let A travels x m when when it meets B in that track. then B travels (44-x) m
so, x/7 =(44-x)/11 => x=17.11
in 5th A travels a total distance =17.11*5 = 85.6 m
in 8th meeting= 17.11*8= 136.9m
so, in 5th meeting A remain (88-85.6)=2.4 m behind the starting point
and in 8 meeting A remain (136.9-132)=4.9 m ahead of the starting point
so the distance between two meetings= 4.9+2.4 = 7.3 ~ 7
Ans: 7
am iright??
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so, x/7 =(44-x)/11 => x=17.11
in 5th A travels a total distance =17.11*5 = 85.6 m
in 8th meeting= 17.11*8= 136.9m
so, in 5th meeting A remain (88-85.6)=2.4 m behind the starting point
and in 8 meeting A remain (136.9-132)=4.9 m ahead of the starting point
so the distance between two meetings= 4.9+2.4 = 7.3 ~ 7
Ans: 7
am iright??
Posted from my mobiledevice
...
