A is more efficient than B and they together can complete a work in 24 days. Had A done 50% of the work and then B, the remaining work, then the work would have been done in 50 days. B alone will complete 40% of the same work in:
Answer & explanation
Correct answer: option 2
Let work done per unit day by A = a unit,
And, work done per unit by B = b unit,
Now, total work = (a + b) x 24 units,
According to the question,
[12(a + b)/a] = [ 12(a+b)/b] = 50,
As, we have 2 variables, and 1 equation, we can assume value for one variable,
Lets, take a = 1,
[12(1 +b)/1] = [12(1 + b)/b] = 50,
⇒ 12 + 12b = 12/b + 12 = 50,
⇒ 12b + 12/b = 26
⇒ 6b + (6/b) = 13,
⇒ 6b2+ 6 = 13b
⇒ 6b2 - 13b + 6 = 0.
⇒ 3b(2b - 3) - 2(2b - 3) = 0
⇒ (3b - 2)(2b - 3) = 0
⇒ b = \(\frac{2}{3}\) or \(\frac{3}{2}\)
As, a > b, b = \(\frac{2}{3}\)
Now, total work = (1 + \(\frac{2}{3}\)) x 24 = 40 units,
40% of total work = 40 x 40% = 16 units,
Hence, Time taken by B = \(\frac{16}{2/3}\) = 24 days.