A and B can do a work in 34/3
days working alternatively with B working first. A can
complete work alone in 48 days .In how many days B does 4 times of the work?
Answers
Answer:
B does 4 times of the work in just 27 days.
Step-by-step explanation:
Given
A and B are doing a work in 34 / 3 days working alternatively with B working first.
Here,
Total number of days = 34 / 3 > 11.
For convenience, making a table which starts from B, as it is given that the work is being started by B.
Work By Day
B 1
A 2
B 3
A 4
B 5
A 6
B 7
A 8
B 9
A 10
B 11
A 34 / 3
Then,
= > Total days in which B worked = 6
= > Total days in which A worked = Total days - days in which B worked
= > Total days in which A worked = 34 / 3 - 6
= > Total days in which A worked = 16 / 3
Given,
A can complete the work alone in 48 days.
It means : A's one day work is 1 / 48 of total work.
= > A's one day work = 1 / 48 of total work
= > A's 16 / 3 day work = 16 / 3 x 1 / 48 of total work
= > A's 16 / 3 day work = 1 / 9 of total work
Here, we can say :
= > Total work = Work done by A and B in 34 / 3 days
= > Total work = work by A in 16 / 3 days + work by B in 6 days
= > Total work = 1 / 9 of total work + B's work in 6 days
= > total work - 1 / 9 of total work = B's work in 6 days
= > 8 / 9 of total work = B's work in 6 days
= > 8 / ( 9 x 6 ) = 4 / 27 = B's work in one day.
It means that B can complete the total work in just 27 / 4 days.
So,
= > B can do the 4 times of total work in : 4 x ( 27 / 4 ) days = = 27 days.
Hence B does 4 times of the work in 27 days.
Looking for a short answer ?
Solution :
A and B worked for 16 / 3 and 6 days.
A's one day work = 1 / 48
A's 16 / 3 days work = 1 / 9
= > 1 = 1 / 9 + 6 / days taken by B to complete the work alone{ let it be a }
= > 8 / 9 = 6 / a
= > a = 27 / 4
Thus,
Days to complete the 4 times of that work = 4 x ( 27 / 4 ) days = 27 days.
Answer:
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Step-by-step explanation:
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