The "swim time", "cycle time", and "run time" are given by the product of the inverse of the velocities on each exercise, by the length they do on each one. The swimming time is given by the inverse of the velocity times 1.5km(distance they have to swim) for example. Using this concept, the equation for the total time is:
[tex]\frac{1.5}{v_s}_{}+t_1+\frac{40}{v_c}_{}+t_2+\frac{10}{v_r}_{}=T_{}[/tex]For Alexandra, we have the following values:
[tex]\begin{gathered} v_s=5\operatorname{km}/h \\ t_1=4\min =\frac{1}{15}h \\ v_c=15\operatorname{km}/h \\ t_2=3\min =\frac{1}{20}h \\ v_r=6\operatorname{km}/h \end{gathered}[/tex]Since the transition times were in minutes, I had to convert them to hours.
Just making the substitution with those values in our equation, we have the following total time for Alexandra
[tex]\begin{gathered} T=\frac{1.5}{5}+\frac{1}{15}+\frac{40}{15}+\frac{1}{20}+\frac{10}{6} \\ T=\frac{45.5}{15}+\frac{1}{20}+\frac{5}{3}=\frac{70.5}{15}+\frac{1}{20} \\ T=4.75 \end{gathered}[/tex]Alexandra took 4.75 hours to complete the triathlon.
For Chi, we have the following values:
[tex]\begin{gathered} v_s=6\operatorname{km}/h \\ t_1=\frac{1}{12}h \\ v_c=15\operatorname{km}/h \\ t_2=\frac{1}{12}h \\ v_r=7.5\operatorname{km}/h \end{gathered}[/tex]Now, solving the total time for Chi:
[tex]\begin{gathered} T=\frac{1.5}{6}+\frac{1}{12}+\frac{40}{15}+\frac{1}{12}+\frac{10}{7.5} \\ T=\frac{5}{12}+\frac{60}{15}=4+\frac{5}{12}\approx4.42 \end{gathered}[/tex]Chitook 4.42 hours to complete the triathlon.