# Question-1

**If**

**is a positive rational number and***a***is a positive integer greater than 1, prove that***n***is a rational number***a*^{n}**.**

**Solution:**

We know that product of two rational number is always a rational number. Hence if

**is a rational number then**

*a*

*a*^{2}=

**x**

*a***is a rational number,**

*a*

*a*^{3}=

*a*^{2}x

**is a rational number.**

*a***.**

^{.}.

*a*^{n}=

*a*^{n-1 }x

**is a rational number.**

*a*# Question-2

**Write the following rational numbers in decimal form:**

(i) (ii) (iii) 3 (iv) (v) (vi) (vii) (viii)

(i) (ii) (iii) 3 (iv) (v) (vi) (vii) (viii)

**Solution:**

(i) =0.42

(ii)

**= 0.654**

(iii)

**3==3.375**

(iv)

**= 0.833â€¦=**

(v) = 0.2

(vi)

**=**

(vii)

**=**

(viii)

**=**

# Question-3

**Give three rational numbers lying between and .**

**Solution:**

The rational number lying between is and .

= = =

Therefore, < <

Now, the rational number lying between and is

= =

Therefore, < <

The rational number lying between and is

===

Therefore, < <

Hence the three rational numbers lying between and are .

# Question-4

**Insert a rational and an irrational number between 2 and 3.**

**Solution:**

A rational number between 2 and 3 = = 2.5

An irrational number between 2 and 3 is .

# Question-5

**How many rational numbers and irrational numbers can be inserted between 2 and 3 ?**

**Solution:**

There are infinite number of rational and irrational numbers between 2 and 3.

# Question-6

**Find three rational numbers lying between 0 and 0.1. Find twenty rational numbers between 0 and 0.1. Give a method to determine any number of rational numbers between 0 and 0.1.**

**Solution:**

The three rational numbers lying between 0 and 0.1 are 0.01, 0.05, 0.09.

The twenty rational numbers between 0 and 0.1 are 0.001, 0.002, 0.003, 0.004, â€¦ 0.011, 0.012, â€¦ 0.099.

To determine any number of rational numbers between 0 and 0.1 insert the square root of its product.

i.e. The rational numbers between

**and**

*a***is**

*b*# Question-7

**Find three rational numbers lying between**

**Solution:**

The rational number lying between - and -

==

Therefore, -<- <-

The rational number lying between - and-

===

Therefore, -< <-

The rational number lying between - and -

===

Therefore, -< <-

Therefore the three rational numbers are

# Question-8

**Complete the following:**

(i) Every point on the number line corresponds to a â€¦â€¦â€¦â€¦â€¦â€¦. number which may be either â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. or â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(ii) The decimal form of an irrational number is neither â€¦â€¦â€¦â€¦â€¦â€¦ nor â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(iii) The decimal representation of the rational number is â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(iv) 0 is â€¦â€¦â€¦â€¦â€¦.. number. [Hint: a rational /an irrational]

(i) Every point on the number line corresponds to a â€¦â€¦â€¦â€¦â€¦â€¦. number which may be either â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. or â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(ii) The decimal form of an irrational number is neither â€¦â€¦â€¦â€¦â€¦â€¦ nor â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(iii) The decimal representation of the rational number is â€¦â€¦â€¦â€¦â€¦â€¦â€¦

(iv) 0 is â€¦â€¦â€¦â€¦â€¦.. number. [Hint: a rational /an irrational]

**Solution:**

(i) Every point on the number line corresponds to a

__real__number which may be either

__rational__or

__irrational.__

(ii) The decimal form of an irrational number is neither

__recurring__nor

__terminating.__

(iii) The decimal representation of the rational number is

__0.296__

(iv) 0 is

__a rational__number.

# Question-9

**Give an example for each, if two irrational numbers, whose**

(i) difference is a rational number.

(ii) difference is a irrational number.

(iii) sum is a rational number.

(iv) sum is an irrational number.

(v) product is an irrational number,

(vi) product is an irrational number.

(vii) quotient is a rational number.

(viii) quotient is an irrational number.

(i) difference is a rational number.

(ii) difference is a irrational number.

(iii) sum is a rational number.

(iv) sum is an irrational number.

(v) product is an irrational number,

(vi) product is an irrational number.

(vii) quotient is a rational number.

(viii) quotient is an irrational number.

**Solution:**

# Question-10

**Which of the following rational numbers have the terminating decimal representation?**

(i) 3/5 (ii) 7/20 (iii) 2/13

(iv) 27/40 (v) 133/125 (vi) 23/7

[Making use of the result that a rational number

(i) 3/5 (ii) 7/20 (iii) 2/13

(iv) 27/40 (v) 133/125 (vi) 23/7

[Making use of the result that a rational number

**where***p/q***and***p***have no common factor(s) will have a terminating representation if and only if the prime factors of***q***are 2's or 5's or both.]***q***Solution:**

(i) The prime factor of 5 is 5. Hence 3/5 has a terminating decimal representation.

(ii) 20 = 4 x 5 = 2

^{2}x 5.

The prime factors of 20 are both 2's and 5's. Hence 7/20 has a terminating decimal.

(iii) The prime factor of 13 is 13. Hence 2/13 has non- terminating decimal.

