**Important class 10 maths questions for chapter 1 real numbers** are provided here to help the students practice 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 practice 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 RD Sharma Solutions For 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.

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

**What are the 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.

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

Substitute, 3q2+2q = m, to get,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)2 + 22 + 2 × 3q × 2

= 9q2 + 4 + 12q

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

Again, substitute, 3q2 + 4q + 1 = m, to 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 = 22 × 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 × 52 × 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

Therefore,

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Hence, LCM(306,657) = 22338

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

Solution: Let 3 + 2√5 be a rational number.

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

3 + 2√5 = x/y

Rearranging, we get,

2√5 = (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 + 2√5 is a rational number is wrong.

Hence, 3 + 2√5 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

Factorizing 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

Factorizing 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

Factorizing 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

Factorizing the denominator, we get,

1600 = 26 × 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 6n = (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.

### 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