Number Systems Test

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Number Systems Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Rationalise the denominator of :
$$\displaystyle\ \frac{2\sqrt{2}}{\sqrt{3}}$$

A
$$\displaystyle\ \frac{2\sqrt{7}}{3}$$

B
$$\displaystyle\ \frac{2\sqrt{6}}{3}$$

C
$$\displaystyle\ \frac{2\sqrt{2}}{3}$$

D
$$\displaystyle\ \frac{2\sqrt{6}}{10}$$

Solution

$$\cfrac{2\sqrt{2}}{\sqrt{3}}$$
$$=\cfrac{2\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$$
$$=\cfrac{2\sqrt{6}}{3}$$
Question 2
1 / -0

Rationalise the denominator of :
$$\displaystyle\ \frac{6}{\sqrt{10}-2}$$

A
$$\sqrt{10}-2$$

B
$$\sqrt{10}+2$$

C
$$2\sqrt{10}$$

D
None of these

Solution

$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 10 } +2:\\ \cfrac{6(\sqrt { 10 } +2)}{(10-4)}=\sqrt { 10 } +2\\ $$
Question 3
1 / -0

Find conjugate of:
3$$\sqrt{2}$$ -1

A
3$$\sqrt{2}$$ +1

B
3$$\sqrt{2}$$ -1

C
3$$\sqrt{2}$$

D
None of these

Solution

$$3\sqrt{2}+1$$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called rationalizing factor (RF) of the other
$$(3\sqrt{2}+1) \times (3\sqrt{2}-1)=(3\sqrt{2})^2-1^2=18-1=17$$
The Rationalizing factor of $$3\sqrt{2}+1$$ is $$3\sqrt{2}-1$$
Question 4
1 / -0

Rationalise the denominator and find the equivalent of : $$\displaystyle\ \frac{2\sqrt{3}}{\sqrt{2}}$$

A
$$\sqrt{12}$$

B
$$\sqrt{3}$$

C
$$\sqrt{6}$$

D
$$\sqrt{2}$$

Solution

$$\cfrac{2\sqrt{3}}{\sqrt{2}}$$
Rationalising the denominator, we get
$$=\cfrac{2\sqrt{3}\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$$
$$=\cfrac{2\sqrt{6}}{2}$$
$$=\sqrt{6}$$
Question 5
1 / -0

Find conjugate of:
$$3-\sqrt{5}$$

A
$$3+\sqrt{5}$$

B
$$3-\sqrt{5}$$

C
$$3\sqrt{5}$$

D
None of these

Solution

$$3-\sqrt { 5 } $$
The procedure of multiplying a surd by another surd to get a rational number is called Rationalization. The operands are called rationalizing factor (RF) of the other
$$(3-\sqrt { 5 } ) \times (3+\sqrt { 5 })=3^2-(\sqrt{5})^2\\=9-5\\=4$$
The Rationalizing factor of $$3-\sqrt { 5 }$$ is $$3+\sqrt { 5 }$$
Question 6
1 / -0

Rationalise the denominator of:
$$\displaystyle\ \frac{2}{\sqrt{5}+\sqrt{3}}$$

A
$$\sqrt{5}-\sqrt{3}$$

B
$$\sqrt{4}-\sqrt{3}$$

C
$$\sqrt{2}-\sqrt{3}$$

D
$$\sqrt{6}-\sqrt{3}$$

Solution

$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 5 } -\sqrt { 3 } :\\\cfrac{ 2(\sqrt { 5 } -\sqrt { 3 } )}{(5-3)}=\sqrt { 5 } -\sqrt { 3 } \\ $$
Question 7
1 / -0

Rationalise the denominator

(i) $$\displaystyle\ \frac{22}{2\sqrt{3}+1}$$

A
$$(2\sqrt{3}-1)$$

B
$$3(2\sqrt{3}-1)$$

C
$$2(2\sqrt{3}-1)$$

D
$$4(2\sqrt{3}-1)$$

Solution

$$\\ Multiplying\quad and\quad dividing\quad by\quad 2\sqrt { 3 } -1:\quad \\ \cfrac{22(2\sqrt { 3 } -1)}{(12-1)}=2(2\sqrt { 3 } -1)\\ $$
Question 8
1 / -0

Write the simplest rationalisation factor of the following surds:
$$\sqrt{32}$$

A
$$\sqrt{2}$$

B
$$\sqrt{7}$$

C
$$\sqrt{5}$$

D
$$\sqrt{3}$$

Solution

$$\sqrt{32}$$
$$=\sqrt{2\times2\times2\times2\times2}$$
$$=4\sqrt2$$
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called Rationalizing factor (RF) of the other.
Here, $$\sqrt 2$$ is the Rationalisation Factor.
Question 9
1 / -0

Write the simplest rationalization factor of the following surds :
$$2\sqrt{2}$$

A
$$\sqrt{3}$$

B
$$\sqrt{8}$$

C
$$\sqrt{2}$$

D
$$\sqrt{4}$$

Solution

$$2\sqrt 2$$
$$=2\sqrt2$$
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called Rationalizing Factor (RF) of the other.
Here, $$\sqrt 2$$ is the Rationalisation Factor.
Question 10
1 / -0

Write the simplest rationalization factor of the following surds :
$$4\sqrt[3]{3}$$

A
$$\sqrt[3]{9}$$

B
$$\sqrt[2]{9}$$

C
$$\sqrt{9}$$

D
$$\sqrt{3}$$

Solution

In $$4\sqrt[3]{3}$$,the simplest rationalization factor is$$\sqrt[3]{3^2}=\sqrt[3]{9}$$

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Number Systems Test - 19

Result

Number Systems Test - 19

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Question 1

$$\displaystyle\ \frac{2\sqrt{7}}{3}$$

$$\displaystyle\ \frac{2\sqrt{6}}{3}$$

$$\displaystyle\ \frac{2\sqrt{2}}{3}$$

$$\displaystyle\ \frac{2\sqrt{6}}{10}$$

Solution

Question 2

$$\sqrt{10}-2$$

$$\sqrt{10}+2$$

$$2\sqrt{10}$$

None of these

Solution

Question 3

3$$\sqrt{2}$$ +1

3$$\sqrt{2}$$ -1

3$$\sqrt{2}$$

None of these

Solution

Question 4

$$\sqrt{12}$$

$$\sqrt{3}$$

$$\sqrt{6}$$

$$\sqrt{2}$$

Solution

Question 5

$$3+\sqrt{5}$$

$$3-\sqrt{5}$$

$$3\sqrt{5}$$

None of these

Solution

Question 6

$$\sqrt{5}-\sqrt{3}$$

$$\sqrt{4}-\sqrt{3}$$

$$\sqrt{2}-\sqrt{3}$$

$$\sqrt{6}-\sqrt{3}$$

Solution

Question 7

$$(2\sqrt{3}-1)$$

$$3(2\sqrt{3}-1)$$

$$2(2\sqrt{3}-1)$$

$$4(2\sqrt{3}-1)$$

Solution

Question 8

$$\sqrt{2}$$

$$\sqrt{7}$$

$$\sqrt{5}$$

$$\sqrt{3}$$

Solution

Question 9

$$\sqrt{3}$$

$$\sqrt{8}$$

$$\sqrt{2}$$

$$\sqrt{4}$$

Solution

Question 10

$$\sqrt[3]{9}$$

$$\sqrt[2]{9}$$

$$\sqrt{9}$$

$$\sqrt{3}$$

Solution

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