 = 2 \times 2 = 2

LCM of

510

and

92 = 2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23, 460

LCM

\times

HCF

= 23, 460 \times 2 = 46, 920

510 \times 92 = 46, 920

So, LCM.HCF

=

product of the two numbers

= 510 \times 92 .

Hence proved.

(iii)

336

and

54

Factor of

336 = 2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 1

Factor of

54 = 2 \times 3 \times 3 \times 3

HCF of

336

and

54 = 2 \times 3 = 6

LCM of

336

and

54 = 2 \times 3 \times 2 \times 2 \times 2 \times 7 \times 3 \times 3 = 3, 024

LCM

\times

HCF

= 3, 024 \times 6 = 18, 144

366 \times 54 = 18, 144

So, LCM.HCF = product of the two numbers

= 336 \times 54 .

Hence proved.

Question 3. Use Euclid’s division algorithm to find the HCF of: $867$ and $225$

Answer:

Step 1: First find which integer is larger.

867 > 255

Step 2: Then apply Euclid’s division algorithm to

867

and

255

to obtain

867 = 255 \times 3 + 102

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

Step 3: Now consider the divisor

225

and the remainder

102,

and apply the division lemma to get

255 = 102 \times 2 + 51

102 = 51 \times 2 = 0

Since the remainder is zero, we cannot proceed further.

Step 4: Hence the divisor at the last process is 51.
So, the H.C.F. of

867

and

255

51 .

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

(i) $135$ and $225$

(ii) $196$ and $38220$

(iii) $867$ and $225$

Answer:

(i) $135$ and $225$

Step 1: First find which integer is larger.

225 > 135

Step 2: Then apply Euclid’s division algorithm to

225

and

135

to obtain

$225 = 135 \times 1 + 90$

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

Step 3: Now consider the divisor

135

and the remainder

90,

and apply the division lemma to get

$135 = 90 \times 1 + 45$

$90 = 2 \times 45 + 0$

Since the remainder is zero, we cannot proceed further.

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

45

So, the H.C.F. of

135

and

225

45 .

(ii)

196

and

38220

Step 1: First find which integer is larger.

$38220 > 196$

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

$38220 = 196 \times 195 + 0$

Since the remainder is zero, we cannot proceed further.

Step 3: Hence, the divisor at the last process is 196.

So, the H.C.F. of

196

and

38220

196 .

(iii)

867

and

225

Step 1: First find which integer is larger.

$867 > 255$

Step 2: Then apply Euclid’s division algorithm to

867

and

255

to obtain

$867 = 255 \times 3 + 102$

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

Step 3: Now consider the divisor

225

and the remainder

102,

and apply the division lemma to get

$255 = 102 \times 2 + 51$

$102 = 51 \times 2 = 0$

Since the remainder is zero, we cannot proceed further.

Step 4: Hence the divisor at the last process is 51.

So, the H.C.F. of

867

and

255

51 .

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NCERT Solutions for Class 10 Maths Chapter 1 Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

CBSE Class 10 Maths Chapter 1 NCERT Solutions

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

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

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

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