NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

Revise Euclid’s division lemma and other concepts with our NCERT Solutions for CBSE Class 10 Mathematics Chapter 1 Real Numbers. TopperLearning’s Maths experts provide accurate steps to apply Euclid’s algorithm to find HCF in Maths questions. In our chapter solutions, you will also find concept insights which will give you clarity on why a particular method was used to solve the problem.

For more CBSE Class 10 Maths questions and answers, you can explore our textbook solutions such as RD Sharma Solutions and RS Aggarwal Solutions. If you feel the need to discuss certain solutions with Maths experts, you can mention your precise question in our ‘UnDoubt’ section.

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Chapter 1 - Real Numbers Ex. 1.1

Solution 1

(i) 135 and 225

Step 1:  Since 225 > 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r   such that 225 = 135q+r, 0 Real Numbers - Less Than or Is Equal To Number Symbol r
On dividing 225 by 135 we get quotient as 1 and remainder as 90
i.e 225 = 135 x 1 + 90

Step 2: Remainder r which is 90 Real Numbers - Not Equal To a Number 0, we apply Euclid's division lemma to b =135 and r = 90 to find whole numbers q and r such that
135 = 90 x q + r,  0 Real Numbers - Less Than or Equal To Sign r<90
On dividing 135 by 90 we get quotient as 1 and remainder as 45
i.e 135 = 90 x 1 + 45

Step 3:  Again remainder r = 45 Real Numbers - Not Equal to Any Number 0 so we apply Euclid's division lemma to b =90 and r = 45 to find q and r   such that
90 = 90 x q + r,    0 Real Numbers - Less Than or Equal to any particular Number r<45
On dividing 90 by 45 we get quotient as 2 and remainder as 0
i.e 90 = 2 x 45 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).  

Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.

(ii)    196 and 38220
Step 1:  Since 38220 > 196, apply Euclid's division lemma
to a =38220 and b=196  to find  whole numbers q and r   such that
 38220 = 196 q + r, 0 Real Numbers - Less Than or Equal to any Symbol r < 196
On dividing 38220 we get quotient as 195 and remainder r as 0
i.e 38220 = 196 x 195 + 0
Since the remainder is zero, divisor at this stage will be HCF
Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.

NOTE: HCF( a,b) = a if a is a factor of  b. Here, 196 is a factor of  38220 so HCF is 196.
(iii)    867 and 255

Step 1: Since 867 > 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r   such that 867 = 255q + r, 0 Real Numbers - Less Than or Equal to any Symbol r<255
On dividing 867 by 255 we get quotient as 3 and remainder as 102
i.e 867 = 255 x 3 + 102

Step 2: Since remainder 102 Real Numbers - Not Equal To Sign 0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that
255 = 102q + r where 0 Real Numbers - Less Than or Is Equal To Sign r<102
On dividing 255 by 102 we get quotient as 2 and remainder as 51
i.e 255 = 102 x 2 + 51

Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51  to find whole numbers q and r such that
102 = 51 q + r where 0  r < 51
 
On dividing 102 by 51 quotient is 2 and remainder is 0
i.e 102 = 51 x 2 + 0
Since the remainder is zero, the divisor at this stage is the HCF
Since the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.

Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0 Real Numbers - Less than or Equal to a number r < b" in the correct order.

Here, a > b.

Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives  a number which divides a and b exactly.

i.e    HCF(a,b) =HCF(b,r)

Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

Solution 2

Let a be any odd positive integer we need to prove that a is of the form 6q + 1 , or  6q + 3 , or 6q + 5 , where q is some integer.

Since a is an integer consider b = 6 another integer applying Euclid's division lemma  we get
a = 6q + r  for some integer q Real Numbers - Greater Than or Is Equal To 0, and r = 0, 1, 2, 3, 4, 5  since
Real Numbers - Less Than or Is Equal To r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 +  1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.

Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer

Concept Insight:  In order to solve such problems  Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must  be of the form 6q + 1, 6q + 3, 6q + 5.

Basic definition of even and odd numbers and the fact that addition and , multiplication of  integers is always an integer are applicable here.

Solution 3

Maximum number of columns in which the Army contingent and the band can march will be given by HCF (616, 32)

We can use Euclid's algorithm to find the HCF.

Step 1: since 616 > 32 so applying Euclid's division lemma to a= 616 and b= 32 we get integers q and r as 32 and 19
i.e 616 = 32 x 19 + 8

Step 2: since remainder r =8 Real Numbers - Not Equal To 0 so again applying Euclid's  lemma to 32 and 8 we get  integers 4 and 0 as the quotient and remainder    
i.e 32 = 8 x 4 + 0
Step 3: Since remainder is zero so divisor at this stage will be the HCF

The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Concept Insight:  In order to solve the word problems first step is to interpret the problem and identify what is to be determined. The key word "Maximum" means we need to find the HCF.Do not forget to write the unit in the answer.

Solution 4

Let a be any positive integer we need to prove that a2 is of the form 3m or 3m + 1 for some integer m.

Let b = 3 be  the other integer so applying Euclid's division lemma  to a and b=3
We get a = 3q + r for some integer q Real Numbers - Greater Than or Is Equal To Symbol 0and r = 0, 1, 2
Therefore, a = 3q or 3q + 1 or 3q + 2
Now Consider a2
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.1 - Solution 4 - Euclid's Division Lemma

Where k1 = 3q2, k2 =3q2+2q   and k3 = 3q2+4q+1  since q ,2,3,1 etc are all integers so is their sum and product.
So k1 k2 k3 are all integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1 for any integer m.

Concept Insight: In order to solve such problems  Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 3 because a must  be of the form 3m or 3m + 1. Do not forget to take a2. Note that variable is just a notation and not the absolute value.

Solution 5

Let a be any positive integer and b = 3
a = 3q + r, where q Real Numbers - Greater Than or Is Equal To Sign 0 and 0Real Numbers - Less Than or Is Equal To Symbol r < 3

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.1 - Solution 5 - Euclid's Division Lemma
Therefore, every number can be represented as these three forms. There are three cases.

Case 1:    When a = 3q,

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.1 - Solution 5 - Euclid's Division Lemma - Case 1

Where m is an integer such that m =  3q3

Case 2:  When a = 3q + 1,
a3 = (3q +1)3
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1

Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3:  When a = 3q + 2,
a3 = (3q +2)3
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8

Where m is an integer such that m = (3q3 + 6q2 + 4q)

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

Concept Insight: In this problem Euclid's division lemma can be applied to integers a and b = 9 as well but using 9 will give us 9 values of r and hence as many cases so solution will be lengthy. Since every number which is divisible by 9 is also divisible by 3 so 3 is used.

Do not forget to take a3 and all the different values of a i.e
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.1 - Solution 5 - Euclid's Division Lemma Example

Chapter 1 - Real Numbers Ex. 1.2

Solution 1

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 1 - Product of Prime Factors

Concept Insight:  Since the given number needs to be expressed as the product of prime factors so in order to solve this problem knowing prime numbers is required. Do not forget to put the exponent in case a prime number is repeating.

Solution 2

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 2 - Example 1 - HCF and LCM

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 2 - Product of Two Numbers

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 2 - Product of Two Numbers - (2)

Concept Insight: HCF is the product of common prime factors raised to least power, while LCM is product of prime factors raised to highest power. HCF is always a factor of the LCM.

Do not skip verification product of two numbers = HCF x LCM as it can help in cross checking the answer.
 

Solution 3

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 3 - HCF and LCM

Concept Insight: HCF is the product of common prime factors of all three numbers  raised to least power, while LCM is product of prime factors of all here  raised to highest power.  Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c)  can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .

Solution 4

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.2 - Solution 4 -  Product of Two Numbers

Concept Insight: This problem must be solved using product of two numbers = HCF x LCM rather then prime factorisation

Solution 5
If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and  5 both
Prime factorisation of 6n = (2 x 3)n

By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.

Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.

Solution 6
Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself  whereas composite numbers have factors other than 1 and itself.

It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6

The given expression has 6 and 13  as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x  4 x 3 x 2 x 1 + 1)
 = 5 x (1008 + 1)
 = 5 x 1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.

Solution 7
It can be observed that Ravi and Sonia does not take same amount of time Ravi takes lesser time than Sonia for completing 1 round of the circular path.

As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
i.e When  Sonia completes one round then ravi completes 1.5 rounds.   So they will meet first time at the time which is a common multiple of the time taken by them to complete 1 round

i.e LCM of 18 minutes and 12 minutes.
Now
18 = 2 x 3 x 3 = 2 x 32
And, 12 = 2 x 2 x 3 = 22 x 3

LCM of 12 and 18 = product of factors raised to highest exponent =
22 x 32 = 36  

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Concept Insight: In order to solve the word problems first step is to interpret the problem and identify what is to be determined. The problem asks for simultaneous reoccurrence of events so we need to find LCM. The key word for simultaneous reoccurrence of events is LCM. Do not forget to write the final answer.

