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A takes 5 days more than B to do a certain job and 9 days more than C; A and B together can do the job in the same time as C. Then A alone can do the job in
Let the day taken by C alone be n so in one day work done by C=1/n Time taken by A to do the same work = n+9 days so in one day work done by A=1/n+9 Also A and B can together do the work in same time as C so work done in one day by A and B together = work done by C in one day=1/n let work done by B in one day be b so 1/n=b+(1/n+9) so b=9/n(n+9) Hence B takes n(n+9)/9 days to finish the work alone. According to Question days taken by A alone = days taken by B alone +5 so n+9=n(n+9)/9+5 solving this n2=36 and n=6 days Hence A alone can do the job in n+9=6+9=15 days time Please approve if correct.
Let the day taken by C alone be n
so in one day work done by C=1/n
Time taken by A to do the same work = n+9 days
so in one day work done by A=1/n+9
Also A and B can together do the work in same time as C
so work done in one day by A and B together = work done by C in one day=1/n
let work done by B in one day be b
so 1/n=b+(1/n+9)
so b=9/n(n+9)
Hence B takes n(n+9)/9 days to finish the work alone.
According to Question days taken by A alone = days taken by B alone +5
so n+9=n(n+9)/9+5
solving this n2=36 and n=6 days
Hence A alone can do the job in n+9=6+9=15 days time
Please approve if correct.
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