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Class 10 maths chapter 1 Real Numbers deals with the foundation of numeric system. Students preparing for their Class 10 exams will be able to clear all their concepts on real numbers class 10 at the root level. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all of the major concepts in detail, allowing students to understand the ideas better.

NCERT solutions for class 10 maths chapter 1 concentrates on the essential concepts such as Euclid's division lemma, Prime Numbers, Composite Numbers, Fundamental Theorem of Arithmetic, HCF and LCM by Prime Factorization Method, and Irrational Numbers, etc.

Maths class 10 chapter 1 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing in-depth analysis of the subjects. The Class 10 Maths NCERT Solutions Chapter 1 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Exercise 1.1

Question 1

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

(i) $135$ and $225$ (ii) $196$ and $38220$ (iii) $867$ and $255$.

(i) $135$ and $225$ (ii) $196$ and $38220$ (iii) $867$ and $255$.

Find the highest HCF among them.

Solution

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$225=135×1+90$

$r=90=0$

$135=90×1+45$

$r=45=0$

$90=45×2+0$

$r=0$

So,H.C.F of $135$ and $225$ is $45$

(ii) By Euclid's division lemma,

$38220=196×195+0$

$r=0$

So, H.C.F of $38220$ and $196$ is $196$

(iii) By Euclid's division lemma,

$867=255×3+102$

$r=102=0$

$255=102×2+51$

$r=51=0$

$102=51×2+0$

$r=0$

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 $6q+1$, or $6q+3$, or $6q+5$, where $q$ is some integer.

Solution

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Using Euclid division algorithm, we know that $a=bq+r,$ $0≤r≤b$ ----(1)

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

Then, by Euclid’s algorithm, $a=6q+r$ for some integer $q≥0$, and $r=0,1,2,3,4,5$ ,$or$ $0≤r<6$.

Therefore, $a=6qor6q+1or6q+2or6q+3or6q+4or6q+5$

$6q+0:6$ is divisible by $2$, so it is an even number.

$6q+1:6$ is divisible by $2$, but $1$ is not divisible by $2$ so it is an odd number.

$6q+2:6$ is divisible by $2$, and $2$ is divisible by $2$ so it is an even number.

$6q+3:6$ is divisible by $2$, but $3$ is not divisible by $2$ so it is an odd number.

$6q+4:6$ is divisible by $2$, and $4$ is divisible by $2$ so it is an even number.

$6q+5:6$ 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 $6q+1or6q+3or6q+5$

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?

Solution

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

**Step 1**: First find which integer is larger.

**Step 2**: Then apply the Euclid's division algorithm to $616$ and $32$ to obtain

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

**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.

$616>32$

$616=32×19+8$

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

$32=8×4+0$

Since the remainder is zero, we cannot proceed further.

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 $3m$ or $3m+1$ for some integer m, but not of the form $3m+2$.

Solution

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Let $a$ be the positive integer and $b=3$.

We know $a=bq+r$, $0≤r<b$

Now, $a=3q+r$, $0≤r<3$

The possibilities of remainder is $0,1,$ or $2$.

Case 1 : When $a=3q$

$a_{2}=(3q)_{2}=9q_{2}=3q×3q=3m$ where $m=3q_{2}$

Case 2 : When $a=3q+1$

$a_{2}=(3q+1)_{2}=(3q)_{2}+(2×3q×1)+(1)_{2}=3q(3q+2)+1=3m+1$ where $m=q(3q+2)$

Case 3: When $a=3q+2$

$a_{2}=(3q+2)_{2}=(3q)_{2}+(2×3q×2)+(2)_{2}=9q_{2}+12q+4=9q_{2}+12q+3+1=3(3q_{2}+4q+1)+1=3m+1$

where $m=3q_{2}+4q+1$

Hence, from all the above cases, it is clear that square of any positive integer is of the form $3m$ or $3m+1$.

Question 5

Use Euclid's division lemma to show that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$

Solution

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

So, we have the following cases :

**Case I : **When *x *= 3*q*.

then, *x*^{3} = (3*q*)^{3} = 27*q*^{3} = 9 (3*q*^{3}) = 9*m*, where *m *= 3*q*^{3}.

