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**Q 1. Consider the number 12 ^{n}, where n is a natural number. Check whether there is any value of n ∈ N for which 12^{n} ends with the digit zero or 5.**

∵ Any number ending with the digit zero is always divisible by 5.

∴ If 12^{n}ends with the digit zero, then it must be divisible by 5.

⇒ Prime factorisation of 12^{n}must contain a prime factor 5.

Now,

12 = 2 × 2 × 3 = 2^{2}× 3

⇒ (12)^{n}= (2^{2}× 3)^{n}= 2^{2}^{n}× 3^{n}

∴ The prime factorisation of 12^{n}does not contain the prime factor 5.

⇒ There is no value of n∈N such that 12^{n}ends with the digit zero.

**Q 2. If ‘p’ is prime, prove that √p is irrational.**

Let

√p be rational in the simplest form a/b, where p is prime.

∴ a and b are integers having no common factor other than 1 and b ≠ 0.

⇒ Now,√p = a/b

⇒ Squaring both sides, we have

⇒ pb^{2 }= a^{2}...(1)

⇒ Since pb^{2}is divisible by p, a^{2}is also divisible by p.

⇒ a is also divisible by p ...(2)

Let a = pc for some integer c.

⇒ Substituting a = pc in (1), we have

pb^{2}= (pc)^{2}

⇒ pb^{2 }= p^{2}c^{2}

⇒ b^{2}= pc^{2}

pc^{2}is divisible by p,

∴ b^{2}is divisible by p

⇒b is divisible by p ...(3)

From (2) and (3),

⇒ p is a common factor of ‘a’ and ‘b’. But this contradicts our assumption that a and b are co-prime.

∴ Our assumption that√p is rational is wrong. Thus,√p is irrational if p is prime.

**Q 3. Find the HCF of 18 and 24 using prime factorisation.**

Using factor tree method, we have:

∴ 18 = 2 × 3 × 3 = 2 × 3^{2}

24 = 2 × 2 × 2 × 3 = 2^{3}× 3

HCF = Product of common prime factors with lowest powers.

⇒ HCF (18, 24) = 3 × 2 = 6

**Q 4. Find the LCM of 10, 30 and 120.**

∴ 10 = 2 × 5 = 2^{1}× 5^{1}

30 = 2 × 3 × 5 = 2^{1}× 3^{1}× 5^{1}

120 = 2 × 2 × 2 × 3 × 5 = 2^{3}× 3^{1}× 5^{1}

LCM = Product of each prime factor with highest powers

⇒ LCM of 10, 30 and 120 = 2^{3}× 3 × 5 = 120.

**Q 5. Express**** as a rational number in the simplest form.**

Let x = 0.666 ..... ...(1)

∴ 10x = 0.666 ..... × 10

= 6.666 .......(2)Subtracting (1) from (2), we have:

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

⇒ 9x = 6

⇒ x = 6/9 = 2/3

Thus,= 2/3

**Q 6. Express **** as a rational number in the simplest form.**

Let x = 1.161616 ........(1)

∴ 100x = (1.161616 .....) × 100

⇒ 100x = 116.161616 ........(2)

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

100x - x = [116.161616 .....] – [1.161616 .....]

⇒ 99x = 115

⇒ x = 115/99

Thus, =115/99

**Q 7. Use Euclid's division algorithm to find HCF of 870 and 225.**

We have 870 = 3 × 225 + 195

225 = 1 × 195 + 30

195 = 6 × 30 + 15

30 = 2 × 15 + 0

∴ HCF (870, 225) is 15.

Note:

LCM of two numbers = Product of the numbers

**Q 8. Find the LCM and HCF of 1296 and 5040 by prime factorisation method:**

and

∴ 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7

= 2^{4}× 3^{2}× 5 × 7 and

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2^{4}× 3^{4}

∴ LCM = Product of each prime factor with highest powers

= 2^{4}× 3^{4}× 5 × 7

= 16 × 81 × 5 × 7 = 45360

HCF = Product of common prime factors with lowest powers

= 2^{4}× 3^{2}

= 16 × 9 = 144

**Q 9. Prove that √3 is irrational.**

Let

√3be rational in the simplest form of P/q.

