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**Ques 1: Consider the number 12 ^{n}, where n is a natural number. Check whether there is any value of n âˆˆ N for which 12^{n} ends with the digit zero or 5.** âˆµ Any number ending with the digit zero is always divisible by 5.

Sol:

âˆ´ If 12

â‡’ Prime factorisation of 12

Now,

12 = 2 Ã— 2 Ã— 3 = 2

â‡’ (12)

âˆ´ The prime factorisation of 12

â‡’ There is no value of n

Sol:

âˆ´ a and b are integers having no common factor other than 1 and b â‰ 0.

Now,

Squaring both sides, we have

â‡’ pb

Since pb

â‡’ a is also divisible by p ...(2)

Let a = pc for some integer c.

Substituting a = pc in (1), we have

pb

â‡’ pb

â‡’ b

pc

âˆ´ b

â‡’b is divisible by p ...(3)

From (2) and (3),

p is a common factor of â€˜aâ€™ and â€˜bâ€™. But this contradicts our assumption that a and b are co-prime.

âˆ´ Our assumption that

âˆ´ 18 = 2 Ã— 3 Ã— 3 = 2 Ã— 3

24 = 2 Ã— 2 Ã— 2 Ã— 3 = 2

HCF = Product of common prime factors with lowest powers.

â‡’ HCF (18, 24) = 3 Ã— 2 = 6

âˆ´ 10 = 2 Ã— 5 = 2^{1} Ã— 5^{1}

30 = 2 Ã— 3 Ã— 5 = 2^{1} Ã— 3^{1} Ã— 5^{1}

120 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 5 = 2^{3} Ã— 3^{1} Ã— 5^{1}

LCM = Product of each prime factor with highest powers

â‡’ LCM of 10, 30 and 120 = 2^{3} Ã— 3 Ã— 5 = 120.**Ques 5: Express **** as a rational number in the simplest form.****Sol:** Let x = 0.666 ..... ...(1)

âˆ´ 10x = 0.666 ..... Ã— 10

= 6.666 .......(2)

Subtracting (1) from (2), we have:

10x - x = 6.666 ..... - 0.666 .....

â‡’ 9x = 6

â‡’ x = 6/9 = 2/3

Thus,= 2/3

**Ques 6: Express **** as a rational number in the simplest form.****Sol:** Let x = 1.161616 ........(1)

âˆ´ 100x = (1.161616 .....) Ã— 100

â‡’ 100x = 116.161616 ........(2)

Subtracting (1) from (2), we have:

100x - x = [116.161616 .....] â€“ [1.161616 .....]

â‡’ 99x = 115

â‡’ x = 115/99

Thus, =115/99**Ques 7: Use Euclid's division algorithm to find HCF of 870 and 225.****Sol: **We have 870 = 3 Ã— 225 + 195

225 = 1 Ã— 195 + 30

195 = 6 Ã— 30 + 15

30 = 2 Ã— 15 + 0

âˆ´ HCF (870, 225) is 15.

Note:

LCM of two numbers = Product of the numbers

**Ques 8: Find the LCM and HCF of 1296 and 5040 by prime factorisation method:****Sol: **

and

âˆ´ 5040 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 7

= 2^{4} Ã— 3^{2} Ã— 5 Ã— 7 and

1296 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 2^{4} Ã— 3^{4}

âˆ´ LCM = Product of each prime factor with highest powers

= 2^{4} Ã— 3^{4} Ã— 5 Ã— 7

= 16 Ã— 81 Ã— 5 Ã— 7 = 45360

HCF = Product of common prime factors with lowest powers

= 2^{4} Ã— 3^{2}

= 16 Ã— 9 = 144**Ques 9: Prove that âˆš3 is irrational.****Sol:** Let **âˆš**3** **be rational in the simplest form of P/q.

i.e., p and q are integers having no common factor other than 1 and q â‰ 0.

Now, **âˆš**3 = p/q

Squaring both sides, we have**(âˆš**3)^{2} = (p/q)^{2}

â‡’ 3 = p^{2}/q^{2}

â‡’ 3q^{2} = p^{2 }............(1)

Since 3q^{2} is divisible by 3

âˆ´ p^{2} is also divisible by 3

â‡’ p is divisible by 3 ..........(2)

Let p = 3c for some integer â€˜câ€™.

