##### Class 10^{th} Mathematics CBSE Solution

**Exercise 1.1**- Use Euclids division algorithm to find the HCF of :(i) 135 and 225(ii) 196 and…
- Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5,…
- An army contingent of 616 members is to march behind an army band of 32 members…
- Use Euclids division lemma to show that the square of any positive integer is…
- Use Euclids division lemma to show that the cube of any positive integer is of…

**Exercise 1.2**- Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii)…
- Find the LCM and HCF of the following pairs of integers and verify that LCM HCF…
- Find the LCM and HCF of the following integers by applying the prime…
- Given that HCF (306, 657) = 9, find LCM (306, 657).
- Check whether 6ncan end with the digit 0 for any natural number n.…
- Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.…
- There is a circular path around a sports field. Sonia takes 18 minutes to drive…

**Exercise 1.3**- Prove that is irrational.
- Prove that 3+2 root 5 is irrational.
- Prove that the following are irrationals: (i) 1/root 2 (ii) 7 root 5 (iii) 6 +…

**Exercise 1.4**

**Exercise 1.1**

- Use Euclids division algorithm to find the HCF of :(i) 135 and 225(ii) 196 and…
- Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5,…
- An army contingent of 616 members is to march behind an army band of 32 members…
- Use Euclids division lemma to show that the square of any positive integer is…
- Use Euclids division lemma to show that the cube of any positive integer is of…

**Exercise 1.2**

- Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii)…
- Find the LCM and HCF of the following pairs of integers and verify that LCM HCF…
- Find the LCM and HCF of the following integers by applying the prime…
- Given that HCF (306, 657) = 9, find LCM (306, 657).
- Check whether 6ncan end with the digit 0 for any natural number n.…
- Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.…
- There is a circular path around a sports field. Sonia takes 18 minutes to drive…

**Exercise 1.3**

- Prove that is irrational.
- Prove that 3+2 root 5 is irrational.
- Prove that the following are irrationals: (i) 1/root 2 (ii) 7 root 5 (iii) 6 +…

**Exercise 1.4**

###### Exercise 1.1

**Question 1.**Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Answer:****Concept used : **

To obtain the HCF of two positive integers, say c and d, with c > d,we follow

the steps below:

Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.

Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

**(i) **We know that,

= 225>135

Applying Euclid’s division algorithm:

(Dividend = Divisor × Quotient + Remainder)

225 = 135 ×1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90×1+45

Here remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence,

HCF of (135, 225) = 45

**(ii)**

We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0 (Dividend = Divisor × Quotient + Remainder)

Remainder = 0

Hence,

HCF of (196, 38220) = 196

(iii)

We know that,

867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102 (Dividend = Divisor × Quotient + Remainder)

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence,

(HCF 0f 867 and 255) = 51

**Question 2.**Show that any positive odd integer is of the form 6*q* + 1, or 6*q* + 3, or 6*q* + 5, where *q* is some integer.

**Answer:**To Prove: Any Positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5

Proof: To prove the statement by Euclid's lemma we have to consider divisor as 6 and then find out the possible remainders when divided by 6

By taking,’ a’ as any positive integer and b = 6.

Applying Euclid’s algorithm

a = 6 q + r

As divisor is 6 the remainder can take only 6 values from 0 to 5

Here, r = remainder = 0, 1, 2, 3, 4, 5 and q ≥ 0

So, total possible forms are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5

6q + 0 , (6 is divisible by 2, its an even number)

6q + 1, ( 6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, ( 6 and 4 both are divisible by 2, its an even number)

6q + 5 , (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 6q + 1, or 6q + 3, or 6q+5

Hence, Proved.

**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?

**Answer:** Suppose, both groups are arranged in 'n' columns, for completely filling each column,

The maximum no of columns in which they can march is the highest common factor of their number of members.

i.e. n = HCF(616, 32)

By using, Euclid’s division algorithm

616 = 32×19+8

Remainder ≠ 0

So, again Applying Euclid’s division algorithm

32 = 8×4+0

HCF of (616, 32) is 8.

