A can do a piece of work in 12 days. He worked for 6 days and left, and then B finished it in 10 days. If both work together, then in how many days will they finish the same work?
Answer & explanation
Correct answer: option 3
A = 12 days,
Lets' take A's efficiency as 1.
Therefore, total work = 1 x 12 = 12 units, ..(Efficiency × Days = Total work)
⇒ A worked for 6 days = 1 x 6 = 6 units,
⇒ Remaining work = 12 - 6 = 6 units,
⇒ B completed the remaining work in 10 days = \(\frac{6}{B}\) = 10 days,
⇒ B = \(\frac{3}{5}\), (Efficiency)
⇒ Time required by A + B to complete work = \(\frac{12}{1+3/5}\) = \(\frac{12}{8/5}\) = (\frac{15}{2}\) = 7.5 days, ..(\(\frac{Work}{Efficiency}\) = Time)