Given : \(3\frac { 1 }{ 3 } \)
Improper fraction = \(\cfrac { (Whole\ number\ X\ Deno\ minator)+Numerator }{ Deominator } \)
= \(\cfrac { \left( 3\times 3 \right) +1 }{ 3 } =\cfrac { 9+1 }{ 3 } \)
(\(\therefore\) W.no = 3, D = 3, N = 1)
= \(\cfrac { 10 }{ 3 } \)
(ii)  Given :\(\cfrac { 45 }{ 7 } \)
Mixed fraction = Quotilent + \(\cfrac { Remainder }{ Divisor } \)
= \(6+\cfrac { 3 }{ 7 } \)
= \(6\cfrac { 3 }{ 7 } \)

(i) Total no. of boxes = 3
shaded portion = 2
\(\therefore\) Fraction = \(\cfrac { 2 }{ 3 } \)
(ii) Total no. of boxes = 4
shaded portion' = 3
\(\therefore\) Fraction = \(\cfrac { 3 }{ 4 } \)
(iii) Total no. of boxes = 5
shaded portion = 4
\(\therefore\) Fraction = \(\cfrac { 4 }{ 5 } \)

Difference between \(1\cfrac { 1 }{ 3 } \) and \(3\cfrac { 1 }{ 6 } \)
= \(3\cfrac { 1 }{ 6 } -1\cfrac { 1 }{ 3 } \)
= \(\cfrac { 19 }{ 6 } -\cfrac { 4 }{ 3 } \)
LCM of 6 and 3 is 6
= \(\cfrac { 19 }{ 6 } -\cfrac { 4\times 2 }{ 3\times 2 } =\cfrac { 19 }{ 6 } -\cfrac { 8 }{ 6 } =\cfrac { 19-8 }{ 6 } \)
= \(\cfrac { 11 }{ 6 } \)
Also given:
Difference between \(4\cfrac { 1 }{ 6 }\ and\ 2\cfrac { 1 }{ 3 } \)
= \(4\cfrac { 1 }{ 6 } -2\cfrac { 1 }{ 3 } \)
= \(\cfrac { 25 }{ 7 } -\cfrac { 7 }{ 3 } \)
LCM of 6 and 3 is 6
= \(\cfrac { 25 }{ 6 } -\cfrac { 7\times 2 }{ 3\times 2 } =\cfrac { 25 }{ 6 } -\cfrac { 14 }{ 6 } =\cfrac { 25-14 }{ 6 } \)
= \(\cfrac { 11 }{ 6 } \)
Adding (1) and (2), we get
= \(\cfrac { 11 }{ 6 } +\cfrac { 11 }{ 6 } =\cfrac { 11+11 }{ 6 } =\cfrac { 22 }{ 6 } \)
= \(\cfrac { 11 }{ 3 } =3\cfrac { 2 }{ 3 } \)

To know the share, we need to find \({1\over2}\div 2\)
\({1\over2}\div 2={1\over2}\times{1\over2}\)(reciprocal of 2 is \(\frac{1}{2}\))
\(={1\over2}\times{1\over2}={1\times1\over 2\times2}={1\over 4}\)

Common multiple of 2 and 4 is 4
Equivalent fraction of \(1\over2\) is
\({1\over2}={1\times2\over 2\times2}={2\over4}\)
Now 
\({3\over 4}-{1\over2}={3\over4}-{2\over4}={3-2\over4}={1\over 4}\)
Therefore, Vani has \(1\over4\) amount of water in the bottle. This can be verified by the following diagram

Equivalent fractions of \(2\over 3\)are \({4\over 6},{6\over 9},{8\over 12},{10\over 15},{12\over 18},.....\)
Equivalent fractions of \(1\over 6\) are \({2\over12},{3\over 18}\) ,...
Equivalent fractions of \(4\over9\) is \({8\over18},....\)
Therefore \({3\over 18}<{8\over18}<{12\over 18}\)
The ascending order of given fractions is \({1\over 6},{4\over 9},{2\over3}\)

