# Question-1

**Use Euclidâ€™s division algorithm to find the HCF of:**

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Solution:**

(i) We start with the larger number 225

By Euclid Division Algorithm, we have

225 = 1 Ã— 135 + 90

135 = 1 Ã— 90 + 45 = 2 Ã— 45 + 0

HCF(225, 135) = HCF(135, 90) = HCF(90, 45) = 45.

Therefore the HCF of 135 and 225 is 45.

(ii) We start with the larger number 38220.

By Euclid Division Algorithm, we have

38220 = 196 Ã— 195 + 0

We apply Euclid Division Algorithm of Divisor 196 and the remainder 0.

Therefore 196 = 196 Ã— 1 + 0

Therefore HCF(38220, 196) = 196.

(iii) Using Euclid Division Algorithm,

867 = 255 Ã— 3 + 102

Applying Euclid Division Algorithm on the divisor 225 and the remainder 102,

We have

255 = 102 Ã— 2 + 51

Again, applying Euclid Division Algorithm on the Divisor 102 and the number 51.

102 = 51 Ã— 2 + 0

Therefore HCF(867, 255) = HCF(255, 102) = HCF(102, 51) = 51.

# Question-2

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

**Solution:**

Using Euclid division Algorithm, we have

a = bq + r { r â‰¤ 0 < b} .. (1)

Substituting b = 6 in equation (1)

a = 6q + r where r â‰¤ 0 < 6 â‡’ r = 0, 1, 2, 3, 4, 5

If r = 0, a = 6q, 6q is divisible by 6 â‡’ 6q is even.

If r = 1, a = 6q + 1, 6q + 1 is not divisible by 2.

If r = 2, a = 6q + 2, 6q + 2 is divisible by 2 â‡’ 6q + 2 is even.

If r = 3, a = 6q + 3, 6q + 3 is not divisible by 2.

If r = 4, a = 6q + 4, 6q + 4 is divisible by 2 â‡’ 6q + 4 is even.

If r = 5, a = 6q + 5, 6q + 5 is not divisible by 2

Therefore the numbers 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5

are either even or odd.

But 6q, 6q + 2, 6q + 4 are even.

â‡’ The remaining terms i.e.** **6q + 1, 6q + 3 are odd.

# Question-3

**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?**

**Solution:**

For the above problem, the maximum number of columns would be the HCF of 616 and 32.

We can find the HCF of 616 and 32 by using Euclid Division algorithm.

Therefore,

616 = 19 Ã— 32 + 8

32 = 4 Ã— 8 + 0

8 = 8 Ã— 1 + 0

Therefore HCF(616, 32) = HCF of (32, 8) = 8.

Therefore the maximum number of columns in which they can march is 8.

# Question-4

**Use Euclidâ€™s division Lemma to show that the square of any positive inter is either of the form 3m, 3m + 1 for some integer m.****Solution:**

According to Euclid algorithm we have a = bq + r .......... (1)

and on substituting b = 3 in (1), we get

a = 3q + r, {r â‰¤ 0 < 3 â‡’ r = 0, 1, 2}

when r = 0, a = 3q or a

^{2}= 9q

^{2}- ( A )

when r = 1, a = 3q + 1 or a

^{2}= 9q

^{2}+ 1 + 6q - - - ( B )

and when r = 2, a = 3q + 2 or a

^{2}= 9q

^{2}+ 4 + 12q - - - ( C )

We can rewrite equation (A) as a

^{2 }= 3(3q

^{2}) say 3m where, m = 3q

^{2}

Also (B) can be rewritten as a

^{2}= 3(3q

^{2}+ 2q) + 1 or a

^{2}= 3m + 1 where, m = 3q

^{2}+ 2q

Also (C) can be written as a

^{2 }= 3(3q

^{2}+ 4q + 1) + 1 or a

^{2}= 3m + 1 where, m = 3q

^{2}+ 4q + 1

Hence, the square of any positive integer is either of the form 3m or 3m+1 for some integer m.

# Question-5

**Use Euclidâ€™s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m + 8.****Solution:**

We know that by using Euclidâ€™s Division Algorithm, a = bq + r.

