Square And Qube

$$\,\,\,\,\,\,1|\overline 1 \,\,\overline {58} \,\,\overline {76} \,(126$$
$$\,\,\,\,\,\,\,\,\,|1$$
$$\,\,\,\,\,\,\,\,\,| - - - - - - - - $$
$$\,\,22|\,\,\,\,\,\,58$$
$$\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44$$
$$\,\,\,\,\,\,\,\,\,| - - - - - - - - $$
$$246\,|\,\,\,\,\,\,\,14\,76$$
$$\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,14\,76$$
$$\,\,\,\,\,\,\,\,\,| - - - - - - - $$
$$\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,x$$
$$\,\,\,\,\,\,\,\,\,| - - - - - - - $$
$$\therefore \sqrt {15876} = 126$$

$$\,\,\,\,\,\,\,\,\,\,1|\overline 3 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.732$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,|1$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - $$
$$\,\,\,\,\,\,27|\,\,2\,\,\,00$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,1\,\,\,89$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - $$
$$\,\,\,343\,|\,\,\,\,\,\,\,\,\,11\,\,00$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,10\,\,29$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - $$
$$3492\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,71\,\,00$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,69\,\,84$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,16$$
$$\therefore \sqrt 3 $$
$$ = 1.73$$
$$ = \frac{{173}}{{100}}$$

The squares of numbers having 1 and 9 as the unit's digit end in the digit 1.
$${\text{Such numbers are,}}$$
$$1,9,11,19,21,29,31,39,41,49{\text{ i}}{\text{.e}}{\text{.,}}$$
$${\text{There are 10 such numbers}}{\text{.}}$$
$$\therefore {\text{Required percentage}}$$
$$ = \left( {\frac{{10}}{{50}} \times 100} \right)\% $$
$$ = 20\% $$

$${\text{Let the given number be }}x$$
$${\text{Then,}}$$
$$ ⇒ {x^2} - 25 = {\left( {x - 25} \right)^2}{\text{ }}$$
$$ ⇒ {x^2} - 25 = {x^2} + 625 - 50x$$
$$ ⇒ 50x = 650$$
$$ ⇒ x = 13$$

$${\left( {11} \right)^2} = 121{\text{ }}$$
$${\text{And }}$$
$${\left( {17} \right)^2} = 289$$
So, the perfect squares between 120 and 300 are the squares of numbers from 11 to 17.
Clearly, these are 7 in number.

$$ ⇒ \sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} $$
$$ ⇒ x + \frac{x}{y} = {x^2}.\frac{x}{y}$$
$$ ⇒ \frac{{xy + x}}{y} = \frac{{{x^3}}}{y}$$
$$ ⇒ xy +×= {x^3}$$
$$ ⇒ y + 1 = {x^2}$$
$$ ⇒ y = {x^2} - 1$$

$$ ⇒ {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}}$$
$$ ⇒ {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}}$$
$$ ⇒ {{\text{n}}^2} = {5^{128}} \times {2^{150}}$$
$$ ⇒ {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}}$$
$$ ⇒ n = {5^{64}} \times {2^{64}} \times {2^{11}}$$
$$ ⇒ n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}}$$
$$ ⇒ n = {10^{64}} \times 2048$$
∴ Sum of digit's of n
= 2 + 0 + 4 + 8
= 14

$$ ⇒ \frac{{0.13}}{{{p^2}}} = 13$$
$$ ⇒ {p^2} = \frac{{0.13}}{{13}}$$
$$ ⇒ {p^2} = \frac{1}{{100}}$$
$$ ⇒ p = \sqrt {\frac{1}{{100}}} $$
$$ ⇒ p = \frac{1}{{10}}$$
$$ ⇒ p = 0.1$$

$$ ⇒ \sqrt {{3^n}} = 729$$
$$ ⇒ \sqrt {{3^n}} = {3^6}$$
$$ ⇒ {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2}$$
$$ ⇒ {3^n} = {3^{12}}$$
$$ ⇒ n = 12$$

$$ ⇒ \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02$$
$$ ⇒ \frac{{\sqrt x }}{{21}} = 0.02$$
$$ ⇒ \sqrt x = 0.02 \times 21$$
$$ ⇒ \sqrt x = 0.42$$
$$ ⇒ x = {\left( {0.42} \right)^2}$$
$$ ⇒ x = 0.1764{\text{ }}$$