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Question
A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.
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Solution
Let B alone takes x days to finish the work. Then, B’s one day’s work = 1/x.
Similarly, A alone can finish it in (x - 10) days to finish the work. Then, A’s one day’s work `=1/(x-10)`.
It is given that
A’s one day’s work + B’s one day’s work = (A + B)’s one day’s work
`1/x+1/(x-10)=1/12`
`(x-10+x)/(x(x-10))=1/12`
`(2x-10)/(x^2-10x)=1/12`
x2 - 10x = 12(2x - 10)
x2 - 10x = 24x - 120
x2 - 10x - 24x + 120 = 0
x2 - 34x + 120 = 0
x2 - 30x - 4x + 120 = 0
x(x - 30) - 4(x - 30) = 0
(x - 30)(x - 4) = 0
x - 30 = 0
x = 30
Or
x - 4 = 0
x = 4
But x = 4 is not correct.
therefore, x = 30 is correct
Hence, the time taken by B to finish the work in x = 30 days
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