(iv) 40 = 2

^{3}x 5.

The prime factors of 40 are both 2's and 5's. Hence 27/40 has a terminating decimal.

(v) 125 = 5

^{3}

The prime factor of 125 is 5's. Hence 133/125 has a terminating decimal.

(vi) The prime factor of 7 is 7. Hence 23/7 has a non-terminating decimal representation.

# Question-11

**Find the decimal representation of 1/7, 2/7. Deduce from the decimal representation of 1/7, without actual calculation, the decimal representation of 3/7, 4/7, 5/7 and 6/7.**

**Solution:**

# Question-12

**Express 0.6666â€¦â€¦â€¦in the form of**

**Solution:**

Let

*x*= 0.6666â€¦â€¦â€¦. â€¦(i)

âˆ´ 10

*x*= ...(ii)

(ii) â€“ (i), 9

*x*= 6 âˆ´

*x*==

âˆ´ 0.6666 =

# Question-13

**If**

**and***a**are two rational numbers, prove that***b***,***a + b****are rational numbers. If***a - b, ab***â‰ 0, show that***b***is also a rational number.***a/b***Solution:**

Let

*where*

**a = p/q**

*q*

*â‰***and**

*0***where**

*b = r/s***â‰ 0 be the rational numbers then**

*s*(i)

**a + b ==**where

**â‰ 0, since**

*q s***â‰ 0 and**

*q***â‰ 0.**

*s*Also

**is an integer.**

*p s + r q*Hence

**is a rational number.**

*a + b*(ii)

*=*

**a - b =**where

**â‰ 0, since**

*q s***â‰ 0 and**

*q***â‰ 0.**

*s*Also

**is an integer**

*p s - r q*Hence

**is a rational number.**

*a - b*(iii)

**=**

*ab*where

**â‰ 0, since**

*q s***â‰ 0 and**

*q***â‰ 0.**

*s*Also

**is an integer.**

*pr*Hence

*is a rational number.*

**ab**(iv) Since

**â‰ 0, we have**

*b***â‰ 0 thus**

*r/s***â‰ 0 and**

*r***â‰ 0.**

*s*where

**â‰ 0 and**

*q**â‰ 0.*

**r**Also

*is an integer*

**ps**Hence

*is a rational number.*

**a/b**# Question-14

**Express 0.272727â€¦â€¦â€¦in the form of**

*.***Solution:**

Let

*x*= 0.272727â€¦â€¦â€¦. â€¦(i) âˆ´ 100

*x*= 27.2727..... â€¦(ii)

(ii) â€“ (i), 99

**= 27 âˆ´**

*x***= =**

*x*âˆ´ 0.272727â€¦â€¦â€¦=

# Question-15

**You have seen that âˆš**

**2 is not a rational number. Show that 2 + âˆš2 is not a rational number.**

**Solution:**

Let 2 + âˆš2 be a rational number say

*.*

**r**Then 2 + âˆš2 =

**r**âˆš2 =

*-2*

**r**But, âˆš2 is an irrational number.

Therefore,

*- 2 is also an irrational number.*

**r**=>

*is an irrational number.*

**r**Hence our assumption

*is a rational number is wrong.*

**r**# Question-16

**Express 3.7777â€¦â€¦â€¦. in the form of**

*.***Solution:**

Let

**= 3.7777.â€¦â€¦â€¦. â€¦(i)**

*x*âˆ´ 10

**= 37. â€¦(ii)**

*x*(ii) â€“ (i), 9

**= 34 âˆ´**

*x***= âˆ´ 3.7777â€¦â€¦â€¦ =**

*x*# Question-17

**Prove that 3âˆš**

**3 is not a rational number.**

**Solution:**

Let 3âˆš3 be a rational number say

*r*.

Then 3âˆš3 =

**r**âˆš3 = (1/3)

**r**(1/3)

*is a rational number because product of two rational number is a rational number.*

**r**=> âˆš3 is a rational number but âˆš3 is not a rational number.

Therefore our assumption 3âˆš3 is a rational number is wrong.

# Question-18

**Express 18.484848â€¦â€¦â€¦â€¦â€¦.in the form of**

*.***Solution:**

Let

**= 18.484848.â€¦â€¦â€¦.........................â€¦(i) âˆ´ 100**

*x***= 1848.4848â€¦â€¦â€¦....................(ii)**

*x*(ii) â€“ (i), 99

**= 1830 âˆ´**

*x***= = âˆ´ 18.484848â€¦â€¦â€¦=**

*x*# Question-19

**Show that**

**is not a rational number.**

**Solution:**

Let

**be a rational number, say where q â‰ 0.**

Then

**=**

Since 1

^{3}= 1 , and 2

^{3}= 8, it follows that 1 < < 2

Then

*> 1 because if*

**q****= 1 then will be an integer, and there is no integer between 1 and 2.**

*q*Now, 6 =

6 =

6

**q**^{2}=

*being an integer, 6q*

**q**^{2}is an integer, and since

*> 1 and*

**q***does not have a common factor with*

**q****and consequently with**

*p*

*p*^{3}. So, is a fraction, different from an integer.

Thus 6

*q*^{2}â‰ .

