设甲骑摩托车带乙从A到D行驶x千米，放下乙后骑摩托车折回，而此时丙已从A地步行至E后与甲在F处相遇，甲骑摩托车带丙径直驶向B，恰好与乙同时到达．
2×[（x-

x

50

×5）÷（50+5）]+（120-x）÷50=（120-x）÷5
                   2×[

9

10

x÷55]+2.4-

x

50

=24-

x

5

                          

9

275

x+2.4-

x

50

=24-

x

5

                           18x+1320-11x=13200-110x
                                7x+110x=13200-1320
                                   117x=11880
                                      x=

11880

117

（120-x）÷5+x÷50
=24-

9

50

x
=24-

9

50

×

11880

117

≈24-18.3
=5.7（小时）
答：从A地到B地最少需要5.7小时．