Maths M.C.Q

A rational number between two given number $a$ and $b$ is given by $\frac{\text{a}+\text{b}}{2},$
Hence, a rational number between two given number $\sqrt{2}=1.414$ and $\sqrt{3}=1.732.$
$=\frac{\sqrt{2}+\sqrt{3}}{2}=\frac{1.414+1.732}{2}=1.5$

$\frac{2^0+7^0}{5^0}$
$=\frac{1+1}{1}$
$=2$

$\Big(\frac{12^{\frac{1}{5}}}{27^{\frac{1}{5}}}\Big)^{\frac{5}{2}}.$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{5}\times\frac{5}{2}}$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{2}}$
$\Rightarrow\frac{2\sqrt{3}}{3\sqrt{3}}$
$\Rightarrow\frac{2}{3}$

We know the formula $a^2-b^2 = (a - b) (a - b)$
Here put $\text{a}=2$ and $\text{b}=\sqrt{3}$
So, $2^2-(\sqrt{3})^2=4-3=1$

Let $\text{x}=0.\overline{3}$
i.e., $\text{x}=0.3333 \ ...(\text{i})​​$
$\Rightarrow10\text{x}=3.3333 \ ...(\text{ii})​​$
On subtracting $(i)$ and $(ii),$ we get
$9\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{9}$
Let $\text{y}=0.\overline{4}$
i.e., $\text{y}=0.4444 \ ...(\text{i})​​$
$\Rightarrow10\text{y}=4.4444 \ ...(\text{ii})​​$
On subtracting $(i)$ and $(ii),$ we get
$9\text{y}=4$
$\Rightarrow\text{y}=\frac{4}{9}$
$\therefore0.\overline{3}+0.\overline{4}=\text{x}+\text{y}=\frac{\text{3}}{\text{9}}+\frac{\text{4}}{\text{9}}=\frac{\text{7}}{\text{9}}$
Hence, the correct answer is option $(b).$

$\frac{1}{\sqrt{3}-\sqrt{2}}$
$\Rightarrow\frac{1}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}$
$\Rightarrow1.732+1.414$
$\Rightarrow3.146$

An irrational number between $a$ and $b$ is given by $\sqrt{\text{ab}}$
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is
$\sqrt{\sqrt{2}+\sqrt{3}}=\big(\sqrt{6}\big)^{\frac{1}{2}}$
$=6^{\frac{1}{2}\times\frac{1}{2}}$
$=6^{\frac{1}{4}}$
Hence, the correct answer is option $(d).$

$(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$
$=(\sqrt{11})^2-(\sqrt{7})^2$
$=11-7$
$=4$

$\text{x}+\sqrt{15}=4$
$=\text{x}=4-\sqrt{15}\Rightarrow\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}$
$\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}\times\frac{4+\sqrt{15}}{4+\sqrt{15}}=\frac{4+\sqrt{15}}{(4)^2-\big(\sqrt{15}\big)^2}\\ \ =\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15}$
Now, $\text{x}+\frac{1}{\text{x}}=4-\sqrt{15}+4+\sqrt{15}=8$
Hence, correct option is $(c).$

Sometimes rational, sometimes irrational The product of any two irrational numbers is sometimes rational and sometimes irrational.
Hence, $(D)$ is the correct answer.
For example:
rational
$(2+\sqrt{3})(2-\sqrt{3})$
$(2)^2-(\sqrt{3})^2$
$4-3=1$
irrational
$(2+\sqrt{3})(1-\sqrt{3})$
$2(1-\sqrt{3})+\sqrt{3}(1-\sqrt{3})$
$2-2\sqrt{3}+\sqrt{3}-3$
$-1-\sqrt{3}$

The decimal expansion of $\sqrt{23}=4.79583152...,$ which is non-terminating, nonrecurring.
Hence, it is an irrational number.
Hence, the correct opion is $(a).$

$x^{a-b} x^{b-c} x^{c-a}$
$\Rightarrow x^{a-b+c+c-a}$
$\Rightarrow x^0$
$=1$

When we divide $2$ by $11$
We have value $= 0.181818$
Which is $0.\overline{18}$

$25^{\frac{1}{3}}\times5^{\frac{1}{3}}$
$=5^{\frac{2}{3}}\times5^{\frac{1}{3}}$
$=(5)^{\frac{2+1}{3}}\Leftrightarrow5$

After rationalizing:
$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}\times\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$

$\sqrt{7+\sqrt{48}}$
$=\sqrt{7+2\sqrt{12}}$
$=\sqrt{4+3+2\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4\big)^2+\big(\sqrt3\big)^2+2\times\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4+\sqrt3\big)^2}$
$=\pm\big(\sqrt4+\sqrt3\big)$
Positive value is $\sqrt4+\sqrt3=2+\sqrt3$
Hence, correct option is $(c).$

