(a) 4
2
2
piece of work in 20 days and B can do it in 12 days. B worked at it for 9 days. In hou
A can do a
(a) 3
(c) 7
(d) 11
twice as fast as B. If B can complete a piece of work independently in 12 days, then
many days A alone can finish the remaining work?
Answers
Answer:
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Answer:
1) Given,
A can do a work in 20 days.
=> In a single day, A will complete 1/20th part of the total work.
B can do a work in 12 days.
=> In a single day, B will complete 1/12th part of the total work.
So, if they both work together,
=> 1/20 + 1/12
=> 3/60 + 5/60
=> 8/60
=> 2/15
In a single day, A and B together will complete 2/15th part of the total work.
Now, let's move forward to what other things are given. A already worked for 5 days. So, the amount of work completed is
=> 5*(1/20)
=> 1/4.
As 1/4th of the work is already completed so now A and B have to complete the rest 3/4th of the work together. So, number of days required by both of them to complete the work will be
=> 1/((Work Left)*(Work completed in a single day))
=> 1/((3/4)*(2/15))
=> 1/(1/10)
=> 10 days.
The question is how long did the work last. As, A has already worked 5 days so the work lasted 5 days + 10 days = 15 days in total.
So, the correct answer will be 15 days.
2)A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in?
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Accordin to problem, A works twice as fast as B.
Then, 2A=B, ………. Eqn.(1).
According to statement,
The work completed by B=12, days
Then, from eqn. (1).
2A=B=12
2A=12, and B=12, days
A=12/2
A=6 days, (A completed this work in 6 days.)
The work time ratio A/B=6/12=1/2,
1 day's work of A=1/6
1 day's work of B=1/12
1 day's work of A and B
=1/6+1/12
=3/12
=1/4,
So, A and B finish the work together
=1/4× work time ratio
=1/4×1/2
=1/8
So, A and B finish the work together in 8 days.
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