Maths M.C.Q

 When we divide $1$ by $999$ it result is $0.001001001001001$

 $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option $(b).$

 An irrational number between $a$ and $b$ is given by $\sqrt{\text{ab}}$
So, an irrational number between $5$ and $6$ is given by $\sqrt{5\times6}$
Hence, the correct answer is option $(c).$

 $\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}\\ \ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
Now, $\text{x}^2+\text{xy}+\text{y}^2$
$=\big(5 -2\sqrt6\big)^2+\big(5-2\sqrt6\big)\big(5+2\sqrt6\big)+\big(5+2\sqrt6\big)^2$
$=\big(25+24-20\sqrt6\big)+(25-24)+\big(25+24+20\sqrt6\big)$
$=49-20\sqrt6+1+49+20\sqrt6$
$=99$
Hence, correct option is $(b).$

 $\left(x^{a-b}\right)^{a+b} \times\left(x^{b-c}\right)^{b+c} \times\left(x^{c-a}\right)^{c+a}$
$\Rightarrow\text{x}^{\text{a}^{2}-\text{b}^2}\times\text{x}^{\text{b}^2-\text{c}^2}\times\text{x}^{\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^0=1$

 We have to find the value of $\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}$
So,
$\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}=\Big(\frac{10^2}{3^2}\Big)^{-\frac{3}{2}}$
$=\frac{10^{2\times\frac{3}{2}}}{3^{2\times\frac{3}{2}}}$
$=\frac{10^{-3}}{3^{-3}}$
$\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}=\frac{\frac{1}{10^3}}{\frac{1}{3^3}}$
$=\frac{1}{10\times10\times10}\times\frac{3\times3\times3}{1}$
$=\frac{3\times3\times3}{10\times10\times10}$
Since, $\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}$ is equal to $\Big(\frac{100}{9}\Big)^{\frac{3}{2}},\ \frac{1}{\Big(\frac{100}{9}\Big)^{\frac{3}{2}}},\ \frac{3\times3\times3}{10\times10\times10}.$
Hence the correct choice is $d$.

 $\sqrt{3-2\sqrt2}$
$=\sqrt{2+1-2\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2-2\big(\sqrt2\big)(1)}$
$=\sqrt{\big(\sqrt2-1\big)^2}$
$=\sqrt2-1$
Hence, correct option is $(a).$

 $(1296)\frac{-1}{4}=(6^4)^{\frac{-1}{4}}$
$=(6)^{-1}=\frac{1}{6}$

 Divide numerator and denominator by $3.$

$\frac{1}{\sqrt9-\sqrt8}$
$=\frac{1}{\sqrt9-\sqrt8}\times\frac{{\sqrt9+\sqrt8}}{{\sqrt9+\sqrt8}}$
$\frac{{\sqrt9+\sqrt8}}{\big(\sqrt9\big)^2-\big(\sqrt8\big)^2}$
$=\frac{{\sqrt9+\sqrt8}}{9-8}$
$={\sqrt9+\sqrt8}$
$=3+2\sqrt2$
Hence, correct option is $(a).$

We have to find the value of $10^{-y}$
Given that $10^{2 y}=25$, therefore,
$10^{2 y}=25$
$\left(10^y\right)^2=5^2$
$(10^\text{y})^{2\times\frac{1}{2}}=5^{2\times\frac{1}{2}}$
$\frac{10^{\text{y}}}{1}=\frac{5}{1}$
$\frac{1}{5}=\frac{1}{10^\text{y}}$
$\frac{1}{5}=10^\text{y}$
Hence the correct option is $d.$

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is $(c).$

$x^{p-q} x^{q-r} x^{r-p}$
$=x^{p-q+q-r+r-p}$
$=x^0$
$=1$

 We have to find the value of $\{2-3(2-3)^3\}^3.$ So,
$\{2-3(2-3)^3\}^3=\{2-3(-1)^3\}^3$
$=\{2(-3\times-1)^3\}^3$
$=\{2+3\}^3$
$=5^3=125$
The value of $\{2-3(2-3)^3\}^3$ is $125$
Hence the correct choice is $b.$

$\frac{5-\sqrt3}{2+\sqrt3}$
$=\frac{5-\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{\big(5-\sqrt3\big)\big(2-\sqrt3\big)}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{10-5\sqrt3-2\sqrt3+3}{4-3}$
$=\frac{13-7\sqrt3}{1}$
$=13-7\sqrt3$
$⇒ x = 13$ and $y = -7$
Hence, correct option is $(a).$

 $9^3+(-3)^3-6^3$
$=729-27-216$
$=729-243$
$=486$

 Let $\text{x}=23.\overline{43}=23.434343...(1)$
Now, $100\text{x}=2343.43333...(2)$
Subtracting equation $(1)$ from $(2),$ we get
$99\text{x}=2320$
$\Rightarrow\text{x}=\frac{2320}{99}$
Hence, option $(a)$ is correct.