This contradiction proves the result.

# Question-20

**Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers.**

(i)

(ii) 3

(iii)

(iv)

(v)

(vi)

(i)

(ii) 3

(iii)

(iv)

(v)

**-**(vi)

**Solution:**

(i) = 2 is rational.

(ii) 3

**=**3

**=**3 Ã— 3

**=**9

**is irrational.**

(iii)

**==**= 1.2 is rational.

(iv)

**= ==**is irrational.

(v)

**- =**-0.8 is rational.

(vi)= 10 is rational.

# Question-21

**Find two irrational numbers between 2 and 2.5.**

**Solution:**

The two irrational numbers between 2 and 2.5 are 2.101001000100001â€¦.. and 2.201001000100001â€¦..

# Question-22

**In the following equations, find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers:**

(i) x

(ii) y

(iii) z

(iv) u

(v) v

(vi) w

(vii) t

(i) x

^{2 }= 5(ii) y

^{2 }= 9(iii) z

^{2 }= 0.04(iv) u

^{2 }=(v) v

^{2 }= 3(vi) w

^{3 }= 27(vii) t

^{2 }= 0.4**Solution:**

(i) x

^{2 }= 5

âˆ´ x = is irrational.

(ii) y

^{2 }= 9

âˆ´ y = 3 is rational.

(iii) z

^{2 }= 0.04

âˆ´ z = 0.2 is rational.

(iv) u

^{2 }=

âˆ´ u =

=

=

**=**is irrational.

(v) v

^{2 }= 3

âˆ´ v =

**is irrational.**

(vi) w

^{3 }= 27 âˆ´ w = = 3 is rational.

(vii) t

^{2 }= 0.4 âˆ´ t =

**==**is irrational.

# Question-23

**Find two irrational numbers between 0.1 and 0.12.**

**Solution:**

The two irrational numbers between 0.1 and 0.12 are 0.1010010001â€¦ and 0.1101001000100001â€¦â€¦

# Question-24

**Give an example to show that the product of a rational number and an irrational number may be a rational number.**

**Solution:**

A rational number 0 multiplied by an irrational number gives the rational number 0.

# Question-25

**Give two irrational number lying between âˆš2 and âˆš3.**

**Solution:**

The two irrational number lying between âˆš2 and âˆš3 are âˆš2.1 and âˆš2.2

# Question-26

**State with reason which of the following are surds and which are not.**

(i) Ã— (ii) Ã— (iii) Ã— (iv) Ã— (v) 5Ã— 2

(vi) Ã— (vii) Ã— (viii) 6Ã— 9 (ix) Ã— (x) Ã—

(i) Ã— (ii) Ã— (iii) Ã— (iv) Ã— (v) 5Ã— 2

(vi) Ã— (vii) Ã— (viii) 6Ã— 9 (ix) Ã— (x) Ã—

**Solution:**

(i) Ã—

**=**Ã—

**=**Ã— Ã— =5is a surd.

(ii) Ã—

**=**Ã—

**=**2 Ã—Ã—Ã—

= 4 is a surd.

(iii) Ã— =Ã— =3Ã— = 9 is not a surd.

(iv) Ã— =4 Ã— 2 = 8 is not a surd.

(v) 5Ã— 2= 5Ã— 2

**=**5Ã— 2Ã— 2Ã—

= 5 Ã— 2Ã— 2 Ã— 2 Ã—

= 40 is a surd.

(vi) Ã—

**=**Ã—=5Ã—=

**5 Ã— 5 = 25 is not a surd.**

(vii) Ã— = 10 is a surd.

(viii) 6Ã— 9= 54 is a surd.

(ix) Ã— =Ã—

=2Ã— 3

=2Ã—Ã—Ã— 3Ã—

= 30 is a surd.

(x) Ã— = Ã—

**=**Ã— Ã— Ã—

= 3

**is a surd.**

# Question-27

**Express the following recurring decimals into vulgar fractions :**

**Solution:**

(a) = 0.666â€¦. .........(1)

10 Ã—

**=**6.666â€¦ ........ (2)

(2) - (1) â‡’9 Ã— = 6

âˆ´ = 6/9 = 2/3

(b) = 0.161616â€¦.. ---(1)

100 Ã— = 16.1616â€¦.. -----(2)

Subtracting (1) from (2) , we get

99 Â´ = 16

âˆ´ = 16/99

(c) 0. = 0.234234â€¦.. ----(i)

1000 Ã— 0. = 234.234234â€¦ -----(ii)

Subtracting (i) from (ii) , we get

999 Ã— 0. = 234

âˆ´ 0. = 234/999.

(d) = 0.125454â€¦. -----(i)

âˆ´ 100 Ã— = 12.545454â€¦. -----(ii)

And, 10000 Â´ = 1254.5454â€¦ -----(iii)

Subtracting (ii) from (iii) , we get

9900 Ã— = 1242

âˆ´ = 1242/9900 = 69/550.

