**1 Marks Questions**

**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 Â´ 2 Â´ 5 Â´ 7 = 22Â´ 5 Â´ 7

(ii) 156 = 2 Â´ 2 Â´ 3 Â´ 13 = 22Â´ 3 Â´ 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 â‰¥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 â‰¤ 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)**

Chapter 3: – Pair of Linear Equations in Two Variables |

chapter 3:- EX – 3.3 |

Chapter 5:- Arithmetic Progressions |

Chapter 13:- Surface Areas and Volumes |

Chapter 13:- EX 13 -1 |

Chapter 13:- EX – 13.2 |

Chapter 13:- E.X 13.3 |

Chapter 14:- Statistics |