Equivalent Fraction of \(3\over 4\)
\({3\over 4}={3\times2\over 4\times 2}={6\over 8}\)
\({3\over 4}={3\times3\over 4\times 3}={9\over 12}\)
\({3\over 4}={3\times4\over 4\times 4}={12\over 16}\)
Equivalent Fraction of \(2 \over 7\)
\({2 \over 7}={2\times2\over 7\times2}={4\over 14}\)
\({2 \over 7}={2\times3\over 7\times3}={6\over 21}\)
\({2 \over 7}={2\times4\over 7\times4}={8\over 28}\)
Equivalent fractions of \({3\over4}:{3\over 4}={6\over 8}={9\over 12}={12\over 16}\)
Equivalent fractions of \({2\over7}:{3\over 7}={4\over 14}={6\over 21}={8\over 28}\)

Given:\(\cfrac { 9 }{ 20 } ,\cfrac { 3 }{ 4 } ,\cfrac { 7 }{ 12 } \)
LCM of 20, 4, 12 is 60
\(\cfrac { 9 }{ 30 } \times \cfrac { 3 }{ 3 } =\cfrac { 27 }{ 60 } ,\cfrac { 3 }{ 4 } \times \cfrac { 15 }{ 15 } =\cfrac { 45 }{ 60 } ,\cfrac { 7 }{ 12 } \times \cfrac { 5 }{ 5 } =\cfrac { 35 }{ 60 } \)
\(\Rightarrow \cfrac { 27 }{ 60 } <\cfrac { 35 }{ 60 } <\cfrac { 45 }{ 60 } \Rightarrow \cfrac { 45 }{ 60 } >\cfrac { 35 }{ 60 } >\cfrac { 27 }{ 60 } \)
\(\therefore\)  Descending order is \(\cfrac { 3 }{ 4 } ,\cfrac { 7 }{ 12 } ,\cfrac { 9 }{ 20 } \)
(i) LCM of 3 and 7 is 21
 \(\Rightarrow \cfrac { 2 }{ 3 } \times \cfrac { 7 }{ 7 } +\cfrac { 5 }{ 7 } \times \cfrac { 3 }{ 3 } \)
\(\cfrac { 14 }{ 21 } +\cfrac { 15 }{ 21 } =\cfrac { 14+15 }{ 21 } =\cfrac { 29 }{ 21 } \)
(ii) LCM of5 and 8 is 40
\(\Rightarrow \cfrac { 3 }{ 5 } \times \cfrac { 8 }{ 8 } -\cfrac { 3 }{ 8 } \times \cfrac { 5 }{ 5 } \)
= \(\cfrac { 24 }{ 60 } -\cfrac { 15 }{ 40 } =\cfrac { 24-15 }{ 40 } =\cfrac { 9 }{ 40 } \)

Given:\(\cfrac { 3 }{ 8 } ,\cfrac { 3 }{ 5 } \)
L. C. M of 8 and 5 is 40
\(\cfrac { 3 }{ 8 } \times \cfrac { 5 }{ 5 } =\cfrac { 15 }{ 40 } ,\cfrac { 3 }{ 5 } \times \cfrac { 8 }{ 8 } =\cfrac { 24 }{ 40 } \)
\(\therefore \ \cfrac { 24 }{ 40 } >\cfrac { 15 }{ 40 } \Rightarrow \cfrac { 3 }{ 5 } >\cfrac { 3 }{ 8 } \)
\(>\cfrac { 3 }{ 8 } \\ \) is greater

(i) \(\cfrac { 1 }{ 3 } \) is greater than \(\cfrac { 1 }{ 5 } \)
That is \(\cfrac { 1 }{ 3 } \ge \cfrac { 1 }{ 5 } \)
(ii) \(\cfrac { 2 }{ 5 } \) is smaller \(\cfrac { 5 }{ 8 } \)
That is \(\cfrac { 2 }{ 5 } \le \cfrac { 1 }{ 5 } \)

Mangai bought = \(6\cfrac { 3 }{ 4 } kg\) of apple 
Kalai bought = \(1\cfrac { 1 }{ 2 } \) times as Mangai bought
= \(\left[ 1\cfrac { 1 }{ 2 } \times 6\cfrac { 3 }{ 4 } \right] kg\) of apples

= \(\cfrac { 3 }{ 2 } \times \cfrac { 27 }{ 4 } \)
= \(\cfrac { 81 }{ 8 } \)
\(\therefore \) Kalai bought = \(10\cfrac { 1 }{ 8 } \) kg of apples.