Simplify $\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=\big(256\big)^{-(2^2)^{-\frac{3}{2}}}$
$=\big(256\big)^{\Big(2^{2\times-\frac{3}{2}}\Big)}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=\big(256\big)^{-(2)^{(-3)}}$
$\big(256\big)^{\Big(-4^{-\frac{3}{2}}\Big)}=\big(256\big)^{\frac{1}{(-2) ^3}}$
$=\big(256\big)^{\frac{1}{-8}}$
$=\big(2^8\big)^{\frac{1}{-8}}$
$=2^{8\times\frac{1}{-8}}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=2^{8\times\frac{1}{-8}}=\frac{1}{2}$
Hence the correct choice is $d.$

$7.\bar{2}=7+0.\bar{2}$
i.e., $7+\frac{2}{9}$
i.e., $\frac{63+2}{9}=\frac{65}{9}$

On number line, we have $-\infty$ to $\infty$ numbers, consisting $-\infty...-4,-3,-2,-1,0,1,2,3,4...\infty, 1.12,1.14$ and $1.41406532, 3.146201286295...$ etc. That means on number line, there are natural nmbers $(1, 2, 3, 4 ...),$ integers, rational numbers $\frac{1}{2},\frac{1}{3},1.3333,$ irrational numbers $1.4148625385...$
But if we see every number as a complete family, it becomes
Real numbers $($any number which can be represent on Real axes$)$
So, every point on the number line reoresent a unique real number which contains every type.
Hence, option $(a)$ is correct.

$0.4014001400014$ is the irrational number.
Because an irrational number is non terminating and non repeation.

For $\frac{\text{x}}{\text{x}^{1.5}}=8\text{x}^{-1}$ we have to find the value of $x.$
So,
$\frac{\text{x}^1}{\text{x}^{1.5}}=8\text{x}^{-1}$
$\text{x}^{1-1.5}8\text{x}^{-1}$
$\text{x}^{-0.5}=2^3\text{x}^{-1}$
$\frac{\text{x}^{0.5}}{\text{x}^{-1}}=2^3$
$\frac{\text{x}^{-\frac{5}{10}}}{\text{x}^{-1}}=2^3$
$\text{x}^{-\frac{1}{2}+1}=2^3$
$\text{x}^{\frac{1}{2}+\frac{2}{2}}=2^3$
$\text{x}^{\frac{-1+2}{2}}=2^3$
$\text{x}^{\frac{1}{2}}=2^3$
By raising both sides to the power $2$ we get
$\text{x}^{\frac{1}{2}\times2}=2^{3\times2}$
$\text{x}^{\frac{1}{2}\times2}=2^6$
$\text{x}^1=64$
The value of $x$ is $64$
Hence the correct alternative is $d.$

Between any two rational number there are many rational number,
Example $- 4$ and $8$
We have $5, 6, 7, 7.5.......$ and many more.

$(7)^{\frac{1}{2}}\times(8)^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}$
Hence, the correct answer is option $(b).$

The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is $2\sqrt{2}+\sqrt{3}$
Hence, the correct option is $(b).$

$0.\overline{57}=\frac{57-5}{90}$
$=\frac{52}{90}=\frac{26}{45}$

We have to find the value of m for $\Bigg[\bigg\{\Big(\frac{1}{7^2}\Big)^{-2}\bigg\}^{-\frac{1}{3}}\Bigg]^{\frac{1}{4}}=7^{\text{m}},$
$\Rightarrow\bigg[\Big\{\frac{1}{7^{2\times-2}}^{-2}\Big\}^{-\frac{1}{3}}\bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\bigg[\Big\{\frac{1}{7^{-4}}\Big\}^{-\frac{1}{3}}\bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\bigg\{\frac{1}{7^{-4\times\frac{-1}{3}}}\bigg\}^{-\frac{1}{3}}\Bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\bigg\{\frac{1}{7^{\frac{4}{3}}}\bigg\}\Bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\frac{1}{7^{\frac{4}{3}\times\frac{1}{4}}}\Bigg]=7^{\text{m}}$
$\Rightarrow\Bigg[\frac{1}{7^{\frac{1}{3}}}\Bigg]=7^{\text{m}}$
By using rational exponents $\frac{1}{\text{a}^{\text{n}}}=\text{a}^{-\text{n}}$
$7^{\frac{-1}{3}}=7^{\text{m}}$
Equating power of exponents we get $-\frac{1}{3}=\text{m}$
Hence the correct choice is a.