A and b completed a work together in 5 days. had a worked at twice the speed and b at half the speed, it would have taken them four days to complete the job. how much time would it take for a alone to do the work?
Answers
Answered by
8
A one day work=1/x
B one day work=1/y
(A+B)=1/x+1/y=1/5
(x+y)/xy=1/5
5(x+y)=xy -------(1)
if A worked at twice =2/x
B worked at half=1/2y
2/x+1/2y=1/4
(4y+x)/2xy=1/4
8y+2x=xy-----(2)
substitute (2) in (1)
8y+2x=5x+5y
3y=3x
y=x that means number days worked by A alone is equal to B alone
substitute in (1)
1/x+1/x=1/5
2/x=1/5
A=10 days
B=10days
B one day work=1/y
(A+B)=1/x+1/y=1/5
(x+y)/xy=1/5
5(x+y)=xy -------(1)
if A worked at twice =2/x
B worked at half=1/2y
2/x+1/2y=1/4
(4y+x)/2xy=1/4
8y+2x=xy-----(2)
substitute (2) in (1)
8y+2x=5x+5y
3y=3x
y=x that means number days worked by A alone is equal to B alone
substitute in (1)
1/x+1/x=1/5
2/x=1/5
A=10 days
B=10days
Answered by
12
Thank you for asking this question. Here is your answer:
A and B's 1 day work would be equal to a+b=1/5
And with twice d speed of A and half of B completes work in 4 days,
So, 2a+b/2 would be equal to 1/4.
When we will solve both the equations we get a = 1/10
So A will complete the whole work in 10 days
If there is any confusion please leave a comment below.
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