$\dfrac{4+\sqrt5}{4-\sqrt5} = \dfrac{(4+\sqrt5)}{(4-\sqrt5)} \dfrac{(4+\sqrt5)}{(4+\sqrt5)}$    ...... rationalizing numerator and the denominator

A complete day has $24$ hours so

$Required\>ratio\>=\>\dfrac{16}{24}\>=\>\dfrac23$

Hence option $'C'$ is the answer.

(i) $\displaystyle \frac{35}{16} = \frac{35 \times 5^4}{2 \times 5^4}= \frac{35 \times 625}{(10)^4}= \frac{21875}{10000}=2.1875$
(ii) $\displaystyle \frac{17}{8} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{(10)^3}=\frac{2125}{1000}=2.125$
(iii) $\displaystyle \frac{327}{500} = \frac{327}{5\times 5 \times 5 \times 2 \times 2}$
$=\displaystyle \frac{327}{5^3 \times 2^2} = \frac{327}{5^3 \times 2^3}= \frac{654}{(10)^3} = 0.654$
(iv) $\displaystyle \frac{14588}{625} = \frac{2^2 \times 7 \times 521}{5^4} = \frac{2^6 \times 7 \times 521}{2^4 \times 5^4}$
$\displaystyle =\frac{233408}{10^4}=23.3408$

Given, $\displaystyle 8\frac{2}{3} + 9\frac{3}{8}$

Let $A=12k,B=15k, C=25k$

$\dfrac{-225}{465}=\dfrac{-45}{93}=-\dfrac{15}{31}$

Given percentage is $ 225 % $

In fraction form, $ 225  % $ $  = \dfrac {225}{100} $

Simplifying it by dividing the numerator and denominator by $ 25  $, we get the fraction $ = \dfrac {9}{4} $

Given, $ 2a-5b = 0 $

$=>2a = 5b $

$ => \dfrac {a}{b} = \dfrac {5}{2} $

Applying componendo and dividendo,

Now, $ \dfrac {a+b}{a-b} = \dfrac{5+2}{5-2} =

\dfrac {7}{3} $

Given, $ 2a-5b = 0 $
$=> 2a = 5b $
$ => \dfrac {a}{b} = \dfrac {5}{2} $
 
Now, $ \dfrac {a-b}{b} = \dfrac {a}{b} - 1 = \dfrac {5}{2} - 1 = \dfrac{5-2}{2} = \dfrac {3}{2} $

$ 5m-n=m+2n \$
$ => 4m = 3n \$
$ => \dfrac {m}{n} = \dfrac {3}{4} \$
$ => m : n = 3:4 $

Let $ m = 3a ; n = 4a $

Given, $ 2a-5b = 0 $
$=>. 2a = 5b $
$ => \frac {a}{b} = \frac {5}{2} $
 
Now, $ \frac {a+b}{b} = \frac {a}{b} + 1 = \frac {5}{2} + 1 = \frac{5+2}{2} = \frac {7}{2} $

In the given ratios "b" is

the common term, and the values of b in both ratios are not equal.

To make them equal, find the L.C.M.

of values corresponding to b i.e., $ 8 $ and $ 12 $.





L.C.M. of $ 8  $ and $ 12 = 24 $





Therefore, an equivalent ratio of $ a:b $ such that $ b = 24 $ is $= 7 \times 3:8 \times 3 = 21:24 $

Similarly, an equivalent ratio of $ b:c$ is $ = 12 \times 2 :7 \times 2 = 24:14 $





Therefore,$ a: c = 21:14 $

Dividing by $ 7 $

$ a:c = 3:2 $

$\left( \sqrt { 169-144 }  \right) \div \left( \sqrt { 64+36 }  \right) \ =\sqrt { 25 } \div \sqrt { 100 } =5\div 10=\displaystyle\frac { 5 }{ 10 } =0.5$

$\displaystyle \frac {2}{5}\, =\, \displaystyle \frac {2}{5}\, \times\, \displaystyle \frac {3}{3}\, =\, \displaystyle \frac {6}{15}$

$\displaystyle \frac {1}{2}\, =\, \displaystyle \frac {1\, \times\, 2}{2\, \times\, 2}\, =\, \displaystyle \frac {2}{4}$

$\displaystyle \frac {1}{2}\, =\, \displaystyle \frac {3}{6}\, =\, \displaystyle \frac {5}{10}\, =\, \displaystyle \frac {9}{18}$

$\displaystyle \frac {15}{45}\, =\, \displaystyle \frac {15\, \div\, 5}{45\, \div\, 5}\, =\, \displaystyle \frac {3}{9}$

${.225}\Rightarrow \frac{225}{1000}=\frac{9}{40}$

$1.\bar{3}$
Let $1.3333...$ be $x$
$1.3333...=x$  {i}
$13.333...=10x$ {ii} [Multiplied by $10$]
                               
$        12  =9x$
$=>x=\frac{12}{9}$
$=>x=\frac{4}{3}$

The fraction for of 0.23$ =\frac{23}{100}$
Hence $B$  is the answer

Consider the given decimal number

$ \Rightarrow -25.6875 $

$ \Rightarrow -\dfrac{256875}{10000} $

$ \Rightarrow -\dfrac{51375}{2000} $

$ \Rightarrow -\dfrac{10275}{400} $

$ \Rightarrow -\dfrac{2055}{80} $

$ \Rightarrow -\dfrac{411}{16} $

Hence, this is the answer.

lowest form of 30/60 is

$79\% = 0.79$ in decimal form. Percent means 'per $100$'. 

