$$\eqalign{ & {\text{Second}}:{\text{Third}} \cr & = {\text{9}}:{\text{16}} \cr & {\text{Third}}:{\text{first}} \cr & = {\text{4}}:{\text{1}} \cr & {\text{ = 16}}:{\text{4}}{\text{}} \cr & \refore {\text{Second}}:{\text{Third}}:{\text{first}} \cr & = {\text{9}}:{\text{16}}:{\text{4}}{\text{}} \cr & {\text{Second number}} \cr & = \left( {{\text{116}} \times \frac{{\text{9}}}{{{\text{29}}}}} \right) \cr & = {\text{36}} \cr} $$

Quantity of milk
$$ = \left( {60 \times \frac{2}{3}} \right)\,{\text{litres}} = 40\,{\text{litres}}$$
Quantity of water in it = (60 - 40) litres = 20 litres
New ratio = 1 : 2
Let quantity of water to be added furr be x litres
$$\eqalign{ & {\text{Then,}}\,{\text{milk : water}} = {\frac{{40}}{{20 + x}}} \cr & {\text{Now}},\, {\frac{{40}}{{20 + x}}} = \frac{1}{2} \cr & \Rightarrow 20 + x = 80 \cr & \Rightarrow x = 60 \cr & \refore {\text{Quantity}}\,{\text{of}}\,{\text{water}}\,{\text{to}}\,{\text{be}}\,{\text{added = 60}}\,{\text{litres}}{\text{.}} \cr} $$

$$\eqalign{ & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3},\,\frac{{\text{B}}}{{\text{C}}} = \frac{2}{4},\,\frac{{\text{C}}}{{\text{D}}} = \frac{2}{5} \cr & \Rightarrow \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \Rightarrow \frac{2}{3} \times \frac{2}{4} \times \frac{2}{5} = \frac{2}{{15}} \cr & \Rightarrow {\text{A}}:{\text{D}} = 2:15. \cr} $$

$$\eqalign{ & {\text{Let}}\,40\% \,{\text{of}}\,A = \frac{2}{3}B \cr & {\text{Then}},\,\frac{{40A}}{{100}} = \frac{{2B}}{3} \cr & \Rightarrow \frac{{2A}}{5} = \frac{{2B}}{3} \cr & \Rightarrow \frac{A}{B} = {\frac{2}{3} \times \frac{5}{2}} = \frac{5}{3} \cr & \refore A:B = 5:3 \cr} $$

$$\eqalign{ & {\text{a}}:{\text{b}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{c}}:{\text{d}} \cr & 5:7\,\,\,\,\,\,\,\,\,\,\,\,\,2{\text{a}}:3{\text{b}} \cr & \frac{a}{b} = \frac{5}{7},\,\frac{c}{d} = \frac{{2a}}{{3b}} \cr & = \frac{2}{3} \times \frac{5}{7} \cr & = \frac{{10}}{{21}} \cr & \refore ac:bd \cr & = \frac{{{\text{ac}}}}{{{\text{bd}}}} \cr & = \frac{5}{7} \times \frac{{10}}{{21}} \cr & = \frac{{50}}{{147}} \cr & = 50:147 \cr} $$

Ratio of students in 2012 in institutes A and B = 7 : 15
Let number of students in institute A in 2012 = 700
And Number of students in institutes B in 2012 = 1500
25% increase in number of students in 2013
Now, number of students in Institute A = 700 + 25% of 700 = 875
Number of students in B in 2013 as 26% students increased in B
= 1500 + 26% of 1500 = 1890
Current Ratio of students,
$$ = \frac{{875}}{{1890}} = 25:54$$

Let missing number be x.
Then,
$$\eqalign{ & = \frac{1}{3}:\frac{1}{2}::27y:x \cr & \Rightarrow \frac{3}{4}x = \frac{1}{2} \times {\text{27}}y \cr & \Rightarrow x{\text{ = }}\frac{{27y}}{2} \times \frac{4}{3} \cr & = 18y \cr} $$

$$\eqalign{ & 20\% \,{\text{of A}} = 30\% \,{\text{of B}} = \frac{1}{6}{\text{of C}} \cr & \Rightarrow \frac{{20{\text{A}}}}{{100}} = \frac{{30{\text{B}}}}{{100}} = \frac{{\text{C}}}{{\text{6}}} \cr & \Rightarrow \frac{{\text{A}}}{5} = \frac{{3{\text{B}}}}{{10}} = \frac{C}{6} = k(say) \cr & \Rightarrow A = 5K,\,B = \frac{{10k}}{3},\,C = 6k \cr & \refore {\text{A}}:{\text{B}}:{\text{C}} = 5k:\frac{{10k}}{3}:6k \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5:\frac{{10}}{3}:6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15:10:18 \cr} $$