## CBSE Class 10 Maths Chapter-1 Real Numbers – Free PDF Download

Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 1 – Real Numbers prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.

CBSE Class 10 Maths Chapter-1 Real Numbers Important Questions

**Study without Internet (Offline)**

**CBSE Class 10 Maths Important Questions Chapter 1 – Real Numbers**

**2 Mark Questions**

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

**Ans .**Let

*a*be any positive integer and

*b*= 6. Then, by Euclid’s algorithm,

*a*= 6

*q*+

*r*for some integer

*q*

**≥**0, and

*r*= 0, 1, 2, 3, 4, 5 because 0

**≤**

*r*< 6.

Therefore,

*a*= 6

*q*or 6

*q*+ 1 or 6

*q*+ 2 or 6

*q +*3 or 6

*q*+ 4 or 6

*q*+ 5

Also, 6

*q*+ 1 = 2 × 3

*q*+ 1 = 2

*k*

_{1}+ 1, where

*k*

_{1}is a positive integer

6

*q*+ 3 = (6

*q*+ 2) + 1 = 2 (3

*q*+ 1) + 1 = 2

*k*

_{2}+ 1, where

*k*

_{2}is an integer

6

*q*+ 5 = (6

*q*+ 4) + 1 = 2 (3

*q*+ 2) + 1 = 2

*k*

_{3}+ 1, where

*k*

_{3}is an integer

Clearly, 6

*q*+ 1, 6

*q*+ 3, 6

*q*+ 5 are of the form 2

*k*+ 1, where

*k*is an integer.

Therefore, 6

*q*+ 1, 6

*q*+ 3, 6

*q*+ 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 6

*q*+ 1, or 6

*q*+ 3,

Or 6

*q*+ 5

**2. 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?**

**Ans.** We have to find the HCF(616, 32) to find the maximum number of columns in which they can march.

To find the HCF, we can use 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.

**3. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.**

**Ans.**Let

*a*be any positive integer and

*b*= 3

*a =*3

*q*+

*r*, where

*q*≥ 0 and 0 ≤

*r*< 3

a = 3q or 3q + 1 or 3q + 2

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

We have three cases.

**Case 1**: When

*a = 3q*,

Where

*m*is an integer such that

*m*=3q

^{3}

**Case 2**: When

*a*= 3

*q*+ 1,

*a*3

^{3 }= (*q +*1

*)*

^{3}*a*27

^{3}=*q*27

^{3 }+*q*9

^{2 }+*q +*1

*a*= 9(3

^{3}*q*3

^{3 }+*q*) + 1

^{2 }+ q*a*9

^{3 }=*m*+ 1

Where

*m*is an integer such that

*m*= (3

*q*3

^{3 }+*q*

^{2 }+ q)**Case 3**: When

*a*= 3

*q*+ 2,

*a*3

^{3 }= (*q +*2

*)*

^{3}*a*27

^{3}=*q*54

^{3 }+*q*36

^{2 }+*q +*8

*a*= 9(3

^{3}*q*6

^{3 }+*q*4

^{2 }+*q*) + 8

*a*9

^{3 }=*m*+ 8

Where

*m*is an integer such that

*m*= (3

*q*6

^{3 }+*q*4

^{2 }+*q)*

Therefore, the cube of any positive integer is of the form 9

*m*, 9

*m*+ 1, or 9

*m +*8.

