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A and B do a work with 5 days, A and C in 6 days, B and C in 20 days.If A do alone how many days it will take?
Answers:
Answer #1I think A will take 9-days.
Answer #2A+B+C=31/2
B+C=20
31/2-20=4.5days A will take if he will work alone
B+C=20
31/2-20=4.5days A will take if he will work alone
Answer #31
Answer #4it should be 120/19 days
Answer #5A will take 11 days.
Answer #6120/19
Answer #7it will be solved by unitary method.Lets say the work is 60 units which is multiple of 5,4 and 3. Now A and B will take 12 units per day,similarly B and C will have 3 rate per day,also A and C will have 10 units per day.now on adding units of (A,B ),(B , C),(A,C).We got 2(A+B+C)=2(25/2) units .Now, substract it from that of rate of B and C,hence, 25/2 -3 =19/2 which is 9 n 1/2 days,
Answer #85a+5b=x
6a+6c=x
20b+20c=x
Now express X in terms of a and you get x=(240/28)a=(120/19)a... Ans = 120/19 days
6a+6c=x
20b+20c=x
Now express X in terms of a and you get x=(240/28)a=(120/19)a... Ans = 120/19 days
Answer #9Assuming A can do the job in x days, meaning A can do 1/x of the job in 1 day. Same goes for B and C (1/y and 1/z per day respectively). So, in 5 days B can do 5/x of the job and B can do 5/y of the job. 5/x+5/y is 100% of the job or 1 [5/x+5/y = 1]. Doing the same thing for all 3 givens, we have 3 equations with 3 unknowns [5/x+5/y=1] ,[6/x+6/z=1] ,[20/y+20/z=1]. Solving for the unknowns, x = 120/19 or approximately 6.3 days