Decimal Arithmetic Operations - Class VI

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

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 $=\displaystyle\frac{5}{8}\times 1600=1000$.

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

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

Performing the arithmetic computation by estimating or multiplying the numerator and dividing by the product of the denominator values yields approximately 12.65. Precise calculation confirms this decimal value.

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

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

The expression is (0.44^2 + 0.06^2 + 0.024^2) / (0.044^2 + 0.006^2 + 0.0024^2). Since 0.44 = 10 * 0.044, 0.06 = 10 * 0.006, and 0.024 = 10 * 0.0024, the numerator is 10^2 times the denominator, which equals 100.

$\frac{(0.22)^{3}+(0.11)^{3}+(0.32)^{3}}{(0.66)^{3}+(0.96)^{3}-(0.33)^{3}}+\frac{(0.32)^{3}+(0.45)^{3}-(0.77)^{3}}{81(0.32)(0.45)(0.77)}$
$=\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.