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

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

**Ans. (i)** 26 and 91

26= 2 × 13

91 = 7 × 13

HCF = 13

LCM = 2 × 7 × 13 = 182

Product of two numbers 26 and 91 = 26 × 91 = 2366

HCF × LCM = 13 × 182 = 2366

Hence, product of two numbers = HCF × LCM

**(ii) **510 and 92

510= 2 × 3 × 5× 17

92 = 2 × 2 × 23

HCF = 2

LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 and 92 = 510 × 92 = 46920

HCF × LCM = 2 × 23460 = 46920

Hence, product of two numbers = HCF × LCM

**(iii) **336 and 54

336= 2 × 2 × 2× 2 × 3 × 7 = 2^{4} × 3 × 7

54 = 2 × 3 × 3 × 3 = 2 × 3^{3}

HCF = 2 × 3 = 6

LCM = 2^{4} × 3^{3} × 7 = 3024

Product of two numbers 336 and 54 = 336 × 54 = 18144

HCF × LCM = 6 × 3024 = 18144

Hence, product of two numbers = HCF × LCM

**5. 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**

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

12 = 2^{2} × 3

15 = 3 × 5

21 = 3 × 7

HCF = 3

LCM = 2^{2} × 3 × 5 × 7 = 420

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

17 =1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

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

8 = 2 × 2 × 2 = 2^{3}

9 = 3 × 3 = 3^{2}

25 = 5 × 5 = 5^{2}

HCF = 1

LCM = 2^{3} × 3^{2} × 5^{2} = 1800

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

**Ans.** 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 factorized further

Therefore, the given expression has 5 and 1009 as its factors.

Hence, it is a composite number.

**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?**

**Ans.** 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 **× **3 And, 12 = 2 **× **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.

**8. Prove that is irrational.**

**Ans.** Let us prove irrational by contradiction.

Let us suppose that is rational. It means that we have co-prime integers ** a and b **(

*b*≠0) such that

⇒

*b*=

*a*

Squaring both sides, we get

5

*b*

^{2}=

*a*

^{2}…

**(1)**

It means that 5 is factor of

*a*

^{2}

Hence,

**5 is also factor of**

**by Theorem. …**

*a***(2)**

If,

**5 is factor of**, it means that we can write

*a***for some integer**

*a*=5*c***.**

*c*Substituting value of

**in**

*a***(1)**,

5

*b*

^{2}=25

*c*

^{2}

⇒

*b*

^{2}=5

*c*

^{2}

It means that 5 is factor of

*b*

^{2}.

Hence,

**5 is also factor of**by Theorem. …

*b***(3)**

From

**(2)**and

**(3)**, we can say that

**5**is factor of both

**.**

*a*and*b*But,

**are**

*a*and*b***co-prime**.

Therefore, our assumption was wrong. cannot be rational. Hence, it is irrational.

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

**Ans. (i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

**(vi) **

**10. 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 factors of q?**

**(i) 43.123456789**

**(ii) 0.1201120012000120000…**

**(iii) **

**A ns. (i)** 43.123456789

It is rational because decimal expansion is terminating. Therefore, it can be expressed in form where factors of *q* are of the form 2* ^{n}* × 5

*where*

^{m}*n*and

*m*are non-negative integers.

**(ii)**0.1201120012000120000…

It is irrational because decimal expansion is neither terminating nor non-terminating repeating.

**(iii)**

It is rational because decimal expansion is non-terminating repeating. Therefore, it can be expressed in form where factors of

*q*

**are not**of the form 2

*× 5*

^{n}*where*

^{m}*n*and

*m*are non-negative integers.

**11. Show that every positive even integer is of the from 2q and that every positive odd integer is of the form 2q + 1 for some integer q.**

**Ans.** Let a = bq + r : b = 2

0 r < 2 i.e., r = 0, 1

a = 2q + 0, 2q + 1,

If a = 2q (which is even)

If a = 2q + 1 (which is odd)

So, every positive even integer is of the form 2q and odd integer is of the form 2q + 1.

**12. Show that any number of the form 4 ^{n}, n e N can never end with the digit 0.**

**Ans.**

It does not contains ‘5’. So

**13. Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.**

**Ans.**

HCF of 12576 and 4052 is ‘4’.

**14. Given that HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.**

**Ans.**

**15. Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.**

**Ans.** Let a = 4q + r : 0

**16. Show that any number of the form 6 ^{x}, xN can never end with the digit 0.**

**Ans.**

**17. Find HCF and LCM of 18 and 24 by the prime factorization method.**

**Ans.**

**18. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.**

**Ans.**

**19. Prove that the square of any positive integer of the form 5g + 1 is of the same form.**

**Ans.**

**20. Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.**

**Ans.**

**21. Find the largest number which divides 245 and 1029 leaving remainder 5 in each case. **

**Ans.** The required number is the HCF of (245 – 5) and (1029 – 5) i.e., 240 and 1024.

**22. A shopkeeper has 120 litres of petrol, 180 litres of diesel and 240 litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin?**

**Ans.** The required greatest capacity is the HCF of 120, 180 and 240