NCERT Solutions for Class 10 Maths Chapter 1 : Real Numbers

Q: Q1. Explain the Graphical Method of Solutions of a Pair of Linear Equations.

Answer: Two lines are represented by the graph of a pair of linear equations with two variables. A system is considered to have a unique solution if the graphs of its two equations intersect at a single point. A system is considered to have no solution if the graphs of two equations are parallel lines. When the graphs of two equations in a system are two coincident lines, the system is said to have an unlimited number of solutions, indicating that it is consistent and dependent.

Q: Q2. Are Real Numbers NCERT Solutions for Class 10 Maths Chapter 1 important for exams?

Answer: Yes, Chapter 1 of NCERT Class 10 Maths Solutions is significant for the exam. Your NCERT Class 10 Maths book contains critical questions. This chapter contains both short-answer and long-answer questions. This is a foundational chapter, and the topics presented will be used in subsequent chapters. This chapter will help you improve your problem-solving abilities.

Q: Q3. What are the Key Topics in NCERT Solutions Class 10 Maths Chapter 1?

Answer: Euclid's Division Lemma and the Fundamental Theorem of Arithmetic are two major subtopics taught in NCERT Solutions Class 10 Maths Chapter 1. Students can utilize Euclid's Division Lemma to compute the HCF of two positive integers, and the Fundamental Theorem of Arithmetic to find the HCF and LCM of two positive integers. Students can practice questions relating to both areas to have a thorough comprehension of the material.

Q: Q4. How many exercises are in NCERT Solutions for Class 10 Maths Chapter 1?

Answer: Real Numbers is Chapter 1 of the Class 10 NCERT Maths textbook. This chapter contains a total of four exercises. Every workout has a unique PDF. The NCERT Solutions Class 10 Maths Chapter 1 Real numbers have a total of 18 well-researched problems. The 18 questions are divided into three categories: long replies, middle-level answers, and easy answers. These solutions will enable you to complete the NCERT syllabus and be prepared for the Class 10 Maths exam.

Question 1

Use Euclid's division algorithm to find the HCF of:
(i)

135

and

225

(ii)

196

and

38220

(iii)

867

and

255

.

Find the highest HCF among them.

Medium

Solution

Verified by Toppr

Answer

(i) By Euclid's division lemma,

225 = 135 \times 1 + 90

r = 90 \neq = 0

135 = 90 \times 1 + 45

r = 45 \neq = 0

90 = 45 \times 2 + 0

r = 0

So,H.C.F of

135

and

225

is

45

(ii) By Euclid's division lemma,

38220 = 196 \times 195 + 0

r = 0

So, H.C.F of

38220

and

196

is

196

(iii) By Euclid's division lemma,

867 = 255 \times 3 + 102

r = 102 \neq = 0

255 = 102 \times 2 + 51

r = 51 \neq = 0

102 = 51 \times 2 + 0

r = 0

So, H.C.F of

867

and

255

is

51

The highest HCF among the three is

196

.

Answer

(i) By Euclid's division lemma,

225 = 135 \times 1 + 90

r = 90 \neq = 0

135 = 90 \times 1 + 45

r = 45 \neq = 0

90 = 45 \times 2 + 0

r = 0

So,H.C.F of

135

and

225

is

45

(ii) By Euclid's division lemma,

38220 = 196 \times 195 + 0

r = 0

So, H.C.F of

38220

and

196

is

196

(iii) By Euclid's division lemma,

867 = 255 \times 3 + 102

r = 102 \neq = 0

255 = 102 \times 2 + 51

r = 51 \neq = 0

102 = 51 \times 2 + 0

r = 0

So, H.C.F of

867

and

255

is

51

The highest HCF among the three is

196

.

Question 2

Show that any positive odd integer is of the form

6 q + 1

, or

6 q + 3

, or

6 q + 5

, where

q

is some integer.

Hard

Solution

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Answer

Using Euclid division algorithm, we know that

a = b q + r,

0 \leq r \leq b

----(1)

Let

a

be any positive integer and

b = 6

.

Then, by Euclid’s algorithm,

a = 6 q + r

for some integer

q \geq 0

, and

r = 0, 1, 2, 3, 4, 5

,

o r

0 \leq r < 6

.

Therefore,

a = 6 q o r 6 q + 1 o r 6 q + 2 o r 6 q + 3 o r 6 q + 4 o r 6 q + 5

6 q + 0 : 6

is divisible by

2

, so it is an even number.

6 q + 1 : 6

is divisible by

2

, but

1

is not divisible by

2

so it is an odd number.

6 q + 2 : 6

is divisible by

2

, and

2

is divisible by

2

so it is an even number.