Chapter 1 - Real Numbers Ex. 1.3

Solution 1

Let us assume, on the contrary that Real Numbers - A number in Root Symbol is a rational number.

Therefore, we can find two integers a,b (b # 0) such that Real Numbers - A number in Root Sign = Real Numbers - Division of two Numbers

Where a and b are co-prime integers.


NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 1 - Co-Prime Integers


Therefore, a2 is divisible by 5 then a is also divisible by 5.

So a = 5k, for some  integer k.

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 1 - Co-Prime Integers - 2


This means that b2 is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 1 - Co-Prime Integers - Proof by Contradiction

Concept Insight: There are various ways of proving in mathematics proof by contradiction is one of them. In this approach we assume something which is contrary to what needs to be proved and arrive at a fact which contradicts something which is true in general. Key result used here is "If P is a prime number and it divides a2 then it divides a as well".

Solution 2

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 2 - Co-Prime Integers - Rational Number

Therefore, we can find two integers a, b (b Real Numbers - Not Equal To Symbol 0) such that

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 2 - Proof by Contradiction - Rational Number


Concept Insight: This problem is solved using proof by contradiction. The key concept used is if p is prime number then NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 2 - Prime Number - Rational Number is irrational. Do not prove this question by assuming sum of rational and irrational is irrational.

Solution 3

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 3 - Rational Number

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 3 - Rational and Irrational Number

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 3 - Example of Rational & Irrational Number

Concept Insight: This problem is solved using proof by contradiction. The key concept used is if p is prime number then NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.3 - Solution 3 - Prime Number - Rational Number - P is irrational.Do not prove this question by assuming sum or product of rational and irrational is irrational.

Chapter 1 - Real Numbers Ex. 1.4

Solution 1

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion

3125 = 55
The denominator is of the form 5m.
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion - 2

8 = 23
The denominator is of the form 2m.
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion - 3

455 = 5 x 7 x 13
Since the denominator is not in the form 2m x 5n, and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Number - Example 4

1600 = 26 52
The denominator is of the form 2m x 5n.
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion Terminating (1)
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Number - Example 5
343 = 73
Since the denominator is not in the form 2m x 5n, and it has 7 as its factor,
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion Non-Termination Repeating
The denominator is of the form 2m x 5n.
Hence, the decimal expansion of  NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion is Terminating is terminating.
(vii)
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion of Non-Terminating Recurring

Since the denominator is not of the form 2m  5n, and it also has 7 as its


NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion Non-Termination Repeating (2)
The denominator is of the form 5n.

Hence the decimal expansion of begin mathsize 12px style 6 over 15 end style is terminating.

(ix)50 = 2 x 5 x 5

The denominator is of the form 2m Cross Sign 5n.
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion Terminating
30 = 2 x 3 x 5
Since the denominator is not of the form 2m 5n, and it also has 3 as its factors,
NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion Non-Termination Repeating (3)

Concept Insight: The  concept used in this problem is that

The decimal expansion of rational number NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 1 - Decimal Expansion of Rational Number when Co-Prime Numbers where p and q are coprime numbers,

terminates if and only if the prime factorisation of q is of the form 2n5m, where n and m are non negative integers. Do not forget that 0 is also a non negative integer so n or m can take value 0.
Generally mistake is committed in identifying terminating decimals  when either of the two prime numbers  2 or 5 is appearing in the prime factorisation.

Solution 2

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 2 -  Express Rational No. in Decimal Form

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 2 - Express Rational Number in Decimal Form

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 2 - Express Rational Number in Decimals

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 2 - Convert Rational Number into Decimals

NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 2 - Convert Rational Number in Decimals

 




Concept Insight: This is  based on performing the long division and expressing the rational number in the decimal form learnt in lower classes.

Solution 3

(i)    43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form  NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 3 - Decimal Expansion of Two Rational Number and q is of the form 2m x 5n,

i.e., the prime factors of q will be either 2 or 5 or both.

(ii)    0.120120012000120000...
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.


NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 3 - Decimal Expansion is Non-Terminating Recurring

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 3 - Decimal Expansion of Two Rational Numberand q is not of the form 2m x 5n  i.e., the prime factors of q will also have a

factor other than 2 or 5.


Concept Insight: The  concept used in this problem is that, 

If the decimal expansion of rational number NCERT Solutions Class 10 Maths Chapter 1 - Real Numbers Exercise Ex 1.4 - Solution 3 - Decimal Expansion of Rational Numbers , [where p and q are coprime numbers] terminates, then  prime factorization of q is of the form 2n5m, where n and m are non negative integers.