3. (10 points) Person A and Person B start riding bicycles from place $A$ to place $B$ at the same time. Person A's speed is 1.2 times that of Person B. After riding 4 kilometers, Person B's bicycle breaks down, and the downtime is equivalent to the time it would take to ride $\frac{1}{6}$ of the entire journey. After fixing the bicycle, Person B's speed increases by 60%, and as a result, both Person A and Person B arrive at place $B$ at the same time. The distance between $A$ and $B$ is $\qquad$ kilometers.
【Analysis】Let the distance between $A$ and $B$ be $X$; the speed of Yi is $V$, then the speed of Jia is $1.2 V$; when Yi has walked 4000, Jia must have walked 4800; assuming the time Yi takes to fix the fault is $t$, then the distance Jia walks during this time is $1.2 V \times\left(\frac{\mathrm{X}}{6} \times \frac{1}{V}\right)$ $=0.2 X$; we assume the time after Yi fixes the fault is $T$, we can set up the equation: $4800+0.2 X+1.2 V T=4000+1.6 V T$; we get $800+0.2 X=0.4 V T$; since $X=4000+1.6 V T$, substituting this in gives: $V T=20000$, then we can calculate $X=36000$ meters $=36$ kilometers.

【Solution】Solution: Let the distance between $A$ and $B$ be $X$; the speed of Yi is $V$, then the speed of Jia is $1.2 V$: when Yi has walked 4000, Jia has walked: $4000 \times 1.2=4800$ (meters); assuming the time Yi takes to fix the fault is $t$, then the distance Jia walks during this time is: $1.2 V \times\left(\frac{\mathrm{X}}{6} \times \frac{1}{V}\right)=0.2 X$; assuming the time after Yi fixes the fault is $T$, we can set up the equation: $4800+0.2 X+1.2 V T=4000+1.6 V T$; we get $800+0.2 X=0.4 V T$; since $X=4000+1.6 V T$, substituting this in gives: $V T=20000$, substituting $V T=20000$ into $4800+0.2 X+1.2 V T=$ 4000+1.6VT,

we get: $X=36000$ meters $=36$ kilometers.
Answer: The distance between $A$ and $B$ is 36 kilometers.
The answer is: 36.