6 q + 3 : 6

is divisible by

2

, but

3

is not divisible by

2

so it is an odd number.

6 q + 4 : 6

is divisible by

2

, and

4

is divisible by

2

so it is an even number.

6 q + 5 : 6

is divisible by

2

, but

5

is not divisible by

2

so it is an odd number.

And therefore, any odd integer can be expressed in the form

6 q + 1 o r 6 q + 3 o r 6 q + 5

Answer

Using Euclid division algorithm, we know that

a = b q + r,

0 \leq r \leq b

----(1)

Let

a

be any positive integer and

b = 6

.

Then, by Euclid’s algorithm,

a = 6 q + r

for some integer

q \geq 0

, and

r = 0, 1, 2, 3, 4, 5

,

o r

0 \leq r < 6

.

Therefore,

a = 6 q o r 6 q + 1 o r 6 q + 2 o r 6 q + 3 o r 6 q + 4 o r 6 q + 5

6 q + 0 : 6

is divisible by

2

, so it is an even number.

6 q + 1 : 6

is divisible by

2

, but

1

is not divisible by

2

so it is an odd number.

6 q + 2 : 6

is divisible by

2

, and

2

is divisible by

2

so it is an even number.

6 q + 3 : 6

is divisible by

2

, but

3

is not divisible by

2

so it is an odd number.

6 q + 4 : 6

is divisible by

2

, and

4

is divisible by

2

so it is an even number.

6 q + 5 : 6

is divisible by

2

, but

5

is not divisible by

2

so it is an odd number.

And therefore, any odd integer can be expressed in the form

6 q + 1 o r 6 q + 3 o r 6 q + 5

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?

Hard

Solution

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Answer

HCF (

616, 32

) is the maximum number of columns in which they can march.

Step 1: First find which integer is larger.

616 > 32

Step 2: Then apply the Euclid's division algorithm to

616

and

32

to obtain

616 = 32 \times 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 \times 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.

Answer

HCF (

616, 32

) is the maximum number of columns in which they can march.

Step 1: First find which integer is larger.

616 > 32

Step 2: Then apply the Euclid's division algorithm to

616

and

32

to obtain

616 = 32 \times 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 \times 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.

Question 4

Use Euclid's division lemma to show that the square of any positive integer is either of the form

3 m

or

3 m + 1

for some integer m, but not of the form

3 m + 2

.

Hard

Solution

Verified by Toppr

Answer

Let

a

be the positive integer and

b = 3

.

We know

a = b q + r

,

0 \leq r < b

Now,

a = 3 q + r

,

0 \leq r < 3

The possibilities of remainder is

0, 1,

or

2

.

Case 1 : When

a = 3 q

a^{2} = (3 q)^{2} = 9 q^{2} = 3 q \times 3 q = 3 m

where

m = 3 q^{2}

Case 2 : When

a = 3 q + 1

a^{2} = (3 q + 1)^{2} = (3 q)^{2} + (2 \times 3 q \times 1) + (1)^{2} = 3 q (3 q + 2) + 1 = 3 m + 1

where

m = q (3 q + 2)

Case 3: When

a = 3 q + 2

a^{2} = (3 q + 2)^{2} = (3 q)^{2} + (2 \times 3 q \times 2) + (2)^{2} = 9 q^{2} + 12 q + 4 = 9 q^{2} + 12 q + 3 + 1 = 3 (3 q^{2} + 4 q + 1) + 1 = 3 m + 1

where

m = 3 q^{2} + 4 q + 1

Hence, from all the above cases, it is clear that square of any positive integer is of the form

3 m

or

3 m + 1

.

Answer

Let

a

be the positive integer and

b = 3

.

We know

a = b q + r

,

0 \leq r < b

Now,

a = 3 q + r

,

0 \leq r < 3

The possibilities of remainder is

0, 1,

or

2

.

Case 1 : When

a = 3 q

a^{2} = (3 q)^{2} = 9 q^{2} = 3 q \times 3 q = 3 m

where

m = 3 q^{2}

Case 2 : When

a = 3 q + 1

a^{2} = (3 q + 1)^{2} = (3 q)^{2} + (2 \times 3 q \times 1) + (1)^{2} = 3 q (3 q + 2) + 1 = 3 m + 1

where

m = q (3 q + 2)

Case 3: When

a = 3 q + 2

a^{2} = (3 q + 2)^{2} = (3 q)^{2} + (2 \times 3 q \times 2) + (2)^{2} = 9 q^{2} + 12 q + 4 = 9 q^{2} + 12 q + 3 + 1 = 3 (3 q^{2} + 4 q + 1) + 1 = 3 m + 1

where

m = 3 q^{2} + 4 q + 1

Hence, from all the above cases, it is clear that square of any positive integer is of the form

3 m

or

3 m + 1

.