So,

They can march in 8 columns each.

**Question 4.**Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3*m* or 3*m* + 1 for some integer *m*.

**Answer:****To Prove:** Square of any number is of the form 3 m or 3 m +1

**Proof:** to prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3m and 3m + 1 as the form

So,

By taking, ’ a’ as any positive integer and b = 3.

Applying Euclid’s algorithm a = bq + r.

a = 3q + r

Here, r = remainder = 0, 1, 2 and q ≥ 0 as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.

So, putting all the possible values of the remainder in, a = 3q + r

a = 3q or 3q+1 or 3q+2

And now squaring all the values,

When a= 3q

Squaring both sides we get,

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

a^{2} = 9q^{2}

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

a^{2} = 3 k_{1}

Where k_{1 }= 3q^{2}

When a=3q+1

Squaring both sides we get,

a^{2} = (3q + 1)^{2}

a^{2} = 9q^{2 }+ 6q + 1

a^{2} =3( 3q^{2 }+ 2q )+ 1

a^{2} = 3k_{2 }+ 1

Where k_{2}_{}= 3q^{2 }+ 2q

When a = 3q+2

Squaring both sides we get,

a^{2} = (3q + 2)^{2}a^{2} = 9q^{2 }+ 12q + 4

a^{2} = 9q^{2 }+ 12q + 3+1

a^{2} = 3( 3q^{2 }+ 4q + 1) +1

a^{2} = 3k_{3 }+ 1

Where k_{3}_{}= 3q^{2 }+ 4q + 1

Where k_{1}, k_{2} and k_{3} 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.**

**Question 5.**Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9*m*, 9*m* + 1 or 9*m* + 8.

**Answer:**Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma:

a = bq + r So, we have the following cases:

__Case I When a = 3q__

In this case, we have

a^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3} ) = 9m, where m = 3q^{3}

__Case II When a = 3q + 1__

In this case, we have

a^{3} = (3q + 1)^{3}

⇒27q^{3} + 27q^{2} + 9q + 1

⇒9q(3q^{2} + 3q + 1) + 1

⇒a^{3} = 9m + 1, where m = q(3q^{2} + 3q + 1)

__Case III When a = 3q + 2__

In this case, we have

a^{3} = (3q + 1)^{3}

⇒27q^{3} + 54q^{2} + 36q + 8

⇒9q(3q^{2} + 6q + 4) + 8

⇒a^{3} = 9m + 8, where m = q(3q^{2} + 6q + 4)

Hence, a^{3} is the form of 9m or, 9m + 1 or, 9m + 8

**Question 1.**

Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Answer:**

**Concept used : **

To obtain the HCF of two positive integers, say c and d, with c > d,we follow

the steps below:

Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.

Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.**(i) **We know that,

= 225>135

Applying Euclid’s division algorithm:

(Dividend = Divisor × Quotient + Remainder)

225 = 135 ×1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90×1+45

Here remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence,

HCF of (135, 225) = 45

**(ii)**

We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0 (Dividend = Divisor × Quotient + Remainder)

Remainder = 0

Hence,

HCF of (196, 38220) = 196

**(iii)**We know that,

867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102 (Dividend = Divisor × Quotient + Remainder)

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence,

(HCF 0f 867 and 255) = 51

**Question 2.**

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

**Answer:**

To Prove: Any Positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5

Proof: To prove the statement by Euclid's lemma we have to consider divisor as 6 and then find out the possible remainders when divided by 6

By taking,’ a’ as any positive integer and b = 6.

Applying Euclid’s algorithm

a = 6 q + r

As divisor is 6 the remainder can take only 6 values from 0 to 5

Here, r = remainder = 0, 1, 2, 3, 4, 5 and q ≥ 0

So, total possible forms are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5

6q + 0 , (6 is divisible by 2, its an even number)

6q + 1, ( 6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, ( 6 and 4 both are divisible by 2, its an even number)

6q + 5 , (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 6q + 1, or 6q + 3, or 6q+5

Hence, Proved.