$\displaystyle \frac { 20 }{ 25 } =\frac { 20\div 5 }{ 25\div 5 } =\frac { 4 }{ 5 }  $

Here, we have to covert fraction in to percentage.

3×3=9 so 3/9=1/3

$=\displaystyle \frac { 3 }{ 2 } -\frac { 1 }{ 4 } -\frac { 2 }{ 4 }=  \frac { 6-1-2 }{ 4 }= \frac{3} {4} $ 

The value of $\dfrac {1}{(1 + \sqrt {3})+\sqrt {2}} + \dfrac {1}{(1 + \sqrt {3}) - \sqrt {2}}$ is
$=\dfrac {(1 + \sqrt {3} - \sqrt {2}) + (1 + \sqrt {3} + \sqrt {2})}{(1 + \sqrt {3})^{2} - (\sqrt {2})^{2}}$
$= \dfrac {2(1 + \sqrt {3})}{1 + 3 + 2\sqrt {3} - 2}$
$= \dfrac {2(1 + \sqrt {3})}{2(1 + \sqrt {3})} = 1$

$\dfrac{144}{36}$

$\dfrac{100}{200}$

$\dfrac{12}{16}$

$\dfrac{125}{625}$

$\dfrac{25}{100}$

$\dfrac{81}{36}$

(A) $\dfrac {3}{6} = \dfrac {3\div 3}{6\div 3} = \dfrac {1}{2}$
(B) $\dfrac {1}{2}$ of an hour $= \dfrac {1}{2}\times 60$ minutes = $30$ minutes
(C) $\dfrac {5}{6} = 0.8333, \dfrac {6}{5} = 1.2$
$\therefore \dfrac {5}{6}\neq \dfrac {6}{5}$
(D) $\dfrac {1}{100}$ of $1$ cm = $\dfrac {1}{100}\times 10$ mm = $\dfrac {1}{10}$ mm.

If the denominator has a prime factor $2$ or $5$ then it is a terminating decimal

the standard form of $ - \dfrac{192}{168}$

Among the given values, the one which is not equivalent to $\dfrac{4}{8}$ is 1.

$\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} -\dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10} + \sqrt{3})(\sqrt{10} - \sqrt{3})} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})} -\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{(\sqrt{15} - 3\sqrt{2})(\sqrt{15} + 3\sqrt{2})}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{10-3} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{6-5} -\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{15-18}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{7} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{1} +\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{3}$

$\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}=\dfrac{(\sqrt{3} + \sqrt{2})^2+(\sqrt{3} - \sqrt{2})^2}{3-2}=\dfrac{3+2+3+2}{1}=10$

$\\(a.)(\frac{36}{144})=(\frac{3}{12})=(\frac{1}{4})\\(b.)(\frac{65}{117})=(\frac{13\cdot 5}{13\cdot 9})=(\frac{5}{9})\\(c.)(\frac{180}{120})=(\frac{60\cdot 3}{60\cdot 2})=(\frac{3}{2})$

$0.35$

Let the common difference be d. As there are $n\;AM's$ between $a$ and $b$ and total number of terms in the sequence is $=n+2$ 

$\displaystyle 5\,\frac{7}{x}\,\times\,y\,\frac{1}{13}\,=\,12$
By Hit and Trial method. 
Let $x = 9, y = 2$
Where the fractions are in their lowest terms, then x should be maximum possible single digit and $y$ is minimum possible single digit.
Putting this value in equ. (1)
$\displaystyle \,5\,\times\,\frac{7}{9}\,\times\,2\,\times\,\frac{1}{13}\,=\,\frac{52}{9}\,\times\,\frac{27}{13}\,=\,12
$
$\therefore \,x\,-\,y\,=7$
Hence, option 'C' is correct.

We have, $ 140:24 $
Dividing by $ 4 $ we get $ 35:6 $
As there is no other common factor  to divide with, so $ 35:6 $ is the simplest form.

Reciprocal of $ -3 = \dfrac {1}{-3} $ or $ \dfrac {-1}{3} $

$\left [ \left ( -2\dfrac{3}{4} \right )- \left ( -1\dfrac{3}{4} \right )\right ]+\left [ \left ( -2\dfrac{3}{4} \right )- \left ( -1\dfrac{3}{4} \right )\right ]+.......$ upto $30$ times

$\dfrac {a^{2} + b^{2} - c^{2} + 2ab}{a^{2} + c^{2} - b^{2} + 2ac} = \dfrac {(a + b)^{2} - c^{2}}{(a + c)^{2} - b^{2}} = \dfrac {(a + b + c)(a + b - c)}{(a + c + b)(a + c - b)}$
$= \dfrac {a + b - c}{a + c - b}$ with $(a + c)^{2} \neq b^{2}$.

Let us first factorize $75$ as shown below: