Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 42145 by Tawa1 last updated on 18/Aug/18

Commented by MrW3 last updated on 18/Aug/18

it moves 2 meters in 8 seconds. for  the first 6 meters it needs 24 seconds.  then it moves 5 steps and falls into the  pit. i.e. it falls into the pit after 29 s.

$${it}\:{moves}\:\mathrm{2}\:{meters}\:{in}\:\mathrm{8}\:{seconds}.\:{for} \\ $$$${the}\:{first}\:\mathrm{6}\:{meters}\:{it}\:{needs}\:\mathrm{24}\:{seconds}. \\ $$$${then}\:{it}\:{moves}\:\mathrm{5}\:{steps}\:{and}\:{falls}\:{into}\:{the} \\ $$$${pit}.\:{i}.{e}.\:{it}\:{falls}\:{into}\:{the}\:{pit}\:{after}\:\mathrm{29}\:{s}. \\ $$

Commented by Tawa1 last updated on 18/Aug/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Necxx last updated on 18/Aug/18

distance covered in 1 step=1m  timetaken=1s  timetaken to move first 5m forward  =5s  timetaken to move 3m backwards  =3s  hence,the net distance travelled=5−3  =2m  net timetaken to cover this 2m=8s    if the drunkard covers 2m in 8s,then  distance covered in 8m=((8×8)/2)=32s  since there is still a remainder of  5m which can be covered in 5s    then the total time taken for the  drunkard to enter a pit 13m away  =32+5=37s

$${distance}\:{covered}\:{in}\:\mathrm{1}\:{step}=\mathrm{1}{m} \\ $$$${timetaken}=\mathrm{1}{s} \\ $$$${timetaken}\:{to}\:{move}\:{first}\:\mathrm{5}{m}\:{forward} \\ $$$$=\mathrm{5}{s} \\ $$$${timetaken}\:{to}\:{move}\:\mathrm{3}{m}\:{backwards} \\ $$$$=\mathrm{3}{s} \\ $$$${hence},{the}\:{net}\:{distance}\:{travelled}=\mathrm{5}−\mathrm{3} \\ $$$$=\mathrm{2}{m} \\ $$$${net}\:{timetaken}\:{to}\:{cover}\:{this}\:\mathrm{2}{m}=\mathrm{8}{s} \\ $$$$ \\ $$$${if}\:{the}\:{drunkard}\:{covers}\:\mathrm{2}{m}\:{in}\:\mathrm{8}{s},{then} \\ $$$${distance}\:{covered}\:{in}\:\mathrm{8}{m}=\frac{\mathrm{8}×\mathrm{8}}{\mathrm{2}}=\mathrm{32}{s} \\ $$$${since}\:{there}\:{is}\:{still}\:{a}\:{remainder}\:{of} \\ $$$$\mathrm{5}{m}\:{which}\:{can}\:{be}\:{covered}\:{in}\:\mathrm{5}{s} \\ $$$$ \\ $$$${then}\:{the}\:{total}\:{time}\:{taken}\:{for}\:{the} \\ $$$${drunkard}\:{to}\:{enter}\:{a}\:{pit}\:\mathrm{13}{m}\:{away} \\ $$$$=\mathrm{32}+\mathrm{5}=\mathrm{37}{s} \\ $$

Commented by MrW3 last updated on 19/Aug/18

the pit is 11 m away, not 13 m!

$${the}\:{pit}\:{is}\:\mathrm{11}\:{m}\:{away},\:{not}\:\mathrm{13}\:{m}! \\ $$

Commented by Necxx last updated on 20/Aug/18

oh....I didnt take note of that.Thanks

$${oh}....{I}\:{didnt}\:{take}\:{note}\:{of}\:{that}.{Thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com