**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?

**Answer:**Suppose, both groups are arranged in 'n' columns, for completely filling each column,

The maximum no of columns in which they can march is the highest common factor of their number of members.

i.e. n = HCF(616, 32)

By using, Euclid’s division algorithm

616 = 32×19+8

Remainder ≠ 0

So, again Applying Euclid’s division algorithm

32 = 8×4+0

HCF of (616, 32) is 8.

So,

They can march in 8 columns each.

**Question 4.**

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3*m* or 3*m* + 1 for some integer *m*.

**Answer:**

**To Prove:** Square of any number is of the form 3 m or 3 m +1**Proof:** to prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3m and 3m + 1 as the form

So,

By taking, ’ a’ as any positive integer and b = 3.

Applying Euclid’s algorithm a = bq + r.

a = 3q + r

Here, r = remainder = 0, 1, 2 and q ≥ 0 as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.

So, putting all the possible values of the remainder in, a = 3q + r

a = 3q or 3q+1 or 3q+2

And now squaring all the values,

When a= 3q

Squaring both sides we get,

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

a^{2} = 9q^{2}

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

a^{2} = 3 k_{1}

Where k_{1 }= 3q^{2}

When a=3q+1

Squaring both sides we get,

a^{2} = (3q + 1)^{2}

a^{2} = 9q^{2 }+ 6q + 1

a^{2} =3( 3q^{2 }+ 2q )+ 1

a^{2} = 3k_{2 }+ 1

Where k_{2}_{}= 3q^{2 }+ 2q

When a = 3q+2

Squaring both sides we get,

a^{2} = (3q + 2)^{2}a^{2} = 9q^{2 }+ 12q + 4

a^{2} = 9q^{2 }+ 12q + 3+1

a^{2} = 3( 3q^{2 }+ 4q + 1) +1

a^{2} = 3k_{3 }+ 1

_{3}

_{}= 3q

^{2 }+ 4q + 1

Where k_{1}, k_{2} and k_{3} 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.**

**Question 5.**

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9*m*, 9*m* + 1 or 9*m* + 8.

**Answer:**

Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma:

a = bq + r So, we have the following cases:

__Case I When a = 3q__

In this case, we have

a^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3} ) = 9m, where m = 3q^{3}

__Case II When a = 3q + 1__

In this case, we have

a^{3} = (3q + 1)^{3}

⇒27q^{3} + 27q^{2} + 9q + 1

⇒9q(3q^{2} + 3q + 1) + 1

⇒a^{3} = 9m + 1, where m = q(3q^{2} + 3q + 1)

__Case III When a = 3q + 2__

In this case, we have

a^{3} = (3q + 1)^{3}

⇒27q^{3} + 54q^{2} + 36q + 8

⇒9q(3q^{2} + 6q + 4) + 8

⇒a^{3} = 9m + 8, where m = q(3q^{2} + 6q + 4)

Hence, a^{3} is the form of 9m or, 9m + 1 or, 9m + 8

###### Exercise 1.2

**Question 1.**Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

**Answer:**(i) 140 = 2×2×5×7 = 2^{2}×5×7

(ii) 156 = 2×2×3×13 = 2^{2}×3×13

(iii) 3825 = 3×3×5×5×17 = 3^{2}×5^{2}×17

(iv) 5005 = 5×7×11×13

(v) 7429 = 17×19×23

**Question 2.**Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =

product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

**Answer:**(i) 26 = 2×13

91 = 7×13

HCF =13

LCM = 2×7×13 =182

Product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

Hence, product of two numbers = HCF×LCM

(ii) 510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460

HCF×LCM = 46920

Hence, product of two numbers = HCF×LCM

(iii) 336 = 2×2×2×2×3×7

336 = 2^{4}×3×7

54 = 2×3×3×3

54 =2×3^{3}

HCF = 2×3 = 6

LCM = 2^{4}×3^{3}×7 = 3024

Product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

Hence, product of two numbers = HCF×LCM

**Question 3.**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

**Answer:**Prime factors of any number is the representation of a number as a product of prime numbers it is composed of

for example: Prime factors of 20 = 2 x 2 x 5

HCF = Highest common factor = The product of the factors that are common to the numbers

