11 cm 52 rnm = 11 cm + 52 rnm
= 11 cm+ \(\cfrac { 52 }{ 10 } \)cm
I·: 1 rnm= \(\cfrac { 1 }{ 10 } \)cm
= 11 cm+ \(\cfrac { 50+2 }{ 10 } cm\)
=11 cm+\(\left( \cfrac { 50 }{ 10 } +\cfrac { 2 }{ 10 } \right) \)
= 11 cm+ \(\left( 5+\cfrac { 2 }{ 10 } \right) \)
= \(\left( 11+5+\cfrac { 2 }{ 10 } \right) cm\)
=\(\left( 16+\cfrac { 2 }{ 10 } \right) \)cm = 16.2 cm

1.Required number
=600+40+9+ \(\cfrac { 2 }{ 10 } \) =649.2
2.Required number
= 300 + 80 + 4 + \(\cfrac { 5 }{ 10 } \) = 384.5
3.Required number
=400+60+5+ \(\cfrac { 7 }{ 10 } \) =465.7

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. 

False, because the place value of a digit at the tenths place is 1/10 times the same digit at the ones place,
e.g. Let a number be 23.37.
Here, place value of 3 at ones place = 3
and place value of 3 at tenths place
\(=\frac { 3 }{ 10 } =3\times \frac { 1 }{ 10 } =\frac { 1 }{ 10 } \times \) Place value of 3 at ones place

Given, numbers can be represented on the number line as given below:
(a) Here, 0.06 is more than 0 but less than 0.1. So, it lies between 0 and 0.1.
(b) Here, 0.45 is more than 0.4 but less than 0.5. So, it lies between 0.4 and 0.5.
(c) Here, 0.19 is more than 0.1 but less than 0.2. So, it lies between 0.1 and 0.2.
(d) Here, 0.66 is more than 0.6 but less than 0.7. So, it lies between 0.6 and 0.7.
(e) Here, 0.92 is more than 0.9 but less than 1. So, it lies between 0.9 and 1.
(f) Here, 0.57 is more than 0.5 but less than 0.6. So, it lies between 0.5 and 0.6.

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:

(i) If we shade 8 squares, then whole square with shaded portion is given.

Here, total number of squares = 100 and number of shaded squares = 8
\(\therefore \) Ordinary fraction = \(\frac { Shaded\quad squares }{ Total\quad squares } =\frac { 8 }{ 100 } \)
and decimal number = \(\frac { 8 }{ 100 } =0.08\)
(ii) If we shade 15 squares, then whole square with shaded portion is given

Here, total number of squares = 100 and number of shaded squares = 15
\(\therefore \) Ordinary fraction=\(\frac { Shaded\quad squares }{ Total\quad squares } =\frac { 15 }{ 100 } \)
and decimal number = \(\frac { 15 }{ 100 } =0.15\)
(iii) If we shade 50 squares, then whole square with shaded portion is given

Here, total number of squares = 100 and number of shaded squares = 50
\(\therefore \quad Ordinary\quad fraction=\frac { Shaded\quad squares }{ Total\quad squares } =\frac { 50 }{ 100 } \)
and decimal number = \(\frac { 50 }{ 100 } =0.50\)
(iv) If we shade 92 squares, then whole square with shaded portion is given.

Here, total number of squares = 100 and number of shaded squares = 92
\(\therefore \quad Ordinary\quad fraction=\frac { Shaded\quad squares }{ Total\quad squares } =\frac { 92 }{ 100 } \)
\(and\quad decimal\quad number=\frac { 92 }{ 100 } \)
= 0.92
Now, we can write it in the form of table as shown below:

(a) Given, length of Ramesh's notebook
= 9 cm 5 mm
We know that, 10 mm = 1 cm
\(\Rightarrow \quad 1\quad mm=\frac { 1 }{ 10 } cm\)
\(\therefore\) Length of Ramesh's notebook = 9 cm 5 mm
\(=9\quad cm+5\times \frac { 1 }{ 10 } cm\)
\(=9\quad cm+\frac { 5 }{ 10 } cm\)
= 9 cm + 0.5 cm
= (9 + 0.5) cm = 9.5 cm
(b) Given, length of young gram plant = 65 mm
We know that, 10 mm = 1 cm
\(\Rightarrow\) 1 mm = 1/10 cm
\(\therefore\) Length of young gram plant = \(65\times \frac { 1 }{ 10 } cm\)
\(=\frac { 65 }{ 10 } cm=6\frac { 5 }{ 10 } cm=\left( 6+\frac { 5 }{ 10 } \right) cm\)
= (6 + 0.5) cm = 6.5 cm  

Yes, we can represent all decimals on the number line.
We know that, 2.3 is greater than 2, but less than 3. So, divide the unit length between 2 and 3 into 10 equal parts and take 3 parts as shown below:

Thus, A represents 2.3 on the number line. There are 2 ones and 3 tenths in 2.3 and it will lie between 2 and 3 on the number line.

Let five numbers between 0 and 1 be 0.1, 0.3, 0.5, 0.7 and 0.9. Now, we will represent them on the number line.
We know that, all these numbers are greater than 0, but less than 1. So, divide a unit length between 0 and 1 into 10 equal parts and each part represents 0.1 (one-tenth).
Now, take 1 part, 3 parts, 5 parts, 7 parts and 9 parts, respectively. Thus, 0.1, 0.3, 0.5, 0.7 and 0.9 are shown by A, B, C, D and E respectively on the number line.