Answer:
[tex]\large \boxed{\text{a) (i) 37,8 Mt; (ii) 1,9 Mt; b) 72 \%}}[/tex]
Step-by-step explanation:
Assume the data in your table are like the one below.
[tex]\begin{array}{ccccc}\textbf{Means of} & \textbf{Percent} & \textbf{of total} & \textbf{Cargo} & \textbf{transported} \\\textbf{transport} & \textbf{transportation} & \textbf{to the port} & \textbf{(in millions} & \textbf{of tons)} \\&\mathbf{2007 } &\mathbf{2008} & \mathbf{2007} & \mathbf{2008 } \\\text{Rail} & 18\, \% & 24\, \% & \text{6,8} & \text{8,8} \\\text{Road} & 77\, \% & & \text{29,1} & \text{ } \\\text{Pipeline} & \text{ } & \text{ } & \text{ } & \text{ } \\\end{array}[/tex]
a) Cargo transported in 2007
(i) Total cargo
Let x = total cargo. Then
0.77x = amount carried by road
[tex]\begin{array}{rcl}0.77x & = & \text{29,1 Mt}\\x & = & \dfrac{\text{29,1 Mt}}{0.77}\\\\& = & \textbf{37,8 Mt}\\\end{array}\\\text{Total cargo carried to the port was $\large \boxed{\textbf{37,8 Mt}}$}[/tex]
(ii) Transported by pipeline
Let p = cargo transported by pipeline
[tex]\begin{array}{rcl}\text{Total} & = & \text{road + rail + pipeline}\\\text{37,8 Mt} & = & \text{(6,8 + 29,1 + p) Mt}\\\text{37,8} & = & \text{35,9 + p}\\p & = & \text{1,9}\\\end{array}\\\text{The amount of cargo carried by pipeline was $\large \boxed{\textbf{1,9 Mt}}$}[/tex]
b) Percentage share of road transport
(i) Total cargo
Let x = total cargo. Then
0.24x = amount carried by road
[tex]\begin{array}{rcl}0.77x & = & \text{29,1 Mt}\\x & = & \dfrac{\text{29,1 Mt}}{\text{0,77}}\\\\& = & \textbf{37,8 Mt}\\\end{array}\\[/tex]
(ii) Cargo transported by road
Cargo = 29,1 Mt - 2,7 Mt = 26,4 Mt
(iii) Percent of total
[tex]\text{Percent} = \dfrac{\text{26,4 Mt}}{\text{36,7 Mt}} \times \, 100 \, \% = \mathbf{72 \, \%}[/tex]