LCF = Least Common Factor = Product of all the factors of numbers without duplicating the factor

(i)

let us write prime factors of the given numbers

12 = 2×2×3 = 2^{2}×3

15 = 3×5

21 = 3×7

Only 3 is common in all the three numbers, therefore

HCF = 3

As 3 is common in all three numbers, it will be taken as 1 time in the product of calculating LCM

LCM = 2^{2}×3×5×7 = 420

(ii) Let us write the prime factors of the given numbers

17 = 1×17

23 = 1×23

29 = 1×29

As only 1 is common from all the three factors

HCF = 1

As nothing except 1 is common from all three numbers, simply multiplying them will give LCM of numbers.

LCM = 17×23×29 = 11339

(iii) Let us write the prime factors of the given numbers

8=2×2×2

9 = 3×3

25 = 5×5

As nothing is common in the numbers

HCF = 1

As nothing is common in factors of numbers, numbers are simply multiplied to obtain LCM

LCM = 2×2×2×3×3×5×5 = 1800

**Question 4.**Given that HCF (306, 657) = 9, find LCM (306, 657).

**Answer:**Given: HCF of (306, 657) = 9

We know that,

LCM × HCF = product of two numbers

LCM×HCF = 306×657

LCM =

LCM = 22338

**Question 5.**- Check whether 6
^{n} can end with the digit 0 for any natural number *n*.

**Answer:**We need to find can 6^{n} end with zero

If any number has last digit 0,

Then, it should be divisible by 10

Factors of 10 = 2×5

So,

Value 6^{n} should be divisible by 2 and 5

Prime factorisation of 6^{n} = (2×3)^{n}

Hence,

6^{n} is divisible by 2 but not by 5.

It can not end with 0.

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

**Answer:**Composite numbers are those numbers, which can be written in the form of the product of two or more integers, and at least one of them should not be 1

**(i)**

7 × 11 × 13 + 13

= (7 × 11 × 13) + (13 × 1)

Taking 13 as common, we get,

= 13 × (7 × 11 + 1)

= 13 × (77 + 1)

= 13 × 78

= 13 × 13 × 6

As the given no is a multiple of two or more integers, one of them being other than 1.Hence, it is a composite number.

Therefore, it is a composite number.

**(ii)**

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

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

Taking 5 as common , we get,

= 5×(7×6×4×3×2×1+1)

= 5× (1008+1)

= 5×1009

As the given no is a multiple of two integers, one of them being other than 1.

Hence, it is a composite number.

**Question 7.**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?

**Answer:**Both of them start from the same point and start moving in same direction

Sonia takes 18 minutes and Ravi takes 12 minutes to complete the circle.

After 12 minutes Ravi will be back to the starting point and Sonia must have covered (12/18) = 2/3 of the rounds

After 12 more minutes Ravi has completed 2 rounds and Sonia must have covered (24/18) = 4/3 of the rounds

After 12 more minutes Ravi has completed 3 rounds and Sonia must have completed (36/18) = 2 rounds

Hence after 36 minutes both will again meet at the starting point.

**Alternate Method:**

They will meet again after LCM of both values at starting point.

18 = 2×3×3

And

12 = 2×2×3

LCM of 12 and 18 = 2×2×3×3 = 36

Therefore,

Ravi and Sonia will meet together at the starting point after 36 minutes.

