$617.00$
     $6.017$
     $0.617$
$+$ $6.0017$
-----------------
  $629.6357$
-----------------

Addition of $3.005$ and $4.7$ gives $3.005+4.7=7.705$

The sum of $1.2$ and $2.3$ will be $1.2+2.3=3.5$

Total member =2[2+(5×2)+(5×4)]=32

$2.\bar{3}-0.\bar{4}=1.\bar{9}$

The time interval between $8:15$ pm and $6:25 am$ = $10\,\, hrs\,\, 10\,\, min$

$\dfrac 23+\dfrac 43+\dfrac 45+\dfrac 65\\dfrac {2+4}3+\dfrac {4+6}5\2+2=4$

$\Rightarrow$  $0.\overline {23}$ means $0.232323....$

    $0.7500$
$-   0.0075$
______
    $0.7425$
______
$\therefore$  Required difference $= 0.7425$.

$x = 301.01 - 0.101 - 198.01$
   $= 301.01 - 198.111 = 102.899$

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

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

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

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

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

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

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 $1849 - 43 = 1806.$

32.000-27.091=4.909

For obtaining the sum of decimal numbers with different decimal values we expand the decimal values of all the numbers with respect to the number with the highest decimal value. 

$\frac {0.1}{0.01}+\frac {0.01}{0.1}=10+\frac {1}{10}$
$=10+0.1$
$=10.1$

To add $3$ decimal numbers of different decimal values we pick the number with the highest decimal value and expand all the other numbers to the same value by adding $0$ after the decimal point.

The sum of two numbers $=31.021$

$5 + \dfrac { 3}{100} + \dfrac{4}{1000}$

$\displaystyle \frac{0.1}{0.01}+\frac{0.01}{0.1}=10+\frac{1}{10}$
=10+0.1
=10.1

As given,

Let $3889 + 12.952 - x = 3854.002$.
Then $x=\left( 3889+12.952 \right) -3854.002$
             $= 3901.952 - 3854.002$
             $= 47.95$.

The value of $2.\overline {6} - 1.\overline {9}$ is
$= \left (2 + \dfrac {6}{9}\right ) - \left (1 + \dfrac {9}{9}\right )$
$= \dfrac {6}{9} = 0.\overline {6}$

$3.\overline { 87 } -2.\overline { 59 } =\left( 3+0.\overline { 87 }  \right) -\left( 2+0.\overline { 59 }  \right) $

We have, $(30.5 - 30.4) + (30.3 - 30.2) + (30.1 - 30) + (29.9 - 29.8) + (29.7 - 29.6) + (29.5 - 29.4) + (29.3 - 29.2) + (29.1 - 29.0)$
$= 0.1+ 0.1 + 0.1 + 0.1 + 0.1+ 0.1 + 0.1 + 0.1$
$= 0.8$.

The sum of $11.006+34+0.72$ is $45.726$.
Now, the number $45.726$ is rounded off as $45.73$.

given that, $617.0000+6.0170+0.6170+6.0017=629.6357$

$5.5-4.17=1.33$

${\left( 1.02 \right)}^{4} + {\left( 0.98 \right)}^{4}$

Width of class $=U.L-L.L=60.5-55.5=5$

The given expression is

$2.002+7.9\left{ 2.8-6.3\left( 3.6-1.5 \right) +15.6 \right} $
$=2.002+7.9\left{ 2.8-6.3\times 2.1+15.6 \right} $
$=2.002+7.9\left{ 2.8-13.23+15.6 \right} $
$=2.002+7.9\left{ 5.17 \right} $
$=2.002+40.843=42.845$

The value of $25 $ $+$ $\displaystyle{\frac{3}{100}}$ $+$ $\displaystyle{\frac{4}{1000}}$ is

$\displaystyle{\frac{0.1}{0.01} + \frac{0.01}{0.1}}$ = 10 + $\displaystyle{\frac{1}{10}}$ $= 10 + 0.1 = 10.1$

3 tens 6 thousandths is

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

$\displaystyle \frac{0.34-0.034}{0.0034\div 34}=\frac{0.306}{0.0001}=\frac{0.3060}{0.0001}$

Given, $\displaystyle \frac {0.25\times0.25-0.24\times0.24}{0.49}$

Using the identity $a^2 - b^2 = (a+b)(a-b)$
$\therefore \displaystyle 0.25\times 0.25 - 0.24\times 0.24 = (0.25+0.24)(0.252-0.24) = (0.49)(0.01)$

Thus, $\displaystyle \frac {0.25\times0.25-0.24\times0.24}{0.49} = \frac {0.49\times 0.01}{0.49}$ = $0.01$

$\displaystyle \frac{42.31-26.43}{42.31+26.43}\div \frac{423.1-264.3}{4.231+2.643}$

Let us first write the mixed fraction $2\dfrac {3}{10}$ in decimals as follows:

We solve the given expression as follows:

$\displaystyle 3.\overline{36}-2.\overline{05}+1.\overline{33}$

$515.15-15.51-1.51-5.11-1.11.$

Given expression:

(I) $715+12.59+685.35=1412.94$
(II) $518-(216.80-115.96)=518-100.84=417.16$
(III) $4.090+0.050+6.500=10.640$
(IV) $36.050+198.05-21.03=213.07$.

binary(10011)= (1)*2^4 + (0)*2^3 + (0)*2^2 + (1)*2^1 + (1)*2^0

from question 

$3.7-1.9$

The sum of $0.5 ,12.56$ and $0.003$ is

Total number of members in family $\rightarrow$ $5$