# Question-28

**Find the values of**

**and***a***if =***b**a + b***Solution:**

= Ã—

**[Multiplying the numerator and the denominator by**

**(-1)]**

= == 2 -

It is given that 2 -=

*a + b*âˆ´

**= 2 and**

*a***= -1.**

*b*# Question-29

**Give two examples to show that the product of two irrational numbers may be a rational number.**

**Solution:**

Take

*= (2+) and*

**a****=(2 -);**

*b**and*

**a****are irrational numbers, but their product = 4 - 3 = 1,**

*b*is a rational number.

Take

**= and**

*c***= -;**

*d***and**

*c***are irrational numbers, but their product = -3,**

*d*is a rational number.

# Question-30

**Find the values of**

**and***a***if =***b**a + b***Solution:**

= Ã—

**[Multiplying the numerator and the denominator**

**by (3 +)]**

=

**=**

It is given that

**=**

*a + b*âˆ´

**= and**

*a***=**

*b*# Question-31

**Find the value of**

**âˆš5 correct to two places of decimal.**

**Solution:**

We know that 2

^{2}= 4 < 5 < 9 = 3

^{2}

Taking positive square roots we get

2 < âˆš5 < 3.

Next, (2.2)

^{2}= 4.84 < 5 < 5.29 = (2.3)

^{2}

Taking positive square roots, we have

2.2 < âˆš5 < 2.3

Again, (2.23)

^{2}= 4.9729< 5< 5.0176 = ( 2.24)

^{2}

Taking positive square roots, we obtain

2.23 < âˆš5 < 2.24

Hence the required approximation is 2.24 as (2.24)

^{2}is nearest to 5 than (2.23)

^{2}.

# Question-32

**Find the values of a and b if =**

*a + b***Solution:**

=Ã—

**[Multiplying the numerator and the denominator by 7 - 4]**

**=**

=

=

It is given that

**=**

**a + b**âˆ´

**= 11 and**

*a***= -6.**

*b*# Question-33

**Prove that âˆš**

**3 - âˆš2 is irrational.**

**Solution:**

Let âˆš3 - âˆš2 be a rational number, say

*r*Then âˆš3 - âˆš2 =

*r*On squaring both sides we have

(âˆš3 - âˆš2)

^{2}=

*r*^{2}

3 - 2âˆš6 + 2 =

*r*^{2}

5 - 2âˆš6 =

*r*^{2}

- 2âˆš6 =

*r*^{2}- 5

âˆš6 = -(

*r*^{2}- 5)/2

Now -(

*r*^{2}- 5)/2 is a rational number and âˆš 6 is an irrational number.

Since a rational number cannot be equal to an irrational number. Our assumption that

âˆš3 - âˆš2 is rational is wrong.

# Question-34

**Find the values of**

**and***a***if =***b**a + b***Solution:**

= Ã—

**[Multiplying the numerator and the denominator by 5 +]**

=

=

=

It is given that

**=**

*a + b*âˆ´

**=**

*a***and**

**=**

*b*# Question-35

**Prove that âˆš**

**3 + âˆš5 is an irrational number.**

**Solution:**

Let âˆš3 + âˆš5 be a rational number, say

*r*Then âˆš3 + âˆš5 =

*r*On squaring both sides,

(âˆš3 + âˆš5)

^{2}=

*r*^{2}

3 + 2âˆš15 + 5 =

*r*^{2}

8 + 2âˆš15 =

*r*^{2}

2âˆš15 =

*r*^{2}- 8

âˆš15 = (

*r*^{2}- 8)/2

Now (

*r*^{2}- 8)/2 is a rational number and âˆš 15 is an irrational number.

Since a rational number cannot be equal to an irrational number. Our assumption that

âˆš3 + âˆš5 is rational is wrong.

# Question-36

**Find the values of**

**and***a***if =***b**a + b***Solution:**

= Ã—

**[Multiplying the numerator and the denominator by 3 + 4]**

**=**

=

=

It is given that =

*a + b***=**

*a+ b***âˆ´**

**= and**

*a***=**

*b*# Question-37

**If -=**

**, find the values of***a + b**and***a***b.***Solution:**

-=

=

=

It is given that

**-=**

*a + b***âˆ´**

**=**

*a + b***âˆ´**

**= 0 and**

*a***=**

*b*

# Question-38

**Give two rational numbers lying between 0.232332333233332â€¦ and 0.21211211121111â€¦..**

**Solution:**

The two rational numbers are 0.222. and 0.221

# Question-39

**Simplify by rationalising the denominator:**

**Solution:**

=Ã—

**[Rationalising the denominator]**

** ==**

# Question-40

**Examine, whether the following numbers are rational or irrational:**

(i) (âˆš

(i) (âˆš

**2 + 2)**

(ii) (2- âˆš2)x(2 + âˆš2)

(iii) (âˆš2 + âˆš3)

(iv)

^{2}(ii) (2- âˆš2)x(2 + âˆš2)

(iii) (âˆš2 + âˆš3)

^{2}(iv)

**Solution:**

(i) (âˆš2 + 2)

^{2}= (âˆš2)

^{2}+ 2âˆš2x2 + (2)

^{2}= 2 + 4âˆš2 + 4 = 6 + 4âˆš2 .

.

^{.}. It is an irrational number.

(ii) (2- âˆš2)x(2 + âˆš2) = (2)

^{2}- (âˆš2)

^{2}= 4 - 2 = 2 .

.

^{.}. It is a rational number.