**Question 1.**

Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

**Answer:**

(i) 140 = 2×2×5×7 = 2^{2}×5×7

(ii) 156 = 2×2×3×13 = 2^{2}×3×13

(iii) 3825 = 3×3×5×5×17 = 3^{2}×5^{2}×17

(iv) 5005 = 5×7×11×13

(v) 7429 = 17×19×23

**Question 2.**

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =

product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

**Answer:**

(i) 26 = 2×13

91 = 7×13

HCF =13

LCM = 2×7×13 =182

Product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

Hence, product of two numbers = HCF×LCM

(ii) 510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460

HCF×LCM = 46920

Hence, product of two numbers = HCF×LCM

(iii) 336 = 2×2×2×2×3×7

336 = 2^{4}×3×7

54 = 2×3×3×3

54 =2×3^{3}

HCF = 2×3 = 6

LCM = 2^{4}×3^{3}×7 = 3024

Product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

Hence, product of two numbers = HCF×LCM

**Question 3.**

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

**Answer:**

Prime factors of any number is the representation of a number as a product of prime numbers it is composed of

for example: Prime factors of 20 = 2 x 2 x 5

HCF = Highest common factor = The product of the factors that are common to the numbers

LCF = Least Common Factor = Product of all the factors of numbers without duplicating the factor

(i)

let us write prime factors of the given numbers

12 = 2×2×3 = 2^{2}×3

15 = 3×5

21 = 3×7

Only 3 is common in all the three numbers, therefore

HCF = 3

As 3 is common in all three numbers, it will be taken as 1 time in the product of calculating LCM

LCM = 2^{2}×3×5×7 = 420

(ii) Let us write the prime factors of the given numbers

17 = 1×17

23 = 1×23

29 = 1×29

As only 1 is common from all the three factorsHCF = 1

As nothing except 1 is common from all three numbers, simply multiplying them will give LCM of numbers.LCM = 17×23×29 = 11339

(iii) Let us write the prime factors of the given numbers

8=2×2×2

9 = 3×3

25 = 5×5

As nothing is common in the numbersHCF = 1

As nothing is common in factors of numbers, numbers are simply multiplied to obtain LCMLCM = 2×2×2×3×3×5×5 = 1800

**Question 4.**

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

**Answer:**

Given: HCF of (306, 657) = 9

We know that,

LCM × HCF = product of two numbers

LCM×HCF = 306×657

LCM =

LCM = 22338

**Question 5.**

- Check whether 6
^{n}can end with the digit 0 for any natural number*n*.

**Answer:**

We need to find can 6^{n} end with zero

If any number has last digit 0,

Then, it should be divisible by 10

Factors of 10 = 2×5

So,

Value 6^{n} should be divisible by 2 and 5

Prime factorisation of 6^{n} = (2×3)^{n}

Hence,

6^{n} is divisible by 2 but not by 5.

It can not end with 0.

**Question 6.**

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

**Answer:**

Composite numbers are those numbers, which can be written in the form of the product of two or more integers, and at least one of them should not be 1 **(i)**

7 × 11 × 13 + 13

Taking 13 as common, we get,

= 13 × (7 × 11 + 1)

= 13 × (77 + 1)

= 13 × 78

= 13 × 13 × 6

Hence, it is a composite number.

Therefore, it is a composite number.

**(ii)**

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= (7 × 6 × 5 × 4 × 3 × 2 × 1) + (5 × 1)Taking 5 as common , we get,

= 5×(7×6×4×3×2×1+1)

= 5× (1008+1)

= 5×1009

As the given no is a multiple of two integers, one of them being other than 1.Hence, it is a composite number.

**Question 7.**

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?

**Answer:**

Both of them start from the same point and start moving in same direction

Sonia takes 18 minutes and Ravi takes 12 minutes to complete the circle.