 $\sqrt[4]{(64)^{-2}}=\sqrt[4]{(8^2)^{-2}}=8^{-4\times\frac{1}{4}}=8^{-1}=\frac{1}{8}$
Hence, the correct answer is option $(a).$

 We have to find $5\sqrt[\text{n}]{64}$ provided $\sqrt{5^\text{n}}=125$
So,
$\sqrt{5^\text{n}}=125$
$5^{\text{n}\times\frac{1}{2}}=5^3$
$\frac{\text{n}}{2}=3$
$\text{n}=3\times2$
$\text{n}=6$
Substitute $\text{n}=6$ in $5^{\sqrt[\text{n}]{64}}$ to get
$5^{\sqrt[\text{n}]{64}}=5^{2^{6\times\frac{1}{6}}}$
$=5\times5$
$=25$
Hence the value of $5^{\sqrt[\text{n}]{64}}$ is $25$
The correct choice is $a.$

 $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
$\Rightarrow\frac{(3+2+2\sqrt{6})+3+2-2\sqrt{6}}{3-2}$
$\Rightarrow10$

 Let $\text{x}=0.\overline{54}$
Then, $\text{x}=54.5454 \ ...(\text{i})$
$\therefore100\text{x}=54.5454 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii),$ we get
$99\text{x}=54$
$\Rightarrow\text{x}=\frac{54}{99}=\frac{6}{11}$
Hence, the correct option is $(c).$

 $\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{5}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{25}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\text{p}=\frac{25}{\sqrt{7}\times\sqrt{7}}=\frac{25}{7}$
Hence, the correct option is $(b).$

 $(32)^{\frac{1}{5}}+(-7)^0+(64)^{\frac{1}{2}}$
$=2+1+8$
$=11$

 $(\sqrt{\text{x}^3})^{\frac{2}{3}}$
$=(\text{x}\frac{3}{2})^{\frac{2}{3}}$
$=\text{x}$

 We have to find the value of $\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$ if $x = 2, y = 4$
Substitute $x = 2, y = 4,$ in $\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$ to get,
$\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$
$=\Big(\frac{2}{4}\Big)^{2-4}+\Big(\frac{4}{2}\Big)^{4-2}$
$=\Big(\frac{2}{4}\Big)^{-2}+\Big(\frac{4}{2}\Big)^{2}$
$=\Big(\frac{1}{2}\Big)^{-2}+(2)^2$
$=\Big(\frac{1}{2^{-2}}\Big)+4$
$\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$
$=\frac{1}{\frac{1}{2^2}}+4$
$=\frac{1}{\frac{1}{4}}+4$
$=1\times\frac{4}{1}+4$
$=4+4$
$=8$
Hence the correct choice is $b.$

 Because it is the square of $14$ and can be written in the form of $\frac{\text{p}}{\text{q}}.$

$ = 2 + 1 + 8$
$= 11$

 $\left(3^3\right)^2=9^x$
$\Rightarrow\left(3^2\right)^3=\left(3^2\right)^x$
$\Rightarrow x=3$
Then $5^x=5^3=125$
Hence, the correct option is $(d).$

 $\sqrt{3}=1.73$ and $\sqrt{5}=2.236$
$2.1$ lies in between these two numbers.

Let $\text{x}=0.\overline{001}=0.001001001...(1)$
Now, $1000\text{x}=001.001001001...(2)$
Subtracting equation $(1)$ from $(2),$ we get
$999\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{999}$
Hence, option $(d)$ is correct.

 Let $\text{x}=1.\overline{6}$
$\Rightarrow\text{x}=1.666 \ ...(\text{i})$
$\therefore10\text{x}=16.666 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii),$ we get
$9\text{x}=15$
$\Rightarrow\text{x}=\frac{15}{9}=\frac{5}{3}$
Hence, the correct option is $(c).$

 $\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$
$=(3^4)^{\frac{1}{2}\times\frac{1}{2}}$
$=(3^4)^{\frac{1}{4}}$
$=3$

 $\sqrt{48}=\sqrt{16\times3}=4\sqrt3$
$\sqrt{32}=\sqrt{16\times2}=4\sqrt2$
$\sqrt{27}=\sqrt{9\times3}=3\sqrt3$
$\sqrt{18}=\sqrt{9\times2}=3\sqrt2$
Now, $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}=\frac{4\sqrt3+4\sqrt2}{3\sqrt3+3\sqrt2}$
$=\frac{4\big(\sqrt{\not3}+\sqrt{\not2}\big)}{3\big(\sqrt{\not3}+\sqrt{\not2}\big)}$
$=\frac{4}{3}$
Hence, correct option is $(a).$

$\sqrt{13-\text{a}\sqrt10}=\sqrt8+\sqrt5$
Squaring both sides, we get
$13-\text{a}\sqrt{10}=8+5+2\sqrt{40}$
$={13}-\text{a}\sqrt{10}-{13}=2\times2\sqrt{10}$
$=-\text{a}\sqrt{10}=4\sqrt{10}$
$\Rightarrow\text{a}=-4$
Hence, correct option is $(c).$