Substituting p = 3c in (1), we have:

3q^{2 } = (3c)^{2}

â‡’ 3q^{2} = 9c^{2}

â‡’ q^{2} = 3c^{2}

3c^{2 }is divisible by 3

âˆ´ q^{2} is divisible by 3

â‡’ q is divisible by 3 ...(3)

From (2) and (3)

3 is a common factor of â€˜pâ€™ and â€˜qâ€™. But this contradicts our assumption that p and q are having no common factor other than 1.

âˆ´ Our assumption that **âˆš**3 is rational is wrong.

Thus, **âˆš**3 is an irrational.**Ques 10: Show that 3âˆš2 is irrational.Sol: **Let 3âˆš2 be a rational number

âˆ´ p/q = 3âˆš2 where p and q are prime to each other and q â‰ 0.

âˆ´ p/3q = âˆš2 ...(1)

Since, p is integer and 3q is also integer (3qâ‰ 0).

âˆ´ p/3q is a rational number.

From (1), âˆš2 is an integer

But this contradicts the fact that âˆš2 is irrational.

â‡’ 3âˆš2 is irrational.

âˆ´ It can be expressed as p/q where p and q are integers (prime to each other) such that q â‰ 0.

âˆ´ 2 - âˆš3 = p/q

â‡’ ...(1)

âˆµ p is an integer}

âˆ´ q is an integer}

â‡’ p/q is a rational number.

âˆ´ is a rational number. ...(2)

From (1) and (2), âˆš3 is a rational number. This contradicts the fact that âˆš3 is an irrational number.

âˆ´ Our assumption that (2 -âˆš3) is a rational number is not correct. Thus, (2 - âˆš3) is irrational.**Ques 12: Using Euclidâ€™s division algorithm, find the HCF of 56, 88 and 404. ****Sol: **Using Euclidâ€™s division algorithm to 88 and 56, we have

3q^{2} = (3c)^{2}

3q^{2} = (3c)^{2}

88 = 56 Ã— 1 + 32

56 = 32 Ã— 1 + 24

32 = 24 Ã— 1 + 8

24 = 8 Ã— 3 + 0

âˆ´ HCF of 88 and 56 is 8

Again, applying Euclidâ€™s division algorithm to 8 and 404,

we have:

404 = 8 Ã— 504 + 4

8 = 4 Ã— 2 + 0

âˆ´ HCF of 404 and 8 is 4

Thus, HCF of 88, 56 and 404 is 4.**Ques 13: Express **** in the decimal form.****Sol:** We have

[multiplying and dividing by 5]

**Ques 14: Express **** in the p/q form.****Sol:** Let x =

or x = 5.4178178178 .....

âˆ´ 10x = 54.178178178 ..... ...(1)

Also 1000 (10x)= 54178.178178178 .....

â‡’ 10000x = 54178.178178178 ..... ...(2)

Subtracting (1) from (2), we have:

10000x - 10x

= [54178.178178 .....] - [54.178178 .....]

â‡’ 9990x = 54124

â‡’

Thus, **Ques 15: State whether **** is a rational number or not.Sol:** = 1.23333 ..... is a non-terminating repeating decimal.

âˆ´ is a rational number.

3/4 is in the form of p/q , where q â‰ 0 [Here 4 â‰ 0]

âˆ´ 3/4 is a rational number.

Since, the sum of two rational numbers is a rational number.

Therefore, is a rational number.

Let the other number = x

Also LCM = 45 (HCF) ...(1)

And LCM + HCF =1150

â‡’ (45 HCF) + HCF = 1150

â‡’ 46 HCF = 1150

â‡’ HCF = 1150/46 = 25

From (1),

LCM = 45 Ã— 25

âˆ´ LCM Ã— HCF = (45 Ã— 25) Ã— 25

Now, LCM Ã— HCF = Product of the numbers

âˆ´ x Ã— 225 = (45 Ã— 25) Ã— 25

â‡’

= 125

Thus, the required number is 125.**Ques 17: Three different containers contain 496 litres, 806 litres and 713 litres of a mixture. What is the capacity of the biggest container that can measure all the different quantities exactly? Sol.**

âˆ´ 496 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 31

= 2

806 = 2 Ã— 13 Ã— 31

713 = 23 Ã— 31

â‡’ HCF = 31

Thus, the required measure of the biggest container = 31 litres.

âˆ´ 3+âˆš2 = a/b such that â€˜aâ€™ and â€˜bâ€™ are co-prime integers and b â‰ 0.

We have

Since a and b are integers,

âˆ´ is rational.

â‡’ âˆš2 is a rational. This contradicts the fact that âˆš2 is irrational.

âˆ´ Our assumption that 3+âˆš2 is rational is not correct.

â‡’ (3+âˆš2) is an irrational number.

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