After 12 minutes Ravi will be back to the starting point and Sonia must have covered (12/18) = 2/3 of the rounds

After 12 more minutes Ravi has completed 2 rounds and Sonia must have covered (24/18) = 4/3 of the rounds

After 12 more minutes Ravi has completed 3 rounds and Sonia must have completed (36/18) = 2 rounds

Hence after 36 minutes both will again meet at the starting point.**Alternate Method:**

They will meet again after LCM of both values at starting point.

18 = 2×3×3

And

12 = 2×2×3

LCM of 12 and 18 = 2×2×3×3 = 36

Therefore,

Ravi and Sonia will meet together at the starting point after 36 minutes.

###### Exercise 1.3

**Question 1.**Prove that is irrational.

**Answer:**Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)^{2} = a^{2}

⇒ 5b^{2} = a^{2}

⇒ a^{2}/5 = b^{2}

Hence, 5 divides a^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

5b^{2} = a^{2}

Putting a = 5c,

⇒ 5b^{2} = (5c)^{2}

⇒ 5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

⸫ b^{2}/5 = c^{2}

Hence, 5 divides b^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational

**Question 2.**Prove that is irrational.

**Answer:**To Prove: 3 + 2√5 is irrational

Proof:

Let is rational.

A number is said to be rational if it can be expressed in the form p/q where q ≠ 0

Therefore,

We can find two integers p & q where, (q ≠ 0) such that

Since p and q are integers, will also be rational and therefore, is rational.

We know that √5 is irrational but according to above statement it has to be rational

So, both the comments are contradictory,

Hence, the number should have been irrational to make the statement correct.

Therefore, is irrational

**Question 3.**Prove that the following are irrationals:

(i)

(ii)

(iii)

**Answer:**(i) Let is rational.

Therefore, we can find two integers p & q where, q ≠0 such that

is rational as p and q are integers.

Therefore, is rational which contradicts to the fact that is irrational.

Hence, our assumption is false and is irrational.

(ii) Let is rational.

Therefore, we can find two integers p & q where, q≠0 such that

for some integers p and q

is rational as p and q are integers.

Therefore, should be rational.

This contradicts the fact that is irrational.

Therefore our assumption that is rational is false.

Hence, is irrational.

(iii) Let be rational.

Therefore, we can find two integers p & q where q≠0, such that

Since p and q are integers, is also rational

Hence, should be rational, this contradicts the fact that is irrational.

So, our assumption is false and hence, is irrational.

**Question 1.**

Prove that is irrational.

**Answer:**

Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)^{2} = a^{2}

⇒ 5b^{2} = a^{2}

⇒ a^{2}/5 = b^{2}

Hence, 5 divides a^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

5b^{2} = a^{2}

Putting a = 5c,

⇒ 5b^{2} = (5c)^{2}

⇒ 5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

⸫ b^{2}/5 = c^{2}

Hence, 5 divides b^{2}

By theorem, if p is a prime number and p divides a^{2}, then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational

**Question 2.**

Prove that is irrational.

**Answer:**

To Prove: 3 + 2√5 is irrational

Proof:

Let is rational.

Therefore,

We can find two integers p & q where, (q ≠ 0) such that

Since p and q are integers, will also be rational and therefore, is rational.

We know that √5 is irrational but according to above statement it has to be rational

So, both the comments are contradictory,

Hence, the number should have been irrational to make the statement correct.

Therefore, is irrational

**Question 3.**

Prove that the following are irrationals:

(i)

(ii)

(iii)

**Answer:**

(i) Let is rational.

Therefore, we can find two integers p & q where, q ≠0 such that

is rational as p and q are integers.

Therefore, is rational which contradicts to the fact that is irrational.

Hence, our assumption is false and is irrational.

(ii) Let is rational.

Therefore, we can find two integers p & q where, q≠0 such that

for some integers p and q

is rational as p and q are integers.

Therefore, should be rational.

This contradicts the fact that is irrational.

Therefore our assumption that is rational is false.

Hence, is irrational.

