\(y=-12 \frac{7}{12}+3 \frac{1}{3}=\frac{-151}{12}+\frac{10}{3}\)

\(=\frac{-151+40}{12}=\frac{-111}{12}=\frac{-37}{4}=-9 \frac{1}{4}\)

Let the other number be x.

Then, \(x \times\left(\frac{-4}{3}\right)=\frac{-9}{16}\)

\(\Rightarrow x=\frac{-9}{16} \times\left(\frac{3}{-4}\right) \Rightarrow x=\frac{27}{64}\)

Total time taken by Rajeev, Simran and Bhuvik to walk around circular park = \(\left(\frac{2}{3}+\frac{3}{5}+\frac{7}{12}\right)\)h = \(\left(\frac{40+36+35}{60}\right)\) hr. \(=\frac{111}{60}\) hr \(=\frac{111}{60} \times 60\) minutes \(=111\) minutes

LCM of \(7,3,3,9=63\)

\(\therefore \frac{4}{7}=\frac{4}{7} \times \frac{9}{9}=\frac{36}{63}, \frac{1}{3}=\frac{1}{3} \times \frac{21}{21}=\frac{21}{63}\), \(\frac{2}{3}=\frac{2}{3} \times \frac{21}{21}=\frac{42}{63}, \frac{5}{9}=\frac{5}{9} \times \frac{7}{7}=\frac{35}{63}\)

So, ascending order is \(\frac{21}{63}, \frac{35}{63}, \frac{36}{63}, \frac{42}{63}\)

i.e., \(\frac{1}{3}<\frac{5}{9}<\frac{4}{7}<\frac{2}{3}\)

Now, the average of \(\frac{5}{9}\) and \(\frac{4}{7}\) is \(\frac{1}{2}\left[\frac{5}{9}+\frac{4}{7}\right]\)

\(=\frac{1}{2}\left[\frac{35+36}{63}\right]=\frac{71}{126}\)

(a) \(\frac{4}{9}+\frac{-5}{9}=\frac{4-5}{9}=\frac{-1}{9}\)

(b) \(\frac{-2}{5}+\frac{13}{20}=\frac{-8+13}{20}=\frac{5}{20}\)

(c) \(\frac{-5}{12}+\frac{11}{-12}=\frac{-5}{12}-\frac{11}{12}=\frac{-5-11}{12}=\frac{-16}{12}\)

(d) \(\frac{-7}{8}+\frac{1}{12}+\frac{2}{3}=\frac{-21+2+16}{24}=\frac{-3}{24}\)

\(\therefore\) Option (a) is in simplest form

\(\frac{3}{13}+\left(\frac{-13}{33}\right)=\frac{3}{13}-\frac{13}{33}=\frac{99-169}{429}=\frac{-70}{429}\)

First term \(\left(-\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=\frac{-3+1+2}{6}=0\)

Now, zero multiplied by any number is zero. This is called mathematical ingenuity.

Otherwise if you evaluate \(\left(36 \frac{7}{15}-91 \frac{8}{19}+1011 \frac{231}{611}\right)\), it is very tedious.

Positive rational numbers are greater than negative rational numbers.

Total length of the rope = 90m.

Length of each piece = \(2 \frac{1}{4}\) m

Number of pieces = \(\frac{90}{2 \frac{1}{4}}=\frac{90}{\frac{9}{4}}\)

\(=90 \times \frac{4}{9}=40\)

\(\frac{-3}{8}=\frac{x}{-24} \Rightarrow \frac{-3}{8} \times \frac{3}{3}=\frac{-x}{24}\) \(\Rightarrow\) \(\frac{-9}{24}=\frac{-x}{24} \Rightarrow x=9\)

\(\frac{x}{y}=\frac{9}{8}\)  (Given)

\(\therefore\left(\frac{6}{7}+\frac{y-x}{y+x}\right)=\left(\frac{6}{7}+\frac{y\left(1-\frac{x}{y}\right)}{y\left(1+\frac{x}{y}\right)}\right)\) \(=\frac{6}{7}+\left(\frac{1-\frac{9}{8}}{1+\frac{9}{8}}\right)=\frac{6}{7}+\left(\frac{\frac{-1}{8}}{\frac{17}{8}}\right)\) \(=\frac{6}{7}+\left(\frac{-1}{8} \times \frac{8}{17}\right)=\frac{6}{7}-\frac{1}{17}=\frac{102-7}{119}=\frac{95}{119}\)

Point \(D \Rightarrow-1,\) Point \(y=(4 \text { division }=3 \text { units })\)

\(\therefore\) 3 divisions \(=\frac{9}{4}\) units

\(\therefore\) Point \(y=1+\frac{9}{4}=\frac{13}{4}\)

\(\therefore\) D + Y = \(-1+\frac{13}{4}=\frac{9}{4}\)

Let the number to be subtracted be 'x'.

Then, \(\frac{-1}{3}-x=\frac{1}{6}\)

\(\Rightarrow \frac{-1}{3}=\frac{1}{6}+x\)

\(\Rightarrow x=\frac{-1}{3}-\frac{1}{6}=\frac{-2-1}{6}=\frac{-3}{6}=\frac{-1}{2}\)

\(a=\frac{20}{7}, b=\frac{1}{4}: C=\frac{11}{19} ; d=\frac{-9}{4}\)

\(a(b-c) \div d=\frac{\frac{20}{7}\left(\frac{1}{4}-\frac{11}{19}\right)}{(-9 / 4)}=\frac{20}{7} \times \frac{-4}{9} \rightarrow \times\left(\frac{19-44}{76}\right)\)

\(=\frac{-80}{63} \times \frac{-25}{76}=\frac{80 \times 25}{63 \times 76}=\frac{500}{1197}\)

Required number \(=\frac{-33}{4} \div \frac{-22}{3}\)\(=\frac{-33}{4} \times \frac{3}{-22}\)\(=\frac{-99}{-88} \times \frac{99}{88}=\frac{9}{8}\)