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Three friends Adam, Eve and Kane who are working together working at different speed where Adam being the slowest and Kane being the Fastest. When Adam and Eve work together they take x days to finish a task. When Eve and Kane work together they take y days to finish a task. One of them, if he worked alone would take thrice as much time as it would take when all three work together. How much time would it take if all three worked together?

Three friends Adam, Eve and Kane who are working together working at different speed where Adam being the slowest and Kane being the Fastest. When Adam and Eve work together they take x days to finish a task. When Eve and Kane work together they take y days to finish a task. One of them, if he worked alone would take thrice as much time as it would take when all three work together. How much time would it take if all three worked together? Correct Answer 4xy/3(y + x)

Adam < Eve < Kane in terms of efficiency.

Adam and Eve together take x days.

Kane and Eve together take y days.

One of them, if he worked alone would take thrice as much time as it would take when all

three work together. This is a crucial statement. Now, if there are three people who are all

equally efficient, for each of them it would take thrice as much time as for all three

together.

Now, this tells us that the person who takes thrice as much time cannot be the quickest one.

If the quickest one is only one-third as efficient as the entire team, the other two cannot

add up to two-thirds. By a similar logic, the slowest one cannot be the person who is one-

third as efficient.

In other words, the person one-third as efficient = Eve

Let Adam, Eve, and Kane together take m days. Eve alone would take 3m days

Adam and Eve together take x days. Or Adam + Eve in 1 day do 1/x of the task      ----(1)

Eve and Kane together take y days. Or, Adam + Eve in 1 day do 1/y of the task      ----(2)

Eve takes 3m days to do the task. Or, Eve, in one day, does 1/3m of the task      ----(3)

Now, if we do (1) + (2) – (3) we get

Adam + Eve + Kane do 1/x + 1/y – 1/3 m in a day. This should be equal to 1/m as all three of them complete the task in m days.

1/x + 1/y – 1/3m = 1/m

⇒ 1/x + 1/y = 4/3m

⇒ (x + y)xy = 4/3m

⇒ 4xy/3(x + y) = m

⇒ m = 4xy/3(y + x)

∴ Time taken by all three if they work together = 4xy/3(y + x)

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