(iii) Let be rational.

Therefore, we can find two integers p & q where q≠0, such that

Since p and q are integers, is also rational

Hence, should be rational, this contradicts the fact that is irrational.

So, our assumption is false and hence, is irrational.

###### Exercise 1.4

**Question 1.**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:

**Answer:**(i)

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 5^{5}

The denominator is of the form 5^{m}

Hence, the decimal expansion of is terminating.

(ii)

Factorize the denominator we get,

8 = 2 × 2 ×2 = 2^{3}

The denominator is of the form 2^{m}

Hence, the decimal expansion of is terminating.

(iii)

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^{m} × 5^{n}, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)

Factorize the denominator we get,

1600 = 2^{6}×5^{2}

The denominator is in the form 2^{m} × 5^{n}

Hence, the decimal expansion of is terminating.

(v)

Factorize the denominator we get,

343 = 7^{3}

Since the denominator is not in the form of 2^{m} × 5^{n}, it has 7 as its factors.

So, the decimal expansion of non-terminating repeating.

(vi)

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(vii)

Since, the denominator is not in the form of 2^{m} × 5^{n}, as it has 7 in denominator.

So, the decimal expansion of is non-terminating repeating.

(viii)

The denominator is in the form 5^{n}

Hence, the decimal expansion of is terminating.

(ix)

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(x)

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2^{m} × 5^{n}, as it has 3 in denominator.

So, the decimal expansion of non-terminating repeating.

**Question 2.**Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

**Answer:**(i)

(ii)

(iv)

(vi)

(viii)

(ix)

**Question 3.**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) 0.120120012000120000...

(iii)

**Answer:**(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form and q is of the form 2^{m}×5^{n}

That is, the prime factor of q will be 2 or 5 or both.

(ii) 0.120120012000120000...

The decimal expansion is neither terminating nor recurring.

Therefore, the given number is an irrational number.

(iii)

Since the decimal expansion is non-terminating but recurring, the given number is a rational number of the form and q is not of the form 2^{m}×5^{n} that is, the prime factors of q will also have a factor other than 2 or 5.

**Question 1.**

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:

**Answer:**

(i)

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 5^{5}

The denominator is of the form 5^{m}

Hence, the decimal expansion of is terminating.

(ii)

Factorize the denominator we get,

8 = 2 × 2 ×2 = 2^{3}

The denominator is of the form 2^{m}

Hence, the decimal expansion of is terminating.

(iii)

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^{m} × 5^{n}, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)

Factorize the denominator we get,

1600 = 2^{6}×5^{2}

The denominator is in the form 2^{m} × 5^{n}

Hence, the decimal expansion of is terminating.

(v)

Factorize the denominator we get,

343 = 7^{3}

Since the denominator is not in the form of 2^{m} × 5^{n}, it has 7 as its factors.

So, the decimal expansion of non-terminating repeating.

(vi)

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(vii)

Since, the denominator is not in the form of 2^{m} × 5^{n}, as it has 7 in denominator.

So, the decimal expansion of is non-terminating repeating.

(viii)

The denominator is in the form 5^{n}

Hence, the decimal expansion of is terminating.

(ix)

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(x)

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2^{m} × 5^{n}, as it has 3 in denominator.

So, the decimal expansion of non-terminating repeating.

**Question 2.**

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

**Answer:**

(i)

(ii)

(iv)

(vi)

(viii)

(ix)

**Question 3.**

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) 0.120120012000120000...

(iii)

**Answer:**

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form and q is of the form 2^{m}×5^{n}

That is, the prime factor of q will be 2 or 5 or both.

(ii) 0.120120012000120000...

The decimal expansion is neither terminating nor recurring.

Therefore, the given number is an irrational number.

(iii)

Since the decimal expansion is non-terminating but recurring, the given number is a rational number of the form and q is not of the form 2^{m}×5^{n} that is, the prime factors of q will also have a factor other than 2 or 5.