Question 5

Use Euclid's division lemma to show that the cube of any positive integer is of the form

9 m, 9 m + 1

or

9 m + 8

Hard

Solution

Verified by Toppr

Answer

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x³ = (3q)³ = 27q³ = 9 (3q³) = 9m, where m = 3q³.

Case II : When x = 3q + 1

then, x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9 q (3q² + 3q + 1) + 1

= 9m + 1, where m = q (3q² + 3q + 1)

Case III. When x = 3q + 2

then, x³ = (3q + 2)³

= 27 q³ + 54q² + 36q + 8

= 9q (3q² + 6q + 4) + 8

= 9 m + 8, where m = q (3q² + 6q + 4)

Hence, x³is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Answer

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x³ = (3q)³ = 27q³ = 9 (3q³) = 9m, where m = 3q³.

Case II : When x = 3q + 1

then, x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9 q (3q² + 3q + 1) + 1

= 9m + 1, where m = q (3q² + 3q + 1)

Case III. When x = 3q + 2

then, x³ = (3q + 2)³

= 27 q³ + 54q² + 36q + 8

= 9q (3q² + 6q + 4) + 8

= 9 m + 8, where m = q (3q² + 6q + 4)

Hence, x³is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Question 6

Express each number as a product of its prime factors:

(i)

140

(ii)

156

(iii)

3825

(iv)

5005

(v)

7429

Medium

Solution

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Answer

(i) $140$

$140 = 2 \times 70$

$= 2 \times 2 \times 35$

$= 2 \times 2 \times 5 \times 7$

$= 2 \times 2 \times 5 \times 7 \times 1$

(ii) $156$

$156 = 2 \times 78$

$= 2 \times 2 \times 39$

$= 2 \times 2 \times 3 \times 13$

$= 2 \times 2 \times 3 \times 13 \times 1$

(iii) $3825$

$3825 = 3 \times 1275$

$= 3 \times 3 \times 425$

$= 3 \times 3 \times 5 \times 85$

$= 3 \times 3 \times 5 \times 5 \times 17$

$= 3 \times 3 \times 5 \times 5 \times 17 \times 1$

(iv) $5005$

$5005 = 5 \times 1001$

$= 5 \times 7 \times 143$

$= 5 \times 7 \times 11 \times 13$

$= 5 \times 7 \times 11 \times 13 \times 1$

(v) $7429$

$7429 = 17 \times 437$

$= 17 \times 19 \times 23$

$= 17 \times 19 \times 23 \times 1$

Answer

(i) $140$

$140 = 2 \times 70$

$= 2 \times 2 \times 35$

$= 2 \times 2 \times 5 \times 7$

$= 2 \times 2 \times 5 \times 7 \times 1$

(ii) $156$

$156 = 2 \times 78$

$= 2 \times 2 \times 39$

$= 2 \times 2 \times 3 \times 13$

$= 2 \times 2 \times 3 \times 13 \times 1$

(iii) $3825$

$3825 = 3 \times 1275$

$= 3 \times 3 \times 425$

$= 3 \times 3 \times 5 \times 85$

$= 3 \times 3 \times 5 \times 5 \times 17$

$= 3 \times 3 \times 5 \times 5 \times 17 \times 1$

(iv) $5005$

$5005 = 5 \times 1001$

$= 5 \times 7 \times 143$

$= 5 \times 7 \times 11 \times 13$

$= 5 \times 7 \times 11 \times 13 \times 1$

(v) $7429$

$7429 = 17 \times 437$

$= 17 \times 19 \times 23$

$= 17 \times 19 \times 23 \times 1$

Question 7

Find the LCM and HCF of the following pairs of integers and verify that LCM

\times

HCF

=

Product of the integers:
(i)

26

and

91

(ii)

510

and

92

(iii)

336

and

54

Easy

Solution

Verified by Toppr

Answer

1)

26 = 2 \times 13

and

91 = 1 \times 91

HCF = 1

and

LCM = 2 \times 13 \times 91 = 2366

verification:

1 \times 2366 = 26 \times 91 \Rightarrow 2366 = 2366

2)

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

and

92 = 2^{2} \times 23

HCF = 2

and

LCM = 2^{2} \times 3 \times 5 \times 17 \times 23 = 23460

verification:

2 \times 23460 = 92 \times 510 \Rightarrow 46920 = 46920

3)

54 = 2 \times 3^{3}

and

336 = 2^{4} \times 3 \times 7

HCF = 6

and

LCM = 2^{4} \times 3^{3} \times 7 = 3024

verification:

