Tag: decimal numbers

$\displaystyle\frac { 10.24 }{ 1.6 } = \displaystyle\frac { 102.4 }{ 16 } = 6.4$
$\therefore \displaystyle\frac { 10.24\div 1.6 }{ 20-19.8 } = \displaystyle\frac { 6.4 }{ 0.2 } = \displaystyle\frac { 64 }{ 2 } = 32$

$\displaystyle \left ( 0.\overline{1} \right )^{2}\left { 1-9\left ( 0.\overline{16} \right )^{2} \right }$

$\displaystyle \left [ 0.9-\left { 2.3-3.2-\left ( 7.1-5.4-3.5 \right ) \right } \right ]$

$\displaystyle 0.34\overline{67}+0.13\overline{33}=\frac{3467-34}{9900}+\frac{1333-13}{9900}$

$\displaystyle 6\frac{1}{4}\times 0.25+0.75-0.3125$
$= 6.25 \times 0.25 + 0.75 - 0.3125$
$= 1.5625 + 0.75 - 0.3125$
$= 2.3125 - 0.3125 = 2$

$\displaystyle 0.\overline{2}+0.\overline{3}+0.\overline{4}+0.\overline{9}+0.\overline{39}$

$\displaystyle :0.\overline{4}+0.\overline{61}+0.\overline{11}-0.\overline{36}=\frac{4}{9}+\frac{61}{99}+\frac{11}{99}-\frac{36}{99}=\frac{4}{9}+\frac{72}{99}-\frac{36}{99}$

The product of 44 and 11 is 484
If base is x then 3411
$3x^3+4x^2+1x^1+4x^0=484$
$3x^3+4x^2+x=480$
This equation is satisfy when x=5
then base is 5
In decimal system number 3111 will be  written
$3\times 5^3+1\times 5^2+1\times 5^1+1\times 5^0=406$

The man born between $1800$ and $1850$ which is a perfect square. 
The perfect square number is $43 = 1849$ (Since$ 42 = 1764$ and $44 = 1936$ ) 
So in the year $1849$ the man was $43$ years old. 
Which shows the year of born is $1849 - 43 = 1806.$