A fisher had to sail $35$ km to the meeting place and the other fisher had to sail $31 \dfrac{3}{7}$ percent less. To arrive at the place at the same time as the other fisher, the first fisher satrted half an hour earlier and sailed with the speed exceeding by $2$ kmph the speed of the second fisher. Find the speed of each fisher and the time it took each of them to cover the distance.
Let $F_1$ and $F_2$ be the two fishers.
Let the fishers meet at a point $P$.
Let fisher $F_1$ sail $35$ km to the meeting place
Then, distance covered by $F_1$ $= F_1P = 35$ km $\;\;\; \cdots \; (1)$
and distance covered by $F_2$ $= F_2P = 35 - 35 \times \dfrac{31 \dfrac{3}{7}}{100} = 35 - \dfrac{35 \times 220}{700} = 24$ km $\;\;\; \cdots \; (2)$
Let time taken by fisher $F_2$ to reach point $P$ $= t$ hours
and the speed of fisher $F_2$ be $= s$ kmph
$\therefore \;$ Distance covered by $F_2$ to reach point $P$ $= F_2 P$
$F_2 P = s \times t = 24$ $\;\;\; \cdots \; (3)$ $\;$ [in view of $(2)$]
or, $\;$ $s = \dfrac{24}{t}$ $\;\;\; \cdots \; (3a)$
Since fisher $F_1$ started half an hour earlier than $F_2$ to reach point $P$,
time taken by $F_1$ $= t_1 = \left(t + 0.5\right)$ hours $\;\;\; \cdots \; (4a)$
$F_1$ sailed with a speed exceeding by $2$ kmph the speed of $F_2$.
$\therefore \;$ Speed of fisher $F_1$ $= s_1 = \left(s + 2\right)$ kmph $\;\;\; \cdots \; (4b)$
Distance covered by $F_1$ to reach point $P$ $= F_1 P$
$F_1 P = s_1 \times t_1 = \left(s + 2\right) \left(t + 0.5\right) = 35$ $\;$ [in view of $(1)$]
i.e. $\;$ $s \; t + 0.5 s + 2t + 1 = 35$
i.e. $\;$ $s \; t + 0.5 s + 2t = 34$ $\;\;\; \cdots \; (5)$
In view of equations $(3)$ and $(3a)$, equation $(5)$ becomes
$24 + 0.5 \times \dfrac{24}{t} + 2t = 34$
i.e. $\;$ $2 t^2 - 10 t + 12 = 0$
i.e. $\;$ $t^2 - 5t + 6 = 0$
i.e. $\;$ $\left(t - 3\right) \left(t - 2\right) = 0$
i.e. $\;$ $t = 3$ $\;$ or $\;$ $t = 2$
When $t = 3$,
$s = \dfrac{24}{3} = 8$ $\;$ [by equation $(3a)$]
$s_1 = 8 + 2 = 10$ $\;$ [by equation $(4b)$]
and $\;$ $t_1 = 3 + 0.5 = 3.5$ $\;$ [by equation $(4a)$]
When $t = 2$,
$s = \dfrac{24}{2} = 12$ $\;$ [by equation $(3a)$]
$s_1 = 12 + 2 = 14$ $\;$ [by equation $(4b)$]
and $\;$ $t_1 = 2 + 0.5 = 2.5$ $\;$ [by equation $(4a)$]
$\therefore \;$ Speed of $F_1$ $= 10$ kmph; $\;$ time taken by $F_1$ to reach $P$ $= 3.5$ hr
and speed of $F_2$ $= 8$ kmph; $\;$ time taken by $F_2$ to reach $P$ $= 3$ hr
OR
Speed of $F_1$ $= 14$ kmph; $\;$ time taken by $F_1$ to reach $P$ $= 2.5$ hr
and speed of $F_2$ $= 12$ kmph; $\;$ time taken by $F_2$ to reach $P$ $= 2$ hr