Given numbers are 12.142, 12.124, 12.104, 12.401 and 12..214.
\(\therefore\) \(12.142=10+2+\frac { 1 }{ 10 } +\frac { 4 }{ 100 } +\frac { 2 }{ 1000 } \)
\(12.124=10+2+\frac { 1 }{ 10 } +\frac { 2 }{ 100 } +\frac { 4 }{ 1000 } \)
\(12.104=10+2+\frac { 1 }{ 10 } +\frac { 0 }{ 100 } +\frac { 4 }{ 1000 } \)
\(12.401=10+2+\frac { 4 }{ 10 } +\frac { 0 }{ 100 } +\frac { 1 }{ 1000 } \)
\(12.214=10+2+\frac { 2 }{ 10 } +\frac { 1 }{ 100 } +\frac { 4 }{ 1000 } \)
Here, whole part of all numbers are same and tenths part of 12.142, 12.124 and 12.104 are same.
Now, tenths part of 12.401 = \(\frac { 4 }{ 10 } \)
and tenths part of 12.214 = \(\frac { 2 }{ 10 } \)
\(\because \quad \frac { 4 }{ 10 } >\frac { 2 }{ 10 } \)
\(\therefore\) 12.401 > 12.214
Again, hundredths part of 12.142 = \(\frac { 4 }{ 100 } \)
\(\therefore\) Hundredths part of 12.104 = \(\frac { 0 }{ 100 } \)
\(\therefore \quad \frac { 4 }{ 100 } >\frac { 2 }{ 100 } >\frac { 0 }{ 100 } \)
\(\therefore\) 12.142 > 12.124 > 12.104
Hence, the ascending order of given number are 12.104 < 12.124 < 12.142 < 12.214 < 12.401. 

(a) Total money that Seem a has = Rs 2000.
Cost of readymade garments = Rs 987.50
Cost of medicines = Rs 210.25
Cost of groceries = + Rs 530.25
                                    __________
             Total cost = Rs 1728.00
                                   ___________
Money donate for charity = Rs 200
                                        =Rs 1728.00
        Total money spent = + Rs 200
                                       ____________
                                           Rs 1928
                                       ____________
Money she had = Rs 2000
Spent money = - Rs 1928
                           _________
Balance =            Rs 072
                           _________
So, Rs 72 are left with her.
(b) Humanity, helpfulness.

Naresh walked in morning
= 2 km 35 m = 2 km + 35 m
\(=2\ km+35\times{1\over 1000}km\)        \(\begin{bmatrix} \because 1m={1\over 1000}km\end{bmatrix}\)
\(=2\ km+{35\over 1000}km\)
= (2 + 0.035) km = 2.035 km
Naresh walked in evening
=1 km 7 m = 1 km + 7 m
\(=1\ m+7\times{1\over1000}km\)   \(\begin{bmatrix} \because 1m={1\over 1000}km\end{bmatrix}\)
\(=1\ km+{7\over1000}km\)
= (1 + 0.007) km = 1.007 km
\(\therefore\)  Total distance
\(\quad 2.035\\+1.007\\\_\_\_\_\_\_\_\\\ \ \ 3.042\\\_\_\_\_\_\_\_\)
Hence, total distance walked by Naresh is 3.042 km.

\(\because\)   Total distance travelled by Namita
= 20 km 50 m = 20 km + 50 m
\(=20\quad km+50\times \frac { 1 }{ 1000 } km\)   \(\left[ \because 1\quad m=\frac { 1 }{ 1000 } km \right] \)
\(=20\quad km+\frac { 50 }{ 1000 } km\) = (20+0.050) km = 20.050 km
and distance travelled by Namita by bus
= 10 km 200 m = 10 km + 200 m
= 10 km + 200 \(\times \frac { 1 }{ 1000 } km\left[ \because \quad 1m=\frac { 1 }{ 1000 } km \right] \)
\(=10\quad km+\frac { 200 }{ 1000 } km=10\quad km+0.200\quad km\)
= ( 10 + 0.200) km = 10.200 km
\(\therefore\)  Distance travelled by auto
= 20.050 km - 10.200 km
= (20.050 - 10.200) km = 9.850 km
Hence, she travels 9.850 km by auto.

The given decimals can be written as
(a) 0.29=\(0+\frac { 2 }{ 10 } +\frac { 9 }{ 100 } \)
(b) 2.08=\(2+\frac { 0 }{ 10 } +\frac { 8 }{ 100 } \)
(c) 19.60=10+9+\(\frac { 6 }{ 10 } +\frac { 0 }{ 100 } \)
(d) 14832=100+40+8+\(\frac { 3 }{ 10 } +\frac { 2 }{ 100 } \)
(e) 200.812=200+00+0+\(\frac { 8 }{ 10 } +\frac { 1 }{ 100 } +\frac { 2 }{ 1000 } \)
Now, the place value table is given below:

Given number line is a follows:

(i) From the figure, it is clear that A = 0.8 as the unit length between 0 and 1 has been divided into 10 equal parts and 8 parts from 0 have been taken.
(ii) From the figure, it is clear that B = 1.3 as the unit length between 1 and 2 has been divided into 10 equal parts and unit length between 0 to 1 and then 3 parts have been taken.
(iii) From the figure, it is clear that C = 2.2 as the unit length between 2 and 3 has been divided into 10 equal parts and unit length between 0 to 1, 1 to 2 and then 2 parts have been taken.
(iv) From the figure, it is clear that D = 2.9 as the unit length between 2 and 3 has been divided into 10 equal parts and unit length between 0 to 1, 1 to 2 and then 9 parts have been taken.