(iii) (âˆš2 + âˆš3)

^{2}= (âˆš2)

^{2}+ 2âˆš2xâˆš3 + (âˆš3)

^{2}= 2 + 2âˆš6 + 3 = 5 + 2âˆš6

.

^{.}. It is an irrational number.

(iv)

**= = âˆš2**

.

^{.}. It is an irrational number.

# Question-41

**Prove that**

(a) 2 + is not a rational number and

(b) is not a rational number.

(a) 2 + is not a rational number and

(b) is not a rational number.

**Solution:**

If possible, let 2 + =

*, where*

**a***is rational.*

**a**Then, (2 + ) =

**a**7 + 4 =

**a**=-------(i)

Now ,

*is rational â‡’ is rational.*

**a**is rational [from (i)]

This is a contradiction.

Hence, 2 + is not a rational number.

(b) If possible, let

**=**

**, where**

*p/q***and**

*p***are integers, having no common factors and**

*q**â‰ 0.*

**q**Then,

**() = (**

**)**

*p/q*â‡’ 7

*= p------(i)*

**q**â‡’

*is a multiple of 7*

**p**â‡’

*is multiple of 7.*

**p**Let

*= 7m, where m is an integer.*

**p**Then,

*= 343 m ------(ii)*

**p**â‡’ 7

*= 343 m [from (i) and (ii)]*

**q**â‡’

*= 49 m*

**q**â‡’

*is a multiple of 7.*

**q**â‡’

*is a multiple of 7.*

**q**Thus,

*and*

**p***are both multiples of 7, or 7 is a factor of*

**q***and*

**p***.*

**q**This contradicts our assumption that p and

*have no common factors.*

**q**Hence

**is not a rational number.**

# Question-42

**Simplify by rationalising the denominator:**

**Solution:**

=Ã—

**[Rationalising the denominator]**

** ===**

# Question-43

**Simplify by rationalising the denominator:**

**Solution:**

=Ã—

**[Rationalisation the denominator]**

** =**

** = **

** =**

** = **

**
**

Examine whether the following numbers are rational or irrational:
(i) ,(ii) (3-)(3+),(iii)

# Question-44

Examine whether the following numbers are rational or irrational:

(i) (3 + ) = 9 + 2 + 6 = 11 +6, which is irrational.

(ii) (3-)(3+) = (3)- () = 9 - 3 = 6 , which is rational.

(iii)

**Solution:**(i) (3 + ) = 9 + 2 + 6 = 11 +6, which is irrational.

(ii) (3-)(3+) = (3)- () = 9 - 3 = 6 , which is rational.

(iii)

**= Ã—**= = , which is irrational.

# Question-45

**Simplify by rationalising the denominator:**

=Ã—

**Solution:**=Ã—

**[Rationalising the denominator]****=****=****=****= 17 - 12**

# Question-46

**Find three irrational numbers between 2 and 2.5 .**

If a and b are any two distinct positive rational numbers such that ab is not a perfect square, then the irrational number lies between a and b.

âˆ´ Irrational number between 2 and 2.5 is , i.e

Irrational number between 2 and is = 2

Irrational number between and 2.5 is =

Thus, the three irrational numbers between 2 and 2.5 are , 2

**Solution:**If a and b are any two distinct positive rational numbers such that ab is not a perfect square, then the irrational number lies between a and b.

âˆ´ Irrational number between 2 and 2.5 is , i.e

Irrational number between 2 and is = 2

^{(1/2)}Ã— 5^{(1/4)}Irrational number between and 2.5 is =

Thus, the three irrational numbers between 2 and 2.5 are , 2

^{(1/2)}Ã— 5^{(1/4)}and (1/2) Ã— 5^{}Ã— 2^{}.

# Question-47

**Simplify by rationalising the denominator:**

=Ã—

**Solution:**=Ã—

**[Rationalising the denominator]****=****=****=****=**()

# Question-48

**Find two irrational numbers lying between and .**

Irrational numbers lying between and is ,i.e = 6

Irrational numbers lying between and 6

Hence two irrational numbers lying between and are 6

**Solution:**Irrational numbers lying between and is ,i.e = 6

^{(1/4)}Irrational numbers lying between and 6

^{(1/4)}is = 2^{(1/4)}Ã— 6^{(1/8)}Hence two irrational numbers lying between and are 6

^{(1/4)}and 2^{(1/4)}Ã— 6^{(1/8)}.

# Question-49

**Simplify by rationalising the denominator: +**

+=

**Solution:**+=

** =**

** = **

# Question-50

**Express as a decimal fraction.**

0.109375

0

Therefore

**Solution:**0.109375

**600**

__576__**240**__192__**480**__448__**320**__320__0

Therefore

**= 0.109375**

# Question-51

**Simplify by rationalising the denominator: -**

-=

=

=

=

**Solution:**-=

=

=

=

# Question-52

**Express as a decimal fraction.**

Therefore = 0.096

**Solution:**Therefore = 0.096

# Question-53

**Express as a decimal fraction.**

0.111

10

10

Therefore

**Solution:**0.111

10

__9__10

__9____1__Therefore

**= 0.111**

# Question-54

**Simplify by rationalising the denominator: +**

+=

=

=

=

=

**Solution:**+=

=

=

=

=

# Question-55

**Represent 0. in the form of .**

= 0. â€¦â€¦â€¦â€¦â€¦(i)