**Case II : **When *x *= 3*q *+ 1

then, *x*^{3} = (3*q *+ 1)^{3}

= 27*q*^{3} + 27*q*^{2} + 9*q *+ 1

= 9 *q *(3*q*^{2} + 3*q *+ 1) + 1

= 9*m *+ 1, where *m *= *q *(3*q*^{2} + 3*q *+ 1)

**Case III. **When *x *= 3*q *+ 2

then, *x*^{3} = (3*q *+ 2)^{3}

= 27 *q*^{3} + 54*q*^{2} + 36*q *+ 8

= 9*q *(3*q*^{2} + 6*q *+ 4) + 8

= 9 *m *+ 8, where *m *= *q *(3*q*^{2} + 6*q *+ 4)

Hence, *x*^{3 }is either of the form 9 *m *or 9 *m *+ 1 or, 9 *m *+ 8.

Exercise 1.2

Question 1

Express each number as a product of its prime factors:

(i) $140$ (ii) $156$ (iii) $3825$ (iv) $5005$ (v) $7429$

Solution

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Question 2

Find the LCM and HCF of the following pairs of integers and verify that LCM $×$ HCF $=$ Product of the integers:

(i) $26$ and $91$ (ii) $510$ and $92$ (iii) $336$ and $54$

(i) $26$ and $91$ (ii) $510$ and $92$ (iii) $336$ and $54$

Solution

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1) $26=2×13$ and $91=1×91$

$HCF=1$ and $LCM=2×13×91=2366$

verification: $1×2366=26×91⇒2366=2366$

2) $510=2×3×5×17$ and $92=2_{2}×23$

$HCF=2$ and $LCM=2_{2}×3×5×17×23=23460$

verification: $2×23460=92×510⇒46920=46920$

3) $54=2×3_{3}$ and $336=2_{4}×3×7$

$HCF=6$ and $LCM=2_{4}×3_{3}×7=3024$

verification: $6×3024=336×54⇒18144=18144$

Question 3

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 $25$

(i) $12,15$ and $21$ (ii) $17,23$ and $29$ (iii) $8,9$ and $25$

Solution

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Using prime factorisation method:

**(i) 12, 15 and 21**

Factor of $12=2×2×3$

**(ii) 17, 23 and 29**

**(iii) 8, 9 and 25**

Factor of $8=2×2×2×1$

Factor of $15=3×5$

Factor of $21=3×7$

HCF** **$(12,15,21)=3$

LCM $(12,15,21)=2×2×3×5×7=420$

Factor of $17=1×17$

Factor of $23=1×23$

Factor of $29=1×29$

HCF** **$(17,23,29)=1$

LCM $(17,23,29)=1×17×23×29=11,339$

Factor of $9=3×3×1$

Factor of $25=5×5×1$

HCF $(8,9,25)=1$

LCM $(8,9,25)=2×2×2×3×3×5×5=1,800$

Question 4

Given that HCF $(306$ and $657)=9$, find LCM of $(306$ and $657)$

Solution

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We know that,

$LCM x HCF = Product of the two numbers$

$LCM=HCFProduct of two numbers $

Product of the two numbers $=306×657=201042$

LCM $=9201042 $

LCM $=22,338$

Question 5

Check whether $6_{n}$ can end with the digit $0$ for any natural number $n$.

Solution

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If any digit has the last digit $10$ that means it divisible by $10.$

The factor of $10=2×5,$

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 $0.$

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Practice more questions

Both rational and irrational numbers are included in real numbers. A rational number is a number that can be stated as a fraction of two integers, with the denominator not equal to zero. Irrational numbers are those that cannot be stated as fractions. Irrational numbers have decimal expansions that don't end and don't become periodic. Integers, decimals, and fractions are examples of rational numbers, whereas irrational numbers include root overs, pi (22/7), and so on. In a nutshell, real numbers are all numbers other than imaginary numbers.

Euclid's division algorithm is a method for calculating the Highest Common Factor (HCF) of two positive numbers. Euclid's Division Lemma states that:

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Using this theorem, we can express every natural number as a prime number multiplication, for example, 35 = 7 × 5, 253 = 11 x 23, etc.

Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2^{n}5^{m}, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2^{n}5^{m}, where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).

Related Chapters

- Chapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

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.

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.

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.

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.

Related Chapters

- Chapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability