Question

P and Q can complete a piece of work in 28 days and 35 days, respectively. P begins to do the work and they work alternatively one at a time for one day each. In how many days will the whole work be completed?

• A. 31
• B. 27
• C. 18
• D. 21

Answer 1

Answer:

A. 31

The whole work will be completed in 31 days.

• Find the total work and individual efficiencies:

  • Total work = LCM of 28 and 35 = 140 units.

  • Efficiency of P = $140 \div 28 = \mathbf{5}$ units/day.

  • Efficiency of Q = $140 \div 35 = \mathbf{4}$ units/day.

• Calculate the work done in cycles:

  • Since they work alternately starting with P, a 2-day cycle looks like:

    • Day 1 (P) = 5 units

    • Day 2 (Q) = 4 units

  • Work completed in 2 days = $5 + 4 = \mathbf{9}$ units.

• Find the number of full cycles:

  • Divide total work by cycle work: $140 \div 9 = 15$ with a remainder of 5.

  • Total work in 15 cycles = $15 \times 9 = \mathbf{135}$ units.

  • Total days for 15 cycles = $15 \times 2 = \mathbf{30}$ days.

• Calculate the remaining work:

  • Remaining work = $140 - 135 = \mathbf{5}$ units.

  • On Day 31, it is P's turn.

  • P can complete exactly 5 units in 1 day.

• Total time taken:

  • $30\text{ days} + 1\text{ day} = \mathbf{31\text{ days}}$.