100 = 57. â€¦â€¦â€¦â€¦â€¦(ii)

(ii) â€“ (i)

99 = 57

Therefore = 57/99

**Solution:**= 0. â€¦â€¦â€¦â€¦â€¦(i)

100 = 57. â€¦â€¦â€¦â€¦â€¦(ii)

(ii) â€“ (i)

99 = 57

Therefore = 57/99

# Question-56

**Simplify by rationalising the denominator: -**

-=

**Solution:**-=

** =**

** = **

** = -8**

# Question-57

**Simplify by rationalising the denominator:**

= Ã—

**Solution:**= Ã—

** =**

** = **

** =**

** =**[Rationalising the denominator]

** = **

** = **()

# Question-58

**Represent 0.2****in the form of .****Solution:**

**= 0.2**............(1)

10000= 2341.2

**..............(2)**

(2)-(1) â‡’ 9999= 2341

Therefore = 2341/9999

# Question-59

**Simplify by rationalising the denominator:**

**Solution:**

=Ã—

** = **

** = **

** =
= **Ã—

** = **

** = **(**+ +**)

** = **(3**+**2**+**)

# Question-60

**Which of the following is surds:**

(i)

(ii)

(iii) 3Ã· 6

(i)

(ii)

(iii) 3Ã· 6

**Solution:**

# Question-61

**Express as a pure surd:**

(i)

(i)

**(ii) (iii)**

(iv) (v) (vi)

(iv) (v) (vi)

**Solution:**

(i) = = =

(ii) =

(iii) =

(iv) ===

(v) = ==

(vi) == =

# Question-62

**Express as a mixed surd in its simplest form:**

(i) (ii) (iii)

(iv) (v) (vi)

(i) (ii) (iii)

(iv) (v) (vi)

**Solution:**

(i) = = =2x2x = 4

(ii) = = =

(iii) = = =

(iv) = = = 3x5x=15

(v) == =

(vi) = = = 5x3 = 15.

# Question-63

**Which is greater?**

**(i) or**

(ii) or

(iii) or

(iv) or

(v) or

(vi) or

(ii) or

(iii) or

(iv) or

(v) or

(vi) or

**Solution:**

(i) L.C.M. of 2 and 3 is 6.

Thus, = =

And = =

Therefore >

Hence, >

(ii) L.C.M. of 1 and 4 is 4.

Thus, =

and =

Therefore >

Hence, >

(iii) L.C.M. of 4 and 3 is 12.

Thus, = =

and ==

Therefore >

Hence, >

(iv) L.C.M. of 3 and 4 is 12.

Thus, ==

and = =

Therefore >

Hence, >

(v) L.C.M. of 8 and 4 is 8.

Thus, =

and = =

Therefore >

Hence, >

(vi) L.C.M. of 3 and 4 is 12

Thus, = =

and = =

Therefore >

Hence, >.

# Question-64

**Arrange in descending order of magnitude**

**(i)**

(ii) ,,

(iii) , ,

(ii) ,,

(iii) , ,

**Solution:**

(i) L.C.M. of 3, 4 and 2 is 12

Thus, = =

= =

= =

Therefore > >

Hence, > >

(ii) L.C.M of 3,4 and 3 is 12

Thus,= =

==

= =

.

^{.}. > >

Hence, > >

(iii) L.C.M. of 4,3 and 2 is 12

Thus, ==

= =

= =

Therefore > >

Hence, >>.

# Question-65

**Simplify by combining similar terms:**

5

5

**+20**

**Solution:**

5+20 = (5+20) = 25

# Question-66

**Simplify by combining similar terms:**

**+**

**Solution:**

Reducing into simplest form

**==**3

Therefore 2 += 2+3=(2+3)

**=**5.

# Question-67

**Simplify by combining similar terms:**

**4-3+2**

**Solution:**

Reducing into simplest form

= =2

= = 5

Therefore 4- 3 + 2= 4- 3 x 2 + 2 x 5

= 4-6 + 10

= (4-6+10)

= 8 .

# Question-68

**Simplify by combining similar terms:**

**+-**

**Solution:**

Reducing into simplest form.

= = 2

**=**= 2 x 2= 4

.

^{.}. + - =2 +4 - =(2 + 4 - 1) =5.

# Question-69

**Simplify by combining similar terms:**

**- 3 +4**

**Solution:**

Reducing into simplest form

= = 3

= = 2

.

^{.}. - 3 + 4= 3 - 3 x 2 + 4

= 3- 6 + 4

=

# Question-70

**Simplify by combining similar terms:**

**4--7**

**Solution:**

Reducing into simplest form

= = 2

= =5

= = 2x2 = 4

.

^{.}. 4- - 7 = 4x 2 -5 - 7 x 4

= 8 - 5 -28

= (8-28) - 5

= -20 - 5

# Question-71

**Simplify by combining similar terms:**

**+-**

**Solution:**

Reducing into simplest form

= =

= =

.

