Important Questions Class 10 Maths Chapter 1 Real Numbers

Important Class 10 maths questions for Chapter 1 real numbers are provided here to help the students practise better and score well in their CBSE Class 10 maths exam. These real numbers extra questions are as per NCERT book. It covers the complete syllabus and will help the students to develop confidence and problem-solving skills for their exams.

Students can practise solving these questions after they have covered the syllabus for this chapter. Solving different types of questions will help them to clear their doubts. Let us see some questions here with their solutions. Get the important questions of Maths Class 10 for all the chapters here. Solving these important questions will boost the confidence level of students. They will also understand the difficulty level of the questions expected to be asked from this chapter.

Students can also get access to Class 10 Maths Chapter 1 Real Numbers MCQs here.

Before we move ahead for important questions let us learn about real numbers.

What are real numbers?

Real numbers are a combination of rational and irrational numbers. They are the numbers upon which we easily perform mathematical operations. All the numbers which are not imaginary are real numbers. For example, 22, -11, 7.99, 3/2, π(3.14), √2, etc.

Also Check:

Class 10 Real Numbers Important Questions and Answers

Q.1: 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:

Let x be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

Therefore,

x = 3q, 3q + 1 and 3q + 2

As per the given question, if we take the square on both the sides, we get;

x2 = (3q)2 = 9q2 = 3.3q2

Let 3q2 = m

Therefore,

x2 = 3m ………………….(1)

x2 = (3q + 1)2

= (3q)2 + 12 + 2 × 3q × 1

= 9q2 + 1 + 6q

= 3(3q2 + 2q) + 1

Substituting 3q2+2q = m we get,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)+ 2+ 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substituting 3q+ 4q + 1 = m, we get,

x2 = 3m + 1…………………………… (3)

Hence, from eq. 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Q.2: Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2× 5 × 7

(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.

Hence, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

(iii) 3825
Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 5× 17

(iv) 5005
Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.

Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429
Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.

Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Q.3: Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

As we know that,

HCF × LCM = Product of the two given numbers

So,

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Therefore, LCM(306,657) = 22338

Q.4: Prove that 3 + 2√5 is irrational.

Solution:

Let 3 + 25 be a rational number.

Then the co-primes x and y of the given rational number where (y ≠ 0) is such that:

3 + 25 = x/y

Rearranging, we get,

25 = (x/y) – 3

5 = 1/2[(x/y) – 3]

Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.

Therefore, 5 is also a rational number. But this confronts the fact that 5 is irrational.

Thus, our assumption that 3 + 25 is a rational number is wrong.

Hence, 3 + 25 is irrational.

Q.5: 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:
(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 

Solution:

Note: If the denominator has only factors of 2 and 5 or in the form of 2m × 5n then it has a terminating decimal expansion.
If the denominator has factors other than 2 and 5 then it has a non-terminating repeating decimal expansion.

(i) 13/3125

Factoring the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Or

= 20 × 55

Since the denominator is of the form 2m × 5n then, 13/3125 has a terminating decimal expansion.

(ii) 17/8

Factoring the denominator, we get,

8 = 2× 2 × 2 = 23

Or

= = 23 × 50

Since the denominator is of the form 2m × 5n then, 17/8 has a terminating decimal expansion.

(iii) 64/455

Factoring the denominator, we get,

455 = 5 × 7 × 13

Since the denominator is not in the form of 2m × 5n, therefore 64/455 has a non-terminating repeating decimal expansion.

(iv) 15/1600

Factoring the denominator, we get,

1600 = 2× 52

Since the denominator is in the form of 2m × 5n, 15/1600 has a terminating decimal expansion.

Q.6: 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, p/q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Solution:

(i) 43.123456789
Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

(ii) 0.120120012000120000. . .
Since it has non-terminating and non- repeating decimal expansion, it is an irrational number.

Q.7: Check whether 6n can end with the digit 0 for any natural number n.

Solution:

If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.

Prime factorization of 6= (2 × 3)n

Therefore, the prime factorization of 6n doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

Q.8: What is the HCF of the smallest prime number and the smallest composite number?

Solution:

The smallest prime number = 2

The smallest composite number = 4

Prime factorisation of 2 = 2

Prime factorisation of 4 = 2 × 2

HCF(2, 4) = 2

Therefore, the HCF of the smallest prime number and the smallest composite number is 2.

Q.9: Using Euclid’s Algorithm, find the HCF of 2048 and 960.

Solution:

2048 > 960

Using Euclid’s division algorithm,

2048 = 960 × 2 + 128

960 = 128 × 7 + 64

128 = 64 × 2 + 0

Therefore, the HCF of 2048 and 960 is 64.

Q.10: Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

Solution:

Prime factorisation of 404 = 2 × 2 × 101

Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

HCF = 2 × 2 = 4

LCM = 25 × 3 × 101 = 9696

HCF × LCM = 4 × 9696 = 38784

Product of the given two numbers = 404 × 96 = 38784

Hence, verified that LCM × HCF = Product of the given two numbers

Extra questions for Class 10 Maths Chapter 1

Q.1: Find three rational numbers lying between 0 and 0.1. Find twenty rational numbers between 0 and 0.1. Give a method to determine any number of rational numbers between 0 and 0.1.

Q.2: Which of the following rational numbers have the terminating decimal representation?

(i) 3/5

(ii) 7/20

(iii) 2/13

(iv) 27/40

(v) 133/125

(vi) 23/7

Q.3: Write the following rational numbers in decimal form:

(i) 42/100

(ii) 27/8

(iii) 1/5

(iv) 2/13

(v) 327/500

(vi) 5/6

(vii) 1/7

(viii) 11/17

Q.4: If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number.

Q.5: Show that 3√6 and 3√3 are not rational numbers.

Q.6: Show that 2 + √2 is not a rational number.

Q.7: Give an example to show that the product of a rational number and an irrational number may be a rational number.

Q.8: Prove that √3 – √2 and √3 + √5 are irrational.

Q.9: Express 7/64, 12/125 and 451/13 in decimal form.

Q.10: Find two irrational numbers lying between √2 and √3.

Q.11: Mention whether the following numbers are rational or irrational:

(i) (√2 + 2)

(ii) (2 – √2) x (2 + √2)

(iii) (√2 + √3)2

(iv) 6/3√2

Q.12: Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.

Q.13: Write the smallest number which is divisible by both 360 and 657.

Q.14: Show that the square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q.

Q.15: Prove that one of every three consecutive positive integers is divisible by 3.

6 Comments

  1. Can you give questions related to ntse on this chapter

  2. I want NTSE questions

  3. I want that chapter related hard questions for doing with enjoy

  4. I want some more questions of this chapter – Real Numbers.

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