Sankari purchased
To stitch a long skirt = \(2\cfrac { 1 }{ 2 } \ m\ cloth=\cfrac { 5 }{ 2 } m\)
To Stitch blouse = \(1\cfrac { 3 }{ 4 } m=\cfrac { 7 }{ 4 } m\)
Total length of clothe = \(\cfrac { 5 }{ 2 } m+\cfrac { 7 }{ 4 } m|\)
= \(\cfrac { (5\times 2)+7 }{ 4 } =\cfrac { 10+7 }{ 4 } \)
= \(\cfrac { 17 }{ 4 } m\)
Cost of 1 m cloth = Rs. 120
\(\therefore \cfrac { 17 }{ 4 } m\ cloth=120\times \cfrac { 17 }{ 4 } \)
= Rs. 510
Cost of cloth purchased by Sankari =  Rs. 510

\(5\cfrac { 4 }{ 9 } \) and \(3\cfrac { 1 }{ 6 } \)
\(\Rightarrow \cfrac { 49 }{ 9 } +\cfrac { 19 }{ 6 } \)
LCM of 9 and 6 is 18
= \(\cfrac { 49\times 2 }{ 9\times 2 } +\cfrac { 19\times 3 }{ 6\times 3 } \)
= \(\cfrac { 98 }{ 18 } +\cfrac { 57 }{ 18 } \)
= \(\cfrac { 98+57 }{ 18 } =\cfrac { 155 }{ 18 } \)
= \(8\cfrac { 11 }{ 18 } \)
(ii) \(7\cfrac { 1 }{ 6 } \Rightarrow \cfrac { 43 }{ 6 } \)
\(12\cfrac { 3 }{ 8 } \Rightarrow \cfrac { 99 }{ 8 } \)
Now to find : \(\cfrac { 99 }{ 8 } -\cfrac { 43 }{ 6 } \)
LCM = 24
= \(\cfrac { 99\times 3 }{ 8\times 3 } -\cfrac { 43\times 4 }{ 6\times 4 } \)
= \(\cfrac { 297 }{ 24 } -\cfrac { 172 }{ 24 } \)
= \(\cfrac { 297-172 }{ 24 } =\cfrac { 125 }{ 24 } \)
= \(5\cfrac { 5 }{ 24 } \)
(iii) To find
\(\left( 9\cfrac { 2 }{ 3 } +2\cfrac { 1 }{ 2 } \right) -\left( 6\cfrac { 1 }{ 6 } +3\cfrac { 1 }{ 5 } \right) \)
\(\Rightarrow \left( \cfrac { 29 }{ 3 } +\cfrac { 25 }{ 2 } \right) -\left( \cfrac { 37 }{ 6 } +\cfrac { 16 }{ 5 } \right) \)
\(\Rightarrow \) LCM of 3 and 2 is 6
LCM of 6 and 5 is 30
\(\Rightarrow \left( \cfrac { 29\times 2 }{ 3\times 2 } +\cfrac { 5\times 3 }{ 2\times 3 } \right) -\left( \cfrac { 37\times 5 }{ 6\times 5 } +\cfrac { 16\times 6 }{ 5\times 6 } \right) \)
\(\Rightarrow \left( \cfrac { 58 }{ 6 } +\cfrac { 15 }{ 6 } \right) -\left( \cfrac { 185 }{ 30 } +\cfrac { 96 }{ 30 } \right) \)
\(\Rightarrow \left( \cfrac { 58+15 }{ 6 } \right) -\left( \cfrac { 185+96 }{ 30 } \right) \)
\(\Rightarrow \cfrac { 73 }{ 6 } -\cfrac { 281 }{ 30 } \)
LCM of 6 and 30 is 30
\(\Rightarrow \cfrac { 73\times 5 }{ 6\times 5 } -\cfrac { 281 }{ 30 } \)
\(\cfrac { 365 }{ 30 } -\cfrac { 281 }{ 30 } =\) 
= \(2\cfrac { 4 }{ 5 } \)

Total distance = \(26\frac { 1 }{ 4 } \) m
= \(\cfrac { 105 }{ 4 } \)
A rabbit covers in 1 jump = \(1\frac { 3 }{ 4 } \)
= \(\cfrac { 7 }{ 4 } \)m
\(\cfrac { 7 }{ 4 } \) m = 1 jump
\(\cfrac { 105 }{ 4 } m=\cfrac { 7 }{ 7/4 } \times \cfrac { 105 }{ 4 } \)
= \(\cfrac { 4 }{ 7 } \times \cfrac { 105 }{ 4 } =15\)