CBSE Class 10 Maths NCERT Solutions for Chapter 1, Real Numbers

Check the CBSE Class 10 Maths NCERT Solutions for Chapter 1, Real Numbers. NCERT books offer a number of problems which students can practice to assess their learning and prepare for the examinations. Practicing a variety of questions helps to strengthen the fundamentals of a topic which makes students efficient in solving different questions asked in exams based on that particular topic. Therefore, students are suggested to solve all the NCERT questions and learn to write perfect solutions which will help them score good marks in exams.

For all the questions given in CBSE Class 10 Mathematics NCERT book we have collated detailed and accurate answers that will help students find the right approach to solve different questions.

Download the Class 10 Maths NCERT solutions for chapter 1, Real Numbers

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CBSE Class 10 Maths NCERT Problems & Solutions

Here, we are providing the NCERT solutions for Class 10 Mathematics chapter 1, Real Numbers. Our subject experts have reviewed these NCERT solutions to provide you the error free content which will make it easy for you to make an effective preparation for the annual exams.

Main topics discussed in Class 10 Mathematics chapter- Real Numbers are:

  • Basics concepts related to real numbers
  • Euclid’s Division Lemma,
  • Fundamental Theorem of Arithmetic
  • Revisiting irrational numbers
  • Revisiting rational numbers and their decimal expansions and
  • Students may download all the NCERT Solutions for CBSE Class 10 Mathematics chapter – Real Numbers, in the form of PDF.

Some of the questions and their solutions from NCERT Solutions for Class 10 Math’s: Real Numbers, are as follows:

  1. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Sol.

By Euclid’s algorithm,

a = 6q + r, and r = 0, 1, 2, 3, 4, 5

Hence, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

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 not exactly divisible by 2.

Hence, these numbers are odd numbers.

  1. 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?

Sol.

Euclid’s algorithm

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

To get the complete solution click on the following

Question 1:

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

ANSWER:

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii)196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii)867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51,

Therefore, HCF of 867 and 255 is 51.

NCERT  You Tube Solutions for Class 10 Maths Chapter 1, Real Numbers

Question 2:

Show that any positive odd integer is of the form , or , or , where q is some integer.

ANSWER:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 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 not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,

or 6q + 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?

ANSWER:

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

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

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

Question 4:

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

ANSWER:

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

Where k1k2, and k3 are some positive integers

Hence, it can be said that the square of any positive integer is either 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, 9+ 1 or 9m + 8.

ANSWER:

Let a be any positive integer and b = 3

a = 3q + r, where ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m = 

Case 2: When a = 3q + 1,

a= (3q +1)3

a3 = 27q27q9q + 1

a3 = 9(3q3q+ q) + 1

a9m + 1

Where m is an integer such that m = (3q3q+ q)

Case 3: When a = 3q + 2,

a= (3q +2)3

a3 = 27q54q36q + 8

a3 = 9(3q6q4q) + 8

a9m + 8

Where m is an integer such that m = (3q6q4q)

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

PAGE:11

Question 1:

Express each number as product of its prime factors:

ANSWER:

Question 2:

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

ANSWER:

Hence, product of two numbers = HCF × LCM

Hence, product of two numbers = HCF × LCM

Hence, product of two numbers = HCF × LCM

Question 3:

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

ANSWER:

Question 4:

Given that HCF (306, 657) = 9, find LCM (306, 657).

ANSWER:

Question 5:

Check whether 6n can end with the digit 0 for any natural number n.

ANSWER:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6= (2 ×3)n

It can be observed that 5 is not in the prime factorisation 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.

Question 6:

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

ANSWER:

Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)

= 13 × 78

= 13 ×13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

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

Question 7:

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:

It can be observed that 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. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

18 = 2 × × 3

And, 12 = 2 × × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

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

PAGE NO 14:

Question 1:

Prove that  is irrational.

ANSWER:

Let  is a rational number.

Therefore, we can find two integers ab (b ≠ 0) such that 

Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and bare co-prime.

Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

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.

Hence,cannot be expressed as or it can be said that is irrational.

Question 2:

Prove that  is irrational.

ANSWER:

Let is rational.

Therefore, we can find two integers ab (b ≠ 0) such that

Since a and b are integers, will also be rational and therefore,is rational.

This contradicts the fact that is irrational. Hence, our assumption that is rational is false. Therefore,  is irrational.

Question 3:

Prove that the following are irrationals:

ANSWER:

Let is rational.

Therefore, we can find two integers ab (b ≠ 0) such that

 is rational as a and b are integers.

Therefore, is rational which contradicts to the fact that  is irrational.

Hence, our assumption is false and is irrational.

Let  is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

 for some integers a and b

is rational as a and b are integers.

Therefore,  should be rational.

This contradicts the fact thatis irrational. Therefore, our assumption that is rational is false. Hence,  is irrational.

Let  be rational.

Therefore, we can find two integers ab (b ≠ 0) such that

Since a and b are integers,  is also rational and hence, should be rational. This contradicts the fact that is irrational. Therefore, our assumption is false and hence,  is irrational.

PAGE NO 17:

Question 1:

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

ANSWER:

(i) 

The denominator is of the form 5m.

Hence, the decimal expansion ofis terminating.

(ii) 

The denominator is of the form 2m.

Hence, the decimal expansion of is terminating.

(iii) 

455 = 5 × 7 × 13

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

(iv) 

1600 = 26 × 52

The denominator is of the form 2m × 5n.

Hence, the decimal expansion of  is terminating.

(v) 

Since the denominator is not in the form 2m × 5n, and it has 7 as its factor, the decimal expansion of  is non-terminating repeating.

(vi) 

The denominator is of the form 2m × 5n.

Hence, the decimal expansion of is terminating.

(vii) 

Since the denominator is not of the form 2m × 5n, and it also has 7 as its factor, the decimal expansion of  is non-terminating repeating.

(viii) 

The denominator is of the form 5n.

Hence, the decimal expansion of  is terminating.

(ix) 

The denominator is of the form 2m × 5n.

Hence, the decimal expansion of  is terminating.

(x) 

Since the denominator is not of the form 2m × 5n, and it also has 3 as its factors, the decimal expansion of  is non-terminating repeating.

PAGE NO 18:

Question 2:

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

ANSWER:

(viii)  

PAGE NO 18:

Question 3:

The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form  , what can you say about the prime factor of q?

(i) 43.123456789 (ii) 0.120120012000120000… (iii) 

ANSWER:

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form and q is of the form 

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.

(iii) 

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form  and q is not of the form  i.e., the prime factors of q will also have a factor other than 2 or 5.

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