A can finish one-third of a work in 5 days, B can finish $\frac{2}{5}th$ of the same work in 10 days and C can finish 75% of the same work in 15 days. They work together for 6 days. The remaining work will be finished by B alone in:
Answer & explanation
Correct answer: option 4
A = \(\frac{1}{3}\) = 5 days = 15 days,
B = \(\frac{2}{5}\) = 10 days = 25 days,
C = \(\frac{3}{4}\) = 15 days = 20 days,
⇒ A + B + C worked for 6 days = (20 + 12 + 15) x 6 = 47 x 6 = 282 units ..(Efficiency × Days = Total work)
⇒ Remaining work = 300 - 282 = 18 units.
⇒ Time required by B to complete 18 units = \(\frac{18}{12}\) = \( { 1}_{12 }^{6 } \) = \( { 1}_{2 }^{1 } \) days. ..(\(\frac{Work}{Efficiency}\) = Time)