M is two times as efficient as N. M can complete a work in 66 days less than N. If both of them work together, then in how many days will the same work be completed?

Let the work to be done be one complete job (unit work). Suppose M alone can finish it in $$T_M$$ days and N alone can finish it in $$T_N$$ days.

Efficiency is inversely proportional to the time taken. “M is two times as efficient as N” therefore means $$\dfrac{1}{T_M} = 2 \times \dfrac{1}{T_N}$$, or equivalently

$$T_N = 2\,T_M \quad -(1)$$

We are also told that M needs 66 days less than N:

$$T_N - T_M = 66 \quad -(2)$$

Substituting $$(1)$$ into $$(2)$$ gives

$$2\,T_M - T_M = 66 \;\Longrightarrow\; T_M = 66\text{ days}$$

From $$(1)$$, $$T_N = 2 \times 66 = 132\text{ days}$$.

Their combined rate of working is the sum of individual rates:

Combined rate $$= \dfrac{1}{T_M} + \dfrac{1}{T_N} = \dfrac{1}{66} + \dfrac{1}{132} = \dfrac{2}{132} + \dfrac{1}{132} = \dfrac{3}{132} = \dfrac{1}{44}$$ (job per day).

Hence, together they will complete the whole job in $$\dfrac{1}{1/44} = 44$$ days.

Therefore, the required time is 44 days.

Option C which is: 44 days