Substituting b = 9, we get a = 9q + r where, {r â‰¤ 0 < 9 â‡’ r = 0, 1, 2â€¦..8}

When r = 0, a = 9q.

a

^{3 }= 729q

^{3}= 9(81q

^{3}) = 9m where, m = 81q

^{3}.

If r = 1, a = 9q + 1.

â‡’ a

^{3}= 729q

^{3}+ 243q

^{2}+ 27q + 1

= 9(81q

^{3}+ 27q

^{2}+ 3q) + 1

= 9m + 1 where, m = 81q

^{3}+ 27q

^{2}+ 3q

If r = 4, a = 9q + 4

â‡’ a

^{3}= 729q

^{3}+ 972q

^{2}+ 432q + 64 = 9(81q

^{3}+ 108q

^{2}+ 48q + 7) + 1

Continuing the process till r = 8 and a = 9q + 8, we get

a

^{3 }= 729q

^{3}+ 1944q

^{2}+ 1728q + 512

= 9(81q

^{3}+ 216q

^{2}+ 192q + 56) + 8

= 9m + 8 where, m = 81q

^{3}+ 216q

^{2}+ 192q + 56.

Hence, it is proved that any positive integer is either of the form 9m, 9m + 1 or 9m + 8.

# Question-6

**Express each number as a product of its prime factors :****(i) 140**

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

**Solution:**

**(i) 140**

140 = 2 Ã— 70

= 2 Ã— 2 Ã— 35

= 2 Ã— 2 Ã— 5 Ã— 7

= 2 Ã— 2 Ã— 5 Ã— 7 Â´

**1**.

**(ii) 156**

156 = 2 Ã— 78

= 2 Ã— 2 Ã— 39

= 2 Ã— 2 Ã— 13 Ã— 3

= 2 Ã— 2 Ã— 13 Ã— 3 Ã— 1.

**(iii) 3825**

3825 = 3 Ã— 1275

= 3 Ã— 3 Ã— 425

= 3 Ã— 3 Ã— 5 Ã— 85

= 3 Ã— 3 Ã— 5 Ã— 5 Ã— 17

= 3 Ã— 3 Ã— 5 Ã— 5 Ã— 17 Ã— 1.

**(iv) 5005**

5005 = 5 Ã— 1001

= 5 Ã— 7 Ã— 143

= 5 Ã— 7 Ã— 11 Ã— 13

= 5 Ã— 7 Ã— 11 Ã— 13 Ã— 1.

**(v) 7429**

7429 = 17 Ã— 437

= 17 Ã— 19 Ã— 23

= 17 Ã— 19 Ã— 23 Ã— 1.

# Question-7

**Find the LCM and HCF of the following pairs of integers and verify that LCM****Ã—****HCF****=****Product of the two numbers.****(i) 26 and 91**

(ii) 510 and 92

(iii) 336 and 54(ii) 510 and 92

(iii) 336 and 54

**Solution:**

**(i) 26 and 91**

26 = 2 Ã— 13 Ã— 1 (Expressing as product of prime factors)

91 = 7 Ã— 13 Ã— 1 (Expressing as product of prime factors)

Therefore HCF of 26 and 91 = 13 Ã— 1 = 13

And LCM of 26 and 91 = 2 Ã— 7 Ã— 13 Ã— 1 = 182.

**Verification:**

LCM Ã— HCF = 13 Ã— 182 = 2366.

Product of 26 Ã— 91 = 2366.

Therefore it is proved that LCM Ã— HCF = Product of the two numbers.

**(ii) 510 and 92**

510 = 2 Ã— 255

= 2 Ã— 3 Ã— 85

= 2 Ã— 3 Ã— 5 Ã— 17

= 2 Ã— 3 Ã— 5 Ã— 17 Ã— 1

Therefore 510 = 2 Ã— 3 Ã— 5 Ã— 17 Ã— 1 ..........(A)

92 = 2 Ã— 46

= 2 Ã— 2 Ã— 23

Hence 92 = 2 Ã— 2 Ã— 23 ..........(B)

From (A) and (B) HCF of 510 and 92 is = 2.

and their LCM is = 2 Ã— 2 Ã— 3 Ã— 5 Ã— 17 Ã— 23 = 23,460.

Product of the LCM and HCF = 2 Ã— 23, 460 = 46, 920.

Product of the two numbers = 510 Ã— 92 = 46, 920.

Therefore it is proved that LCM Ã— HCF = Product of the two numbers.

**(iii) 336 and 54**

336 = 2 Ã—168

= 2 Ã— 2 Ã— 84

= 2 Ã— 2 Ã— 2 Ã— 42

= 2 Ã— 2 Ã— 2 Ã— 2 Ã— 21

= 2 Ã— 2 Ã— 2 Ã— 2 Ã— 7 Ã— 3 Ã— 1

Therefore 336 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 7 Ã— 3. ..........(A)

54 = 2 Ã— 27

= 2 Ã— 3 Ã— 9

= 2 Ã— 3 Ã— 3 Ã— 3

Therefore 54 = 2 Ã— 3 Ã— 3 Ã— 3 ..........(B)

From (A) and (B) HCF of 336 and 54 = 2 Ã— 3 = 6

LCM of 336 and 54 = 2 Ã— 3 Ã— 2 Ã— 2 Ã— 2 Ã— 7 Ã— 3 Ã— 3 = 2

^{4}Ã— 3

^{3}Ã— 7 = 3024

Product of 336 and 54 = 18, 144.

Product of LCM and HCF = 6 Ã— 3024 = 18, 144

Therefore it is proved that LCM Ã— HCF = Product of the two numbers.

# Question-8

**Find the LCM and HCF of the following integers by applying the prime factorization method.**

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

(ii) 17, 23 and 29

(iii) 8, 9 and 25.

(ii) 17, 23 and 29

(iii) 8, 9 and 25.

**Solution:**

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

12 = 2 Ã— 2 Ã— 3

15 = 5 Ã— 3

21 = 7 Ã— 3

From the above, HCF(12, 15, 21) = 3 and LCM(12, 15, 21) = 420

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

17 = 1 Ã— 17

23 = 1 Ã— 23

29 = 1 Ã— 29

From the above, HCF(17, 23, 29) = 1 and LCM(17, 23, 29) = 11339

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

8 = 2^{3} Ã— 1

9 = 3^{2} Ã— 1

25 = 5^{2} Ã— 1

From the above HCF(8, 9, 25) = 1 and LCM(8, 9, 25) = 1800.

# Question-9

**Given that HCF(306, 657) = 9. Find LCM(306, 657).**

**Solution:**

By the property that LCM Ã— HCF = Product of the two numbers

LCM Ã— 9 = 306 Ã— 657

LCM = ** **

Therefore LCM(306, 657) = 22338.

# Question-10

**Check whether 6**

^{n}can end with the digit 0 for any natural number n.**Solution:**

If the number 6

^{n}ends with the digit zero, then it is divisible by 5. Therefore the prime factorization of 6

^{n}contains the prime 5. This is not possible because the only prime in the factorisation of 6

^{n}is 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the factorization of 6

^{n}.

Hence it is very clear that there is no value of n in natural numbers for which 6^{n} ends with the digit zero.

# Question-11

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

**Solution:**

We have,

7 Ã— 11 Ã— 13 + 13 = 1001 + 13 = 1014

1014 = 2 Ã— 3 Ã— 13 Ã— 13.

â‡’ It is the product of prime factors

2 Ã— 3 Ã— 13 Ã— 13.

Hence it is a composite number.

7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5 = 2520 + 5 = 2525

and 2525 = 5 Ã— 5 Ã— 101

It is the product of prime factors 5 Ã— 5 Ã— 101.

Hence it is a composite number.

# Question-12

**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?****Solution:**

Since they both move in the same direction and at the same time, to find the time when will they both meet together again, it is enough if we find the LCM of 12 and 18.

12 = 2

^{2}Ã— 3 Ã— 1

18 = 3

^{2}Ã— 2 Ã— 1

LCM(18, 12) = 2 Ã— 3 Ã— 3 Ã— 2 Ã— 1 = 36

Therefore Sonia and Ravi will meet again after 36 minutes.

# Question-13

**Prove that is irrational****Solution:**

Let us assume, to the contrary, that irrational.

That is, we can find integers a and b(â‰ 0) such that .

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are co-prime.

Therefore b = a.

Squaring on both sides 5b

^{2}= a

^{2}..........(1)

The above implies that a

^{2}is divisible by 5 and also a is divisible by 5.

