Question:
A takes 10 days less than the time taken by B to finish a piece of work. If A and B together finish it in 12 days. Find the work done by A and B.
Answer:
Let B can alone finish the work = x days
So, A can alone finish the work = (x - 10) days
Now, one day work of A = 1/(x - 10)
and one day work of B = 1/x
Now, given, (A + B) can finish the ork = 12 days
So, one day work of (A + B) = 1/12
=> one day work of A + one day work of B = 1/12
=> 1/(x - 10) + 1/x = 1/12
=> (x + x - 10)/{x*(x - 10)} = 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, 4
if x = 4, then A alone can finish the rok = 4 - 10 = -6, which is not possible.
So, x = 30
Hence, B can alone finish the work = 30 days
So, A can alone finish the work = 30 - 10 = 20 days
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