^{.}. + -= 2+ 7x2 -

= 2+ 14-

= (2+14 -5)

= 11

# Question-72

**Simplify by combining similar terms:**

**+ -**

**Solution:**

Reducing into simplest form

==2

== 5

= =2x2=4

.

^{.}. 2 +3 - = 2+3x5-4x4

= 4 +15-16

= (4 +15 -16)

= 3

# Question-73

**Simplify by combining similar terms:**

3 - +

3 - +

**Solution:**

Reducing into simplest form

= = 7

= =

.

^{.}. 3 - + = 3 x 7 - + 7 x

= 21- +

=

=

= .

# Question-74

**Simplify and express the result in its simplest form:**

**x**

**Solution:**

x = ==

= .

# Question-75

**Simplify and express the result in its simplest form:**

**x**

**Solution:**

x ==.

# Question-76

**Simplify and express the result in its simplest form:**

x

x

**Solution:**

x= ===2.

# Question-77

**Simplify and express the result in its simplest form:**

x

x

**Solution:**

x=28=28= 28x6

= 168.

# Question-78

**Simplify and express the result in its simplest form:**

**x**

**Solution:**

L.C.M. of 3, 2 is 6

==

==

x=x= =

# Question-79

**Simplify and express the result in its simplest form:**

**x**

**Solution:**

L.C.M. of 3, 4 is 12

= =

**= =**

x = x = =

# Question-80

**Simplify and express the result in its simplest form:**

Ã· 3

Ã· 3

**Solution:**

**Ã· 3**

**=**

= = =.

= = =

# Question-81

**Simplify and express the result in its simplest form:**

(1) Ã· x

(2) x x

(1) Ã· x

(2) x x

**Solution:**

1) L.C.M of 2 and 3 is 6

**= =**

= =

x = x ==

= =

**Ã· (x )=Ã· = = =**

2) L.C.M. of 2,3 and 4 is 12

**=**

=

= ==

x x = x x =

=

= ==

**= = = =**

# Question-82

**Write the simplest rationalising factor of**

(i) (ii) (iii) (iv) (v)

(i) (ii) (iii) (iv) (v)

**Solution:**

(i) 2 x = 2 x = 4

.

^{.}. is the simplest rationalising factor of 2

(ii) x = = = 10

.

^{.}. is the simplest rationalising factor of

(iii) = ==

x = = 5x3 = 15

.

^{.}. is the simplest rationalising factor of

(iv) x = =

= 2x5=10

.

^{.}. is the simplest rationalising factor of

(v) =

Now x == 6

.

^{.}.is the simplest rationalising factor of .

# Question-83

**Express with a rational denominator the following surds:**

**(i) (ii) (iii) (iv) (v) (vi)**

**Solution:**

(i)

The simplest Rational factor of isitself.

.

^{.}. x

(ii)

The simplest R.F of is itself.

.

^{.}.x =

(iii)

=

The simplest R.F of is itself.

.

^{.}. =

(iv)

The simplest R.F of is itself.

.

^{.}. x

(v) The simplest R.F of is itself.

(vi) The simplest R.F of is itself as

.

^{.}. = x=

# Question-84

**(a) Which is greater or ?**

(b) Arrange in ascending order of magnitude , , .

(b) Arrange in ascending order of magnitude , , .

**Solution:**

(a) L.C.M of 2 and 3 is 6

Thus

**=**

Now

(b) L.C.M of 4,3, and 2 is 12

Thus,

**=**

**=**

==

= =

Now,

==

= =

**< <**

.

.

^{.}.**<**

**< .**

# Question-85

**Find the value to three places of decimals, of each of the following. It is given that**

**= 1.414, = 1.732, = 3.162 and = 2.236 (approx.) (i) (ii) (iii) (iv) (v) (vi)**

**Solution:**

(i)

= = = 0.707

(ii)

= x = 0.5773= 0.577

(iii)

= = = 0.3162

(iv)

= = = = =1.0796 = 1.080

(v)

= x = = == = 0.1546= 0.155

(vi) = x =

= = = == 0.655

# Question-86

**If both of**

*and***a****are rational numbers, find the values of a and b in each of the following***b***equalities:**

(i) (ii) (iii)

(iv)

(i) (ii) (iii)

(iv)

**Solution:**

(i)

\

2 -

*a + b*Hence,

**= 2 ,**

*a**= -1*

**b**(ii) 3 + is the conjugate of 3-

\ =

=

=

On comparing both sides

We have

**=**

*a***=**

*b*(iii) 7-4 is the conjugate of 7 + 4

âˆ´

=

=

= 35 -20

On comparing both sides

11-6

We have,

**= 11,**

*a***= -6.**

*b*(iv) is the conjugate of the denominator 3âˆš2 - 2âˆš3

\

=

=

= 2 +

On comparing both sides

2 +

we have,

**= 2 and**

*a***= -**

*b*# Question-87

**Simplify each of the following by rationalizing the denominator**

**:**

**(i)** (ii) (iii) **Solution:**

(i) \

5+ is the rationalizing factor of 5 -

=

(ii)

\

= .

(iii) 2

=

# Question-88

**Rationalize the denominator of**

(i)

(i)

**(ii) (iii)**

**Solution:**

(i) The denominator is a trinomial surd.