Therefore we can write that a = 5f for some integer f.

Substituting in (1) 5b

^{2}= (5f)

^{2}

5b

^{2}= 25f

^{2}(or) b

^{2}= 5f

^{2}

â‡’ b

^{2}is divisible by 5 which means b is also divisible by 5.

Therefore a and b have 5 as a common factor.

This contradicts the fact that a and b are co prime.

We arrived at the contradictory statement as above since our assumption is not correct. Hence, we can conclude that is irrational.

# Question-14

**Prove that 3 + 2 is irrational.****Solution:**

If possible let a = 3 + 2

**be a rational number.**

Squaring a

^{2}= (3 + 2)

^{2}

a

^{2}= 29 + 12

**= -------**(1)

Since a is a rational number the expression is also rational number.

â‡’

**is a rational number.**

This is a contradiction. Hence, 3 + 2

**is irrational.**

Hence proved.

# Question-15

**Prove that the following are irrationals:****(i)**

(ii)(ii)

**(iii) 6 +****Solution:**

**(i)**

Ã—=

Let a = be a rational number.

â‡’ 2a =

2a is a rational number since product of two rational number is a rational number.

Which will imply that is a rational number. But it is a contradiction since is an irrational number.

Therefore 2a is irrational or a is irrational.

Therefore is irrational. Hence proved.

**(ii)**

Let a =

**be a rational number.**

â‡’ .

Now, is a rational number since product of two rational number is a rational number.

The above will imply that

**is a rational number. But**

**is an irrational number.**

This contradicts our assumption. Therefore we can conclude that is an irrational number and hence the result.

**(iii) 6 +**

If possible let a = 6 +

**be a rational number.**

Squaring a

^{2}= (6 + )

^{2}

a

^{2}= 38 + 12

**= -**(1)

Since a is a rational number the expression is also rational number.

â‡’

**is a rational number.**

This is a contradiction. Hence, 6 +

**is irrational.**

Hence proved.

# Question-16

**Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:****(i) (ii) (iii) (iv) (v)**

(vi)(vi)

**(vii)****(viii)****(ix)****(x)****Solution:**

**(i)**

**=**

Since the denominator has only 5 as its factor, it is a terminating decimal expansion.

**(ii)**

**=**

Since the denominator has only 2 as its factor, it is a terminating decimal expansion.

**(iii)**

**=**

Since the denominator has factors other than 2 and 5 it is a non terminating decimal.

**(iv)**

**=**

**=**

Since the denominator has only 2 and 5 as its factors it is a terminating decimal.

**(v)**

**=**

Since the denominator has factors other than 2 and 5 it is a non-terminating decimal.

**(vi)**

Since the denominator has only 2 and 5 as its factors it is a terminating decimal.

**(vii)**

Since the denominator has factors other than 2 and 5 it is a non-terminating decimal.

**(viii)**

**=**

Since the denominator has only 5 as its factor it is a terminating decimal.

**(ix)**

**=**

**=**

Since the denominator has only 2 and 5 as its factors it is a terminating decimal.

**(x)**

**=**

=

=

Since the denominator has factors other than 2 and 5 it is a non-terminating decimal.

# Question-17

**Write down the decimal expansion of those rational numbers, which have terminating decimal expansion.**

**(a)**

(b)

(c)

(d)

(e)

(f)

(b)

(c)

(d)

(e)

(f)

**Solution:**

The rational numbers,which has the terminating decimals, are

(a) ** **(b) ** **(c) ** **(d) ** **(e)** **(f)** **

(a) ** = **0.00416

(b) ** = **2.125

(c) ** = = **0.009375

(d) **= = **0.115

(e) **= = **0.4

(f) **= = **0.70.

# Question-18

**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) 0.120120012000120000â€¦

(iii) 43.123456789(ii) 0.120120012000120000â€¦

(iii) 43.123456789

**Solution:**

**(i) 43.123456789**

It is a rational number and q will have factors of 2 or 5 only.

**(ii) 0.120120012000120000â€¦**

It is an irrational number as it is non-terminating and non repeating.

**(iii) 43.123456789**

It is a rational number, as it is non-terminating but repeating. The factors of q will be, apart from 2 or 5, some other factor also.