Suppose, race has been conducted for distance $S$ Km and speed of cars $X,Y,Z$ are $v_1,v_2,v_3$ respectively in km per unit time.

“car X beats car Y by 36 km” means for the same time, X is ahead of Y by $36$ km. Similar meaning is for other two statements. So, we will be having three equations as:

$\frac{S}{v_1} = \frac{S- 36}{v_2}$ $(1)$

$\frac{S}{v_2} = \frac{S- 72}{v_3}$ $(2)$

$\frac{S}{v_1} = \frac{S- 90}{v_3}$ $(1)$

From $(1),$ $\frac{v_2}{v_1} = \frac{S- 36}{S}$

From $(2),$ $\frac{v_2}{v_3} = \frac{S}{S – 72}$

After dividing both equations, we get, $\frac{v_3}{v_1} = \frac{(S- 36)(S – 72)}{S^2}$ $(3)$

From $(3),$ $\frac{v_3}{v_1} = \frac{S- 90}{S}$ $(4)$

From $(3)$ and $(4),$ $\frac{S- 90}{S} = \frac{(S- 36)(S – 72)}{S^2}$, $S \neq 0$

$S= 144$ km = $144000 m$

“car X beats car Y by 36 km” means for the same time, X is ahead of Y by $36$ km. Similar meaning is for other two statements. So, we will be having three equations as:

$\frac{S}{v_1} = \frac{S- 36}{v_2}$ $(1)$

$\frac{S}{v_2} = \frac{S- 72}{v_3}$ $(2)$

$\frac{S}{v_1} = \frac{S- 90}{v_3}$ $(1)$

From $(1),$ $\frac{v_2}{v_1} = \frac{S- 36}{S}$

From $(2),$ $\frac{v_2}{v_3} = \frac{S}{S – 72}$

After dividing both equations, we get, $\frac{v_3}{v_1} = \frac{(S- 36)(S – 72)}{S^2}$ $(3)$

From $(3),$ $\frac{v_3}{v_1} = \frac{S- 90}{S}$ $(4)$

From $(3)$ and $(4),$ $\frac{S- 90}{S} = \frac{(S- 36)(S – 72)}{S^2}$, $S \neq 0$

$S= 144$ km = $144000 m$