A car travels from point $A$ to point $B$ with a constant speed. If the driver increased the speed of the car by $6$ kmph, it would take him $4$ hours less to cover that distance. And traveling with the speed $6$ kmph lower than the initial speed, it would take him $6$ hours more. Find the distance between $A$ and $B$.
Let the distance between the points $A$ and $B$ be $= d$ km
Let the time taken to cover the distance $AB = t$ hours
Let the speed of the car be $= s$ kmph
Then, $\;$ $d = s \times t$ $\;\;\; \cdots \; (1)$
Case 1: $\;$ Speed of car $= s_1 = \left(s + 6\right)$ kmph
Time taken to cover the distance $AB$ $= t_1 = \left(t - 4\right)$ hours (given)
Then, distance $AB = d = s_1 \times t_1 = \left(s + 6\right) \left(t - 4\right)$ km $\;\;\; \cdots \; (2)$
Case 2: $\;$ Speed of car $= s_2 = \left(s - 6\right)$ kmph
Time taken to cover distance $AB$ $= t_2 = \left(t + 6\right)$ hours (given)
Then, distance $AB = d = s_2 \times t_2 = \left(s - 6\right) \left(t + 6\right)$ km $\;\;\; \cdots \; (3)$
$\therefore \;$ We have from equations $(2)$ and $(3)$
$\left(s + 6\right) \left(t - 4\right) = \left(s - 6\right) \left(t + 6\right)$
i.e. $\;$ $s \cdot t - 4s + 6t - 24 = s \cdot t + 6s - 6t - 36$
i.e. $\;$ $10 s - 12 t - 12 = 0$
i.e. $\;$ $5 s - 6t = 6$ $\;\;\; \cdots \; (4)$
We have from equations $(1)$ and $(2)$
$s \times t = \left(s + 6\right) \left(t - 4\right)$
i.e. $\;$ $s \cdot t = s \cdot t - 4s + 6t - 24$
i.e. $\;$ $-4s + 6t = 24$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously (adding the two equations) gives
$s = 30$ $\implies$ the speed of the car $= 30$ kmph
Substituing the value of $s$ in equation $(4)$ gives
$5 \times 30 - 6t = 6$
i.e. $\;$ $150 - 6t = 6$
i.e. $\;$ $6t = 144$ $\implies$ $t = 24$ $\;$ i.e. $\;$ time taken to cover distance $AB = 24$ hours
$\therefore \;$ By equation $(1)$, distance $AB = 30 \times 24 = 720$ km