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

Now,√3 = p/q

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

⇒ 3 = p^{2}/q^{2}

⇒ 3q^{2}= p^{2 }............(1)

Since 3q^{2}is divisible by 3

∴ p^{2}is also divisible by 3

⇒ p is divisible by 3 ..........(2)

Let p = 3c for some integer ‘c’.

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

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

⇒ 3q^{2}= 9c^{2}

⇒ q^{2}= 3c^{2}

3c^{2 }is divisible by 3

∴ q^{2}is divisible by 3

⇒ q is divisible by 3 ...(3)

From (2) and (3)

3 is a common factor of ‘p’ and ‘q’. But this contradicts our assumption that p and q are having no common factor other than 1.

∴ Our assumption that√3 is rational is wrong.

Thus,√3 is an irrational.

**Q 10. Show that 3√2 is irrational.**

Let 3√2 be a rational number

∴ p/q = 3√2 where p and q are prime to each other and q ≠ 0.

∴ p/3q = √2 ...(1)

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

∴ p/3q is a rational number.

From (1), √2 is an integer

But this contradicts the fact that √2 is irrational.

⇒ 3√2 is irrational.

**Q 11. Show that 2 - √3 is an irrational number.**

Let 2 - √3

is rational.

∴ It can be expressed as p/q where p and q are integers (prime to each other) such that q ≠ 0.

∴ 2 - √3 = p/q

⇒ ...(1)∵ p is an integer}

∴ q is an integer}

⇒ p/q is a rational number.

∴ is a rational number. ...(2)

From (1) and (2), √3 is a rational number. This contradicts the fact that √3 is an irrational number.

∴ Our assumption that (2 -√3) is a rational number is not correct. Thus, (2 - √3) is irrational.

**Q 12. Using Euclid’s division algorithm, find the HCF of 56, 88 and 404. **

Using Euclid’s division algorithm to 88 and 56, we have

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

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

88 = 56 × 1 + 32

56 = 32 × 1 + 24

32 = 24 × 1 + 8

24 = 8 × 3 + 0

∴ HCF of 88 and 56 is 8

Again, applying Euclid’s division algorithm to 8 and 404, we have:

404 = 8 × 504 + 4

8 = 4 × 2 + 0

∴ HCF of 404 and 8 is 4

Thus, HCF of 88, 56 and 404 is 4.

**Q 13. Express **** in the decimal form.**

We have

[multiplying and dividing by 5]

**Q 14. Express **** in the p/q form.**

Let x =

or x = 5.4178178178 .....

∴ 10x = 54.178178178 ..... ...(1)

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

⇒ 10000x = 54178.178178178 ..... ...(2)

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

10000x - 10x

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

⇒ 9990x = 54124⇒

Thus,

**Q 15. State whether **** is a rational number or not.**

= 1.23333 ..... is a non-terminating repeating decimal.

∴ is a rational number.

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

∴ 3/4 is a rational number.

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

Therefore, is a rational number.

**Q 16. The LCM of two numbers is 45 times their HCF. If one of the numbers is 225 and sum of their LCM and HCF is 1150, find the other number.**

One of the numbers = 225

Let the other number = x

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

And LCM + HCF =1150

⇒ (45 HCF) + HCF = 1150

⇒ 46 HCF = 1150

⇒ HCF = 1150/46 = 25

From (1),

LCM = 45 × 25

∴ LCM × HCF = (45 × 25) × 25

Now, LCM × HCF = Product of the numbers

∴ x × 225 = (45 × 25) × 25

⇒= 125

Thus, the required number is 125.

**Q 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?**

∴ 496 = 2 × 2 × 2 × 2 × 31

= 2^{4}× 31

806 = 2 × 13 × 31

713 = 23 × 31

⇒ HCF = 31

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

**Q 18. Prove that 3+√2 is an irrational number. **

Let 3 + √2 is a rational number.

∴ 3 + √2 = a/b such that ‘a’ and ‘b’ are co-prime integers and b ≠ 0.

We have

Since a and b are integers,

∴ is rational.

⇒ √2 is a rational. This contradicts the fact that √2 is irrational.

∴ Our assumption that 3 + √2 is rational is not correct.

⇒ (3 + √2) is an irrational number.

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