Number Systems Test

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Number Systems Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following numbers are rational ?

A
$$1$$

B
$$ -6$$

C
$$3\dfrac{1}{2}$$

D
All above are rational

Solution

$$\Rightarrow$$ A rational number is a type of real numbers which can be expressed in the form of $$\dfrac pq, $$ where $$ q \neq0.$$
$$\Rightarrow$$ All the numbers are rational as they are in the form of $$\dfrac pq, $$ where $$ q \neq0.$$

i.e, $$\dfrac{1}{1}\ , \dfrac{-6}{1}\ ,\dfrac{7}{2}$$
Question 2
1 / -0

Find the nine rational numbers between $$0$$ and $$1$$.

A
$$0.1,0.2,0.3, ... ,0.9$$

B
$$ 1.1,0.2 ,10.3, ... ,0.9$$

C
$$0.1,0.2,0.3, ... ,20.9 $$

D
$$0.1 , 0.2 , 10.3 , ... , 0.9 $$

Solution

$$0 < (0+0.1)=0.1 <(0.1+0.1)= 0.2 < (0.2+0.1)=0.3 < ... < (0.8+1)=0.9 <(0.9+0.1)= 1$$
$$0 < 0.1 < 0.2 < 0.3 < ... < 0.9 < 1$$

$$\therefore$$ The nine rational numbers between $$0$$ and $$1$$ are
$$0.1,0.2,0.3, ... ,0.9$$
Question 3
1 / -0

Simplify:
$$3\sqrt{3} + 10\sqrt{3}$$

A
$$13\sqrt{3}$$

B
$$10\sqrt{3}$$

C
$$12\sqrt{3}$$

D
$$11\sqrt{3}$$

Solution

$$3 \sqrt{3}+ 10 \sqrt{3} =13 \sqrt{3}$$
Question 4
1 / -0

The value of $$5\sqrt{3} - 3\sqrt{12} + 2\sqrt{75} $$ on simplifying is :

A
$$5\sqrt{3}$$

B
$$6\sqrt{3}$$

C
$$\sqrt{3}$$

D
$$9\sqrt{3}$$

Solution

The given expression is
$$5\sqrt { 3 } -3\sqrt { 12 } +2\sqrt { 75 } \\ =5\sqrt { 3 } -6\sqrt { 3 } +10\sqrt { 3 } \\ =9\sqrt { 3 }$$.
Therefore, option $$D$$ is correct.
Question 5
1 / -0

Find the five rational numbers between $$\displaystyle \frac{1}{2}$$ and $$\displaystyle \frac{3}{2}$$

A
$$0.5 < 0.6 < 0.7 < 0.8 ... < 1.1 < ... < 1.15 < 1.50$$

B
$$0.5 < 0.6 < 1.7 < 3.8 ... < 1.8< ... < 1.15 < 1.50$$

C
$$0.5 < 0.6 < 0.7 < 2.8 ... < 1.1 < ... < 1.15 < 1.50$$

D
$$0.5 < 0.6 < 0.7 < 0.8 ... < 3.1 < ... < 1.15 < 1.50$$
Question 6
1 / -0

Which of the following order is correct for rational numbers between $$\displaystyle \frac{4}{11}$$ and $$\displaystyle \frac{9}{16}?$$

A
$$\dfrac{4}{11}$$, $$\dfrac{67}{176}$$, $$\dfrac{79}{176}$$, $$\dfrac{9}{16}$$

B
$$\dfrac{4}{11}$$, $$\dfrac{67}{276}$$, $$\dfrac{79}{176}$$, $$\dfrac{9}{16}$$

C
$$\displaystyle \frac{4}{11}, \frac{17}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$$

D
$$\displaystyle \frac{4}{11}, \frac{27}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$$

Solution

$$L.C.M$$ of $$11$$ and $$16$$ is $$11\times 16 =176$$

$$\dfrac{4}{11}=\dfrac{4\times 16}{11\times 16}=\dfrac{64}{176}$$

$$\dfrac{9}{16}=\dfrac{9\times 11}{16\times 11}=\dfrac{99}{176}$$

rational numbers between $$\dfrac{4}{11}$$ and $$\dfrac{9}{16}$$ are

$$\dfrac{67}{176}$$, $$\dfrac{79}{176}$$
Question 7
1 / -0

Rationalise the denominator of:
(i) $$\displaystyle\ \frac{1}{2-\sqrt{3}}$$

A
$$2+\sqrt{7}$$

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

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

D
$$2+\sqrt{5}$$

Solution

$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad 2+\sqrt { 3 } :\\ \cfrac{(2+\sqrt { 3 } )}{(4-3)}=2+\sqrt { 3 } \\ $$
Question 8
1 / -0

