$\begin{array}{l} \dfrac { { 0.304\times 20 } }{ { 3.04\times 2 } }  \ =\dfrac { { 304\times 20\times 100 } }{ { 304\times 2\times 1000 } }  \ =1 \ Hence, \ B\, is\, the\, correct\, answer. \end{array}$

Required decimal = $\dfrac{1}{60 \times 60}=\dfrac{1}{3600}=0.00027$

$\displaystyle\frac { 1 }{ 0.0004175 } = \displaystyle\frac { 1 }{ \displaystyle\frac { 4.175 }{ 10000 }  } = \displaystyle\frac { 10000 }{ 4.175 } = \displaystyle\frac { 10000 }{ \displaystyle\frac { 1 }{ 0.2395 }  } $

$\left( 78.34+96.68-14.44 \right) \div 4=160.58\div 4=40.145$

$\dfrac{(0.35)^{2}-(0.03)^{2}}{0.19}$=$\dfrac{(0.35+0.03)(0.35-0.03)}{0.19}$    $[a^2-b^2=(a+b)(a-b)]$

$\displaystyle 0.\overline{09}\times7.\overline{3}=\frac{9}{99}\times7\frac{3}{9}=\frac{9}{99}\times\frac{66}{9}=\frac{6}{9}=0.\overline{6}$

given that 

$58+\cfrac{3}{100}+\cfrac{7}{1000}$
$=58+\cfrac{0}{10}+\cfrac{3}{100}+\cfrac{7}{1000}=58.037$

Place value chart of decimal number

$\displaystyle \frac { 142.0 }{ 14.2 } = \frac { 142.0 }{ 14.2 }\times \frac { 10 }{ 10 } = \frac { 1420 }{ 142 }=10 $

(68.12)*(2.5)=170.3

$\frac {280.5}{25.5}=\frac {280.5}{25.5}\times \frac {10}{10}\times \frac {10}{10}$
$=\frac {2805}{2.55}\times \frac {1}{100}=\frac {1100}{100}=11$

By BODMAS rule,
$2\times 0.5+9\div 0.3+10\times 0.92$
$=2\times 0.5+30+10\times 0.92$
$=1.0+30+9.2$
$=40.2$

$0.29\rightarrow 2$ decimal places
$0.27\rightarrow 2$ decimal places
$\therefore 0.29\times 0.27=0.0783$
$(2+2=4$ decimal places)

$1.000$
$+0.100$
$+0.010$
$\underline {+0.001}$
$\underline {1.111}$
$\Rightarrow 1.111\times 1000=1111$

Given expression $=\dfrac { { \left( 0.1 \right)  }^{ 3 }+{ \left( 0.02 \right)  }^{ 3 } }{ { 2 }^{ 3 }\left[ { \left( 0.1 \right)  }^{ 3 }+{ \left( 0.02 \right)  }^{ 3 } \right]  } =\dfrac { 1 }{ 8 } =0.125$

Given Expression $=\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ a-b } =\dfrac { \left( a+b \right) \left( a-b \right)  }{ \left( a-b \right)  } =\left( a+b \right) =\left( 2.39+1.61 \right) =4$

Given expression $=\dfrac { \left( 0.3333 \right)  }{ \left( 0.2222 \right)  } \times \dfrac { \left( 0.1667 \right) \left( 0.8333 \right)  }{ \left( 0.6667 \right) \left( 0.1250 \right)  } $

$\dfrac { 144 }{ 0.144 } =\dfrac { 14.4 }{ x } $

Let the required number be $x$
According to question, we have
$0.75$ of $x$ $=1200$
$\Rightarrow \displaystyle\frac{75}{100}\times x=1200$
$\Rightarrow \displaystyle x=1200\times \frac{100}{75}=1600$
Therefore, required number is $1600$.
Now, $\displaystyle\frac{5}{8}$ of the number $=\displaystyle\frac{5}{8}\times 1600=1000$.

$\\35.453\times 10=34.968x+36.956(100-x)\\ 3545.3=(34.968-36.956)x+3695.6\\\therefore x=(\frac{3545.3-3695.6}{34.968-36.956})\\ (\frac{-150.3}{-1.988})=75.6$

$\dfrac{337}{125}=2.696$

Option c is correct answer.

$\displaystyle \frac{3.6\times 0.48\times 2.50}{0.12\times 0.09\times 0.5}=\frac{36\times 48\times 250}{12\times 9\times 5}=800 $

Let us first write the decimals as fractions as follows:

We will first convert the decimals into fraction in the given fraction and then solve it as follows:

Consider the given fraction $\dfrac { 0.342\times 0.684 }{ 0.000342\times 0.000171 }$ and find the square root as follows:

The expression approximately =$\displaystyle \frac{3.2\times 4126\times 3.2}{64\times 2835}=0.232=0.2 (approx)$

$\displaystyle \frac{0.6\times 0.6\times 0.6+0.5\times 0.5\times

0.5+0.1\times 0.1-0.09}{0.6\times 0.6+0.5\times 0.5+0.1\times

0.1-0.41}$

$\frac {(.75)^3+(1)^3-(.75)^3}{1-(0.75)}=\frac {1}{.25}=4$
square root $=2$

Given x+y=1  So 100x+100y=100 and 200x+200y =200

Given exp =$\displaystyle \frac{(6.4+5.4)(6.4-5.4)}{(8.9+1.1)^{2}}=\frac{11.8\times1 }{100}=0.118$

$\displaystyle 12.28\times 1.5-\frac{36}{2.4}=18.42-15=3.42$

$\frac{(0.22)^{3}+(0.11)^{3}+(0.32)^{3}}{(0.66)^{3}+(0.96)^{3}-(0.33)^{3}}+\frac{(0.32)^{3}+(0.45)^{3}-(0.77)^{3}}{81(0.32)(0.45)(0.77)}$
$=\frac { 8(0.11)^{ 3 }+(0.11)^{ 3 }+(0.32)^{ 3 } }{ 216(0.11)^{ 3 }+27(0.32)^{ 3 }+27(0.11)^{ 3 } } -\frac { (0.32)^{ 3 }+(0.45)^{ 3 }-(0.32+0.45)^{ 3 } }{ 81(0.32)(0.45)(0.77) } $
$=\frac { 9(0.11)^{ 3 }+(0.32)^{ 3 } }{ 243(0.11)^{ 3 }+27(0.32)^{ 3 } } -\frac { (0.32)^{ 3 }+(0.45)^{ 3 }-(0.32+0.45)^{ 3 } }{ 81(0.32)(0.45)(0.77) } $
$=\frac { 9(0.11)^{ 3 }+(0.32)^{ 3 } }{ 27{ 9(0.11)^{ 3 }+(0.32)^{ 3 }}  } -\frac { (0.32)^{ 3 }+(0.45)^{ 3 }-[(0.32)^{ 3 }+(0.45)^{ 3 }+3(0.32)(0.45)(0.32)+(0.45) }{ { 81(0.32)(0.45)(0.77) } } $
$=\frac{1}{27}-\frac{1}{27}$
$=0$

$(7.5 \times 7.5 + 37.5 + 2.5 \times 2.5) $

Multiplying 0.02468 by a positive power of 10 will shift the decimal point to the right. Simply shift the decimal point to the right until the result is greater than 10,000. Keep track of how many times you shift the decimal point. Shifting the decimal point 5 times results in 2,468. This is still less than 10,000. Shifting one more place yields 24,680, which is greater than 10,000.