Step 1: Understanding the Concept:
Time and Work problems are based on "Efficiency," which is the amount of work done in a single unit of time (e.g., one day).
The fundamental relationship is: \(\text{Efficiency} = \frac{\text{Total Work}}{\text{Time Taken}}\).
If someone finishes work in \(D\) days, their daily efficiency is \(\frac{1}{D}\) of the total work.
When multiple people work together, their combined efficiency is simply the sum of their individual efficiencies. The total time taken for the group to finish is the reciprocal of this combined efficiency.
Step 2: Key Formula or Approach:
For two people A and B taking \(x\) and \(y\) days respectively, the total time \(T\) is:
\[ T = \frac{\text{Product of days}}{\text{Sum of days}} = \frac{x \times y}{x + y} \]
Alternatively, we can use the LCM (Least Common Multiple) method to define a concrete value for "Total Work" in terms of units.
Step 3: Detailed Explanation:
Let's solve this using the standard fractional efficiency method first.
A's 1-day work = \(\frac{1}{15}\)
B's 1-day work = \(\frac{1}{10}\)
Combined 1-day work (A + B) = \(\frac{1}{15} + \frac{1}{10}\)
To add these fractions, find the common denominator (which is \(30\)):
\[ \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} \]
Simplifying the combined daily work:
\[ \frac{5}{30} = \frac{1}{6} \]
Since they do \(\frac{1}{6}\) of the work in one day, it will take them exactly \(6\) days to complete the full task.
Alternative LCM Method:
1. Find the LCM of \(15\) and \(10\), which is