Find conjugate of:
$$\sqrt{3}+\sqrt{2}$$

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

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

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

D
None of these

Solution

$$\sqrt{3}+\sqrt{2}$$
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
$$(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=9-4=5$$
The Rationalizing factor of $$\sqrt{3}+\sqrt{2}$$ is $$\sqrt{3}-\sqrt{2}$$
Question 9
1 / -0

Rationalise the denominator of :
$$\displaystyle\ \frac{5}{\sqrt{7}+\sqrt{2}}$$

A
$$\sqrt{7}+\sqrt{2}$$

B
$$\sqrt{7}-\sqrt{2}$$

C
$$\sqrt{7}\sqrt{2}$$

D
None of these

Solution

$$\quad \\ Multiplying\quad and\quad dividing\quad by\quad \sqrt { 7 } -\sqrt { 2 } :\\ \cfrac{5(\sqrt { 7 } -\sqrt { 2 } )}{(7-2)}=\sqrt { 7 } -\sqrt { 2 } \\ $$
Question 10
1 / -0

Rationalise the denominator of :
$$\displaystyle\ \frac{\sqrt{6}}{\sqrt{12}}$$

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

B
$$\displaystyle\ \frac{\sqrt{2}}{5}$$

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

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

Solution

$$\cfrac{\sqrt6}{\sqrt{12}} $$=$$\cfrac {\sqrt6\times\sqrt{12}}{\sqrt{12}\times\sqrt{12}}$$=$$\cfrac{6\sqrt2}{12}$$=$$\cfrac{\sqrt2}{2}$$

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Number Systems Test - 18

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Number Systems Test - 18

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Question 1

$$1$$

$$ -6$$

$$3\dfrac{1}{2}$$

All above are rational

Solution

Question 2

$$0.1,0.2,0.3, ... ,0.9$$

$$ 1.1,0.2 ,10.3, ... ,0.9$$

$$0.1,0.2,0.3, ... ,20.9 $$

$$0.1 , 0.2 , 10.3 , ... , 0.9 $$

Solution

Question 3

$$13\sqrt{3}$$

$$10\sqrt{3}$$

$$12\sqrt{3}$$

$$11\sqrt{3}$$

Solution

Question 4

$$5\sqrt{3}$$

$$6\sqrt{3}$$

$$\sqrt{3}$$

$$9\sqrt{3}$$

Solution

Question 5

$$0.5 < 0.6 < 0.7 < 0.8 ... < 1.1 < ... < 1.15 < 1.50$$

$$0.5 < 0.6 < 1.7 < 3.8 ... < 1.8< ... < 1.15 < 1.50$$

$$0.5 < 0.6 < 0.7 < 2.8 ... < 1.1 < ... < 1.15 < 1.50$$

$$0.5 < 0.6 < 0.7 < 0.8 ... < 3.1 < ... < 1.15 < 1.50$$

Question 6

$$\dfrac{4}{11}$$, $$\dfrac{67}{176}$$, $$\dfrac{79}{176}$$, $$\dfrac{9}{16}$$

$$\dfrac{4}{11}$$, $$\dfrac{67}{276}$$, $$\dfrac{79}{176}$$, $$\dfrac{9}{16}$$

$$\displaystyle \frac{4}{11}, \frac{17}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$$

$$\displaystyle \frac{4}{11}, \frac{27}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$$

Solution

Question 7

$$2+\sqrt{7}$$

$$2+\sqrt{2}$$

$$2+\sqrt{3}$$

$$2+\sqrt{5}$$

Solution

Question 8

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

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

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

None of these

Solution

Question 9

$$\sqrt{7}+\sqrt{2}$$

$$\sqrt{7}-\sqrt{2}$$

$$\sqrt{7}\sqrt{2}$$

None of these

Solution

Question 10

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

$$\displaystyle\ \frac{\sqrt{2}}{5}$$

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

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

Solution

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