CBSE Class 10 Mathematics Important Questions Chapter 1
Real Numbers

1 Marks Questions

1. Use Euclid ‘ s division lemma to show that the square of any positive integer is eit
   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.]

Ans. 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 k1, k2, 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.

2. Express each number as product of its prime factors

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Ans. (i) 140 = 2 ´ 2 ´ 5 ´ 7 = 22´ 5 ´ 7

(ii) 156 = 2 ´ 2 ´ 3 ´ 13 = 22´ 3 ´ 13

(iii) 3825 = 3 ´ 3 ´ 5 ´ 5 ´ 17 = 32´ 52´ 17

(iv) 5005 = 5 ´ 7 ´ 11 ´ 13

(v) 7429 = 17 ´ 19 ´ 23

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

Ans. HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

\ LCM × HCF =306 × 657

LCM = 22338

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

Ans. If any number ends with the digit 0, it should be divisible by 10

In other words, it will also be divisible by 2 and 5 as 10 = 2 Ã x 5

Prime factorisation of 6n = (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

5. Prove that (3+2 ) is irrational

Ans. We will prove this by contradiction

Let us suppose that (3+2 ) is rational.

It means that we have co-prime integers a and b (b≠0) such tha

a and b are integers.

It means L.H.S of (1) is rational but we know tha is irrational. It is not possible.

Therefore, our supposition is wrong. (3+2 ) cannot be rational.

Hence, (3+2 ) is irrational

6. 7 x 11 x 13 x 15 + 15+ is a

(a) Composite number
(b) Whole number
(c) Prime number
(d) None of these

Ans. (a)and(b)both

7. For what least value of ‘ n’ a natural number is divisible by 8?

(a) 0
(b) -1
(c) 1
(d) No value of ‘n’ is possible

Ans. (c) 1

8. The sum of a rational and an irrational number is

(a) Rational
(b) Irrational
(c) Both (a) & (c)
(d) Either (a) or (b)

Ans. (b) Irrational

9 .HCF of two numbers is 113, their LCM is 56952. If one number is 904, then other number is:

(a) 7719
(b) 7119
(c) 779
(d) 7911

Ans. (b) 7119

10 . A lemma is an axiom used for proving

(a) other statement
(b) no statement
(c) contradictory statement
(d) none of these

Ans. a) other statement

11. If HCF of two numbers is 1, the two numbers are called relatively or

(a) prime, co-prime
(b) composite, prime
(c) Both (a) and (b)
(d) None of these

Ans. (a) prime, co-prime

(a) a terminating decimal number
(b) a rational number
(c) an irrational number
(d) Both (a) and (b)

Ans. (b) a rational number

13. 2.13113111311113……is

(a) a rational number
(b) a non-terminating decimal number
(c) an irrational number
(d) Both (a) & (c)

Ans. (c) an irrational number

14. The smallest composite number is

(a) 1
(b) 2
(c) 3
(d) 4

Ans. (c) 3

(a) an integer
(b) an irrational number
(c) a rational number
(d) None of these

Ans. (c) a rational number

(a) a rational number
(b) an irrational number
(c) both (a) & (b)
(d) neither rational nor irrational

Ans. (b) an irrational number

(a) a rational number
(b) an irrational number
(c) an integer
(d) not real number

Ans. (b) an irrational number

2 Marks Questions

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

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

a = 6q + r for 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

2 .An army contingent of 616 members is to march behind an army band of 32 member 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 Ã X 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each

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 = 22 X 3
15 = 3 X 5
21 = 3 × 7

HCF = 3

LCM = 22 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

17 =1 × 17
23 = 1X 23
29 = 1 × 29

HCF = 1

LCM = 17 X 23 X 29 = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 = 23
9 = 3 × 3 = 32
25 = 5 × 5 = 52

HCF = 1

LCM = 23 X 32 X 52 = 1800

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 × 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 X (7 X 6 X 4 X 3 X 2 X 1 + 1)
= 5 X (1008 + 1)
= 5 X 1009

1009 cannot be factorized further

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

Hence, it is a composite number.

You Tube Link :- Class – 10 (Important Questions)