Time and Work - Mathematics (SSC CGL CHSL)

In many competitive exam questions, we encounter scenarios where two individuals do not work together every day, but instead work on alternate days. This changes how we approach the solution compared to when they work together daily.

Let us understand this concept:
Suppose person A works on day 1, B on day 2, then A again on day 3, and so on — this alternation continues until the work is completed. In such a case, we calculate the work done by both A and B in one full cycle (2 days), and then see how many such full cycles can fit into the total work.

Example:

Question: A can complete a work in 12 days, and B can complete the same work in 18 days. If they work on alternate days starting with A, in how many days will the work be completed?

Step 1: Assume Total Work
Let us take the LCM of 12 and 18 to assume total work = 36 units.

       A         B         
     (12)      (18)     
      |         |      
    3 |         | 2 (B's Efficiency)   
       \       /        
        \     /           
         \   /             
        LCM = 36 (Total Work = 36 units)

Step 2: Calculate Efficiency
Efficiency of A = \( \frac{36}{12} = 3 \) units/day
Efficiency of B = \( \frac{36}{18} = 2 \) units/day (*Also can be seen in the work efficiency figure)

Working Pattern

              A B A B A B A B ....
  Day         1 2 3 4 5 6 7 8 ....
  Work Done   3 2 3 2 3 2 3 2 ....
 

Step 3: Work in One Cycle (2 days)
In 2 days (A+B), work done = \( 3 + 2 = 5 \) units

What is happening here!
A will work for 1 day = 3 unit work will be done, (ie work left : 36 - 3 = 33)
B will work on day 2 = 2 unit work will be done, (ie work left : 33 - 2 = 31)
Which means 5 unit work done in 2 days - which we are refering as 1 cycle here...

Step 4: Divide Total Work
Total work = 36 units
Work per cycle = 5 units

Number of complete cycles = \( \left\lfloor \frac{36}{5} \right\rfloor = 7 \) cycles
Work done in 7 cycles = \( 7 \times 5 = 35 \) units
Remaining work = \( 36 - 35 = 1 \) unit

Step 5: Who works next?
Since A started, and 7 cycles mean 14 days (7 of A and 7 of B), the 15^th day will be A's turn again.
A's efficiency = 3 units/day. So, to complete 1 unit more: \[ \text{Time} = \frac{1}{3} \text{ day} \]

Final Step: Total Time
Total time = 14 days (7 full cycles) + \( \frac{1}{3} \) day
Answer: \( \boxed{14\frac{1}{3} \text{ days}} \)

Summary:

• Use LCM to assume total work in units.
• Alternate days mean summing work of A and B over 2 days = 1 cycle.
• Use division to count full cycles, and add any remaining time separately.
• Carefully check whose turn it is for the leftover part.