(1) \(\rightarrow\) Additive Inverse
(2) \(\rightarrow\) Rational Inverse
(3) \(\rightarrow\) Product
(4) \(\rightarrow\) Reciprocal
(5) \(\rightarrow\) Additive
(6) \(\rightarrow\) Inverse
(7) \(\rightarrow\) Zero

We have
A rational number between \(\frac { 1 }{ 2 } \)
= \(\left[ \frac { 1 }{ 2 } +(-2) \right] \div 2=\left[ \frac { 1-4 }{ 2 } \right] \div 2\)
= \(\left[ \frac { -3 }{ 2 } \right] \times \frac { 1 }{ 2 } =\frac { -3 }{ 4 } \)
A rational number between \(\frac { 1 }{ 2 } \) and \(\left( \frac { -3 }{ 4 } \right) \)
= \(\left[ \frac { 1 }{ 2 } +\left( \frac { -3 }{ 4 } \right) \right] \div 2\)
\(\left[ \frac { 2-3 }{ 4 } \right] \times \frac { 1 }{ 2 } =\frac { -1 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -1 }{ 8 } \)
A rational number between \(\left( \frac { -3 }{ 4 } \right) \) and (-2)
= \(\left[ \left( \frac { -3 }{ 4 } \right) +(-2) \right] \div 2=\left[ \frac { (-3)+(-8) }{ 4 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -11 }{ 4 } \times \frac { 1 }{ 2 } =\frac { -11 }{ 8 } \)
Thus, the three rational numbers
\(\left( \frac { -3 }{ 4 } \right) ,\left( \frac { -1 }{ 8 } \right) \) and \(\left( \frac { -11 }{ 8 } \right) \) are between \(\frac { 1 }{ 2 } \) and (-2)

Given, sugar required for one fruit salad =\({1\over2}\) cup
Sugar required for another salad = 2\(\times{1\over16}={2\over16}cup\)
\(\therefore\) Required sugar =\({1\over2}-{2\over16}={18-2\over16}={6\over16}={3\over8}cup\)

\((a) \because {1\over2}={1\over2}\times{11\over11}={11\over22} and {5\over11}={5\over11}\times{2\over2}={10\over22}\)
So, paper recycled is less than \({1\over2}\)
\((b) Similarly ,{5\over8}is \ greater \ than {1\over2}(={4\over8})\)
\(\\ Also, {2\over5}={2\times2\over5\times2}={4\over10}<{1\over2}(={5\over10})\)
\(\\ and {3\over4}>{1\over2}(={2\over4}) \)
So, the quantity of paper and glass recycled is less than \({1\over2}\)
(c) Quantity of cans = \({5\over8}(={10\over16})\) is more than half of the quantity of aluminium cans
\(={5\over8}\times {1\over2}={5\over16}\)
(d) Scrap> Aluminium cans> Paper> Glass
(e) Recycling is the process, which reduce our  wastage and create the same matter from the used part of the matter and hence, it is very useful

Seema gets allowance = \(1\over2\)of Huma's share 
Hubna gets allowance =\(1{2\over3}\) of Seema's share
\(={5\over3}of \ seema's \ share\)
\(\\={5\over3}of {1\over2}of \ Huma's \ sh are \)
\(\\ [\because Seema's \ share ={1\over2}of \ Huma's share]\)
\(\\ ={5\over3}\times{1\over2}={5\over6}of \ Huma's \ share\)
But Huma, Hubna and Seema get total share
= Rs 2016
\(\therefore\) Huma's share + Hubna's share + Seema's share
= Rs 2016
1 of Huma's share + \(5\over6\)of Huma's share +\(1\over2\) of Huma's share =Rs 2016
\(So,({1+{5\over6}+{1\over2}}) of \ huma's \ share=Rs2016\)
\(\\ \Rightarrow ({6+5+3\over6})of \ Huma's s\ share \)
\(\\ \Rightarrow {14\over6}of \ Huma's \ share = RS2016\)
\(\\ \therefore Huma's \ share =Rs2016\div {14\over6}\)
\(\\ =Rs2016\times{6\over14}=144\times6=864=432 and \ Hubna's \ share ={5\over6}of 864=5\times144=720\)

\({11\over5},{12\over5},{13\over5},{14\over5}\)

\(51\over46\)

\(6{7\over8}\)