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.
Answers
Answer:
Let's B alone take x days to finish the work.
B’s 1 days work = 1/x
A alone can finish the work in = x - 10
A’s 1 days work= 1/(x-10)
Given: A and B together can finish the same work in 12 days.
(A and B)'s 1 days work = 1/12
A’s 1 days work + B’s 1 days work = 1/x + 1/(x-10)
ATQ
1/x + 1/(x-10) = 1/12
(x-10) + x / x(x-10) = 1/12
x-10 + x / x² -10x = 1/12
2x -10 / x² -10x = 1/12
12(2x -10) = x² -10x
24x -120 = x² -10x
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 or (x-4)= 0
x = 30 or x= 4
The value of x cannot be less than 10, so x= 30
Hence, the time taken by B to finish the work in 30 days.
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