6 \times 3024 = 336 \times 54 \Rightarrow 18144 = 18144

Answer

1)

26 = 2 \times 13

and

91 = 1 \times 91

HCF = 1

and

LCM = 2 \times 13 \times 91 = 2366

verification:

1 \times 2366 = 26 \times 91 \Rightarrow 2366 = 2366

2)

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

and

92 = 2^{2} \times 23

HCF = 2

and

LCM = 2^{2} \times 3 \times 5 \times 17 \times 23 = 23460

verification:

2 \times 23460 = 92 \times 510 \Rightarrow 46920 = 46920

3)

54 = 2 \times 3^{3}

and

336 = 2^{4} \times 3 \times 7

HCF = 6

and

LCM = 2^{4} \times 3^{3} \times 7 = 3024

verification:

6 \times 3024 = 336 \times 54 \Rightarrow 18144 = 18144

Question 8

Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i)

12, 15

and

21

(ii)

17, 23

and

29

(iii)

8, 9

and

25

Medium

Solution

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Answer

Using prime factorisation method:

(i) 12, 15 and 21

Factor of

12 = 2 \times 2 \times 3

Factor of

15 = 3 \times 5

Factor of

21 = 3 \times 7

HCF

(12, 15, 21) = 3

LCM

(12, 15, 21) = 2 \times 2 \times 3 \times 5 \times 7 = 420

(ii) 17, 23 and 29

Factor of

17 = 1 \times 17

Factor of

23 = 1 \times 23

Factor of

29 = 1 \times 29

HCF

(17, 23, 29) = 1

LCM

(17, 23, 29) = 1 \times 17 \times 23 \times 29 = 11, 339

(iii) 8, 9 and 25

Factor of

8 = 2 \times 2 \times 2 \times 1

Factor of

9 = 3 \times 3 \times 1

Factor of

25 = 5 \times 5 \times 1

HCF

(8, 9, 25) = 1

LCM

(8, 9, 25) = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 1, 800

Answer

Using prime factorisation method:

(i) 12, 15 and 21

Factor of

12 = 2 \times 2 \times 3

Factor of

15 = 3 \times 5

Factor of

21 = 3 \times 7

HCF

(12, 15, 21) = 3

LCM

(12, 15, 21) = 2 \times 2 \times 3 \times 5 \times 7 = 420

(ii) 17, 23 and 29

Factor of

17 = 1 \times 17

Factor of

23 = 1 \times 23

Factor of

29 = 1 \times 29

HCF

(17, 23, 29) = 1

LCM

(17, 23, 29) = 1 \times 17 \times 23 \times 29 = 11, 339

(iii) 8, 9 and 25

Factor of

8 = 2 \times 2 \times 2 \times 1

Factor of

9 = 3 \times 3 \times 1

Factor of

25 = 5 \times 5 \times 1

HCF

(8, 9, 25) = 1

LCM

(8, 9, 25) = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 1, 800

Question 9

Given that HCF

(306

and

657) = 9

, find LCM of

(306

and

657)

Medium

Solution

Verified by Toppr

Answer

We know that,

LCM x HCF = Product of the two numbers

L C M = \frac{Product of two numbers}{H C F}

Product of the two numbers

= 306 \times 657 = 201042

LCM

= \frac{2 0 1 0 4 2}{9}

LCM

= 22, 338

Answer

We know that,

LCM x HCF = Product of the two numbers

L C M = \frac{Product of two numbers}{H C F}

Product of the two numbers

= 306 \times 657 = 201042

LCM

= \frac{2 0 1 0 4 2}{9}

LCM

= 22, 338

Question 10

Check whether

6^{n}

can end with the digit

0

for any natural number

n

.

Medium

Solution

Verified by Toppr

Answer

If any digit has the last digit

10

that means it divisible by

10 .

The factor of

10 = 2 \times 5,

So value of

6^{n}

should be divisible by

2

and

5 .

Both

6^{n}

is divisible by

2

but not divisible by

5 .

So, it can not end with

0 .

Answer

If any digit has the last digit

10

that means it divisible by

10 .

The factor of

10 = 2 \times 5,

So value of

6^{n}

should be divisible by

2

and

5 .

Both

6^{n}

is divisible by

2

but not divisible by

5 .

So, it can not end with

0 .

NCERT Solutions for Class 10 Maths Chapter 1 : Real Numbers

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NCERT Solutions for Class 10 Maths Chapter 1 : Real Numbers

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers – Brief Overview

Euclid’s Division Lemma

The Fundamental Theorem of Arithmetic

Rational and Irrational Numbers and their Decimal Expansions

Frequently Asked Questions on NCERT Class 10 Maths Chapter 1 : Real Numbers