We proceed as with a binomial surd by grouping two of the terms. Thus,

= 9 + 6 âˆš5 + 5 - 8

= 6 + 6 âˆš5

= 6(1+ âˆš5)

=

=

=

(ii) The denominator is a trinomial surd.

We proceed as with a binomial surd by grouping two of the terms.

Thus,

= 3 â€“

=

(iii) The denominator is a trinomial surd.

We proceed as with a trinomial surd, by grouping two of the terms.

Thus,

\

# Question-89

**Simplify each of the following:**

(i) (ii)

(i) (ii)

**Solution:**

(i) Let us rationalise the denominator of each term:

\

(ii) Let us rationalize the denominator of each term:

# Question-90

**Taking** and (approx.). Find the value to three places of decimals of each of the following: (i) (ii)

**Solution:**

(i)

=

=

=

=

= (2.236 - 2.449)

= (-0.213)

= - 0.213

(ii)

= 14.268

= 16 - 1.732

= 14.268.

# Question-91

**Simplify (a)**

**+ - (b)**

**Solution:**

(a)

**= =**

**= =**

.

^{.}.

**+ - = +- =**(1+2-3)

**= 0**

(b) = =

= =

.

^{.}.

**=**x = = 10 x 2 =20

# Question-92

Given that = 1.7321, find correct to 3 places of decimals, the value of

**Solution:**

= = 2 x 2 x 2

**=**8

**= =**2 x

**=**

= =

= =

**=**x =

= (8-2-5)

**= =**1.732.

# Question-93

**Simplify by rationalizing the denominator :**

**(i) (ii)**

**Solution:**

(i)

(ii) =

# Question-94

**If** = 2.236 (approx.), evaluate correct to three places of decimals.

**Solution:**

# Question-95

**Simplify :**

**Solution:**

# Question-96

**Simplify : (i) (ii)**

**Solution:**

(ii)

**=**

**=**

# Question-97

**Express each of the following as a mixed surd in the simplest form:**

**Solution:**

# Question-98

**Express each of the following as a mixed surd in the simplest form:**

(i) (ii) 5

(i) (ii) 5

**Solution:**

(i) = = Ã— = =

(ii) 5= 5

**=**5 Ã— 3 Ã— = 15 Ã—

# Question-99

**Arrange in descending order of magnitude:**

, and .

, and .

**Solution:**

The given surds are of order 3,6 and 9 respectively.

The L.C.M. of 3, 6 and 9 is 18.

Reducing each surd to a surd of order 18,

= =

^{ = }

= =

^{ = }

= =

^{ = }

Hence , > >

> > .

# Question-100

**Simplify : +**

**Solution:**

Reducing to the simplest form, we get

= = a.

= = b.

+ = a + b

= (a+b)

# Question-101

**Simplify :**

- 8+ 15+

- 8+ 15+

**Solution:**

- 8 + 15 +

= = 3 ;

= = 6 ;

= = 2 ;

= = 15 ;

- 8 + 15 + = (3 â€“ 8 Ã— 6 + 15 Ã— 2 + 15)

= (3 â€“ 48 + 30 + 15) = 0

# Question-102

**Simplify:**

(3 - 5)(4 + 3)

(3 - 5)(4 + 3)

**Solution:**

# Question-103

**Simplify:**

( Ã· )

( Ã· )

**Solution:**

L.C.M of 6,2 and 3 is 6.

( Ã· ) =

# Question-104

**Find the rationalising factor of**

**Solution:**

*= Â´ Â´ =*

*() Â´ () = abc*Hence the required R.F = =

# Question-105

**Find the rationalising factor of**

**Solution:**

= = Ã— = 2Â´ = 2 Ã— 5

^{(1/5)}

2 Ã— 5

^{(1/5)}Ã— 5

^{(4/5)}= 2 Ã— 5 = 10

Hence the required R.F is 5

^{(1/5)}

**= =**

# Question-106

**Convert and into surds of the same but smallest order.**

**Solution:**

Since = 3

^{1/4 }and = 2

^{1/6}, it follows that the given surds are of order 4 and 6 respectively. The L.C.M. of 4 and 6 is 12.So,we shall convert each one of the given surds into a surd of order 12.

= 3^{1/4 }= 3^{(1/4) (3/3)} = 3^{(3/12)} = ( 3^{3})^{1/12} = (27)^{1/12} =

= 2^{1/6 }= 2^{(1/6) (2/2)} = 2^{(2/12)} = (2^{2})^{1/12 }= (4)^{1/12} =

# Question-107

**If = 2.236 and = 3.162, evaluate**

**Solution:**

The rationalising factor of is .

= Ã— =

=

= - (1/2) = 2.236 â€“(1/2)(3.162)

= (2.236 â€“ 1.581) = 0.655

# Question-108

**If =**

**, find the values of***a - b***and***a**b.***Solution:**

= Ã—

=

=

=

# Question-109

**Express with a rational denominator.**

**Solution:**

= Ã—

** = **** ** Ã— ** = **

** =**Ã— ** = **

** = = **

# Question-110

**Find the value of correct to three places of decimal, it being given that = 1.4142.**

**Solution:**

= Ã— =

= =

= Ã—

= =

=