Firstly, draw the number line and divide the unit length between two whole numbers into 10 equal parts, each of these equal parts represents 0.1 or 1/10. Now, locate the given decimals on this line.

(a) We have, 0.8
From the above figure, it is clear that number 0.8 lies between the whole numbers 0 and 1.
Hence, number 0.8 is nearer to number 1.
(b) We have, 5.1
From the above figure, it is clear that number 5.1 lies between the whole numbers 5 and 6.
Hence, number 5.1 is nearer to number 5.
(c) We have, 2.6
From the above figure, it is clear that number 2.6 lies between the whole numbers 2 and 3.
Hence, number 2.6 is nearer to number 3.
(d) We have, 6.4
From the above figure, it is clear that number 6.4 lies between the whole numbers 6 and 7.
Hence, number 6.4 is nearer to number 6.
(e) We have, 9.1
From the above figure, it is clear that number 9.1 lies between the whole numbers 9 and 10.
Hence, number 9.1 is nearer to number 9.
(f) We have, 4.9
From the above figure, it is clear that number 4.9 lies between the whole numbers 4 and 5.
Hence, number 4.9 is nearer to number 5.

(i) \(\frac { 3 }{ 2 } =\frac { 3\times 5 }{ 2\times 5 } \)
[multiplying numerator and denominator by 5 to make denominator 10]
\(=\frac { 15 }{ 10 } =1\frac { 5 }{ 10 } =1+\frac { 5 }{ 10 } =1+0.5=1.5\)
Therefore, \(\frac { 3 }{ 2 } \) is 1.5 in decimal notation.
(ii) \(\frac { 4 }{ 5 } =\frac { 4\times 2 }{ 5\times 2 } \)
[multiplying numerator and denominator by 2 to make denominator 10]
\(=\frac { 8 }{ 10 } =0.8\)
Therefore, \(\frac { 4 }{ 5 } \) is 0.8 in decimal notation.
(iii) \(\frac { 8 }{ 5 } =\frac { 8\times 2 }{ 5\times 2 } \)
[multiplying numerator and denominator by 2 to make denominator 10]
\(=\frac { 16 }{ 10 } =1\frac { 6 }{ 10 } =1+\frac { 6 }{ 10 } =1+0.6=1.6\)
Therefore, \(\frac { 8 }{ 5 } \) is 1.6 in decimal notation.

Given numbers can be represented in decimals as follows:
(i) 2 \(\times\) hundreds + 1 \(\times\) ten + 6 \(\times\) ones + 3 \(\times\) tenths
= 2 \(\times\) 100 + 1 \(\times\) 10 + 6 \(\times\) 1+ 3 \(\times\) \(\frac { 1 }{ 10 } \)
= 200 + 10 + 6 + \(\frac { 3 }{ 10 } \) = 216 + 0.3 = 216.3
(ii) 4 \(\times\) hundreds + 5 \(\times\) tens + 4 \(\times\) ones + 2 \(\times\) tenths
= 4 \(\times\) 100 + 5 \(\times\) 10 + 4 \(\times\) 1+ 2 \(\times\) \(\frac { 1 }{ 10 } \)
= 400 + 50 + 4 + \(\frac { 2 }{ 10 } \) = 454+0.2= 454.2
(iii) 7 \(\times\)hundreds + 3 \(\times\) tens + 2 \(\times\) ones + 1 \(\times\) tenth
 = 7 \(\times\) 100 + 3 \(\times\) 10 + 2 \(\times\) 1+ 1 \(\times\) \(\frac { 1 }{ 10 } \)
= 700 + 30 + 2 + \(\frac { 1 }{ 10 } \) = 732 + 0.1 = 732.1

Given numbers can be written in decimals as follows:
(i) 5 hundreds + 3 tens + 8 ones + 1 tenth
 = 5 \(\times\) 100 + 3 \(\times\) 10 + 8 \(\times\) 1+ 1 \(\times\) \(\frac { 1 }{ 10 } \)
 = 500 + 30 + 8 + \(\frac { 1 }{ 10 } \)  = 538 + 0.1 = 538.1
(ii) 2 hundreds + 7 tens + 3 ones + 4 tenths
= 2 \(\times\) 100 + 7 \(\times\) 10 + 3 \(\times\) 1+ 4 \(\times\) \(\frac { 1 }{ 10 } \)
= 200 + 70 + 3 + \(\frac { 4 }{ 10 } \) = 273 + 0.4 = 273.4
(iii) 3 hundreds + 5 tens + 4 ones + 6 tenths
= 3 \(\times\) 100 + 5 \(\times\) 10 + 4 \(\times\) 1+ 6 \(\times\) \(\frac { 1 }{ 10 } \)
= 300 + 50 + 4 + \(\frac { 6 }{ 10 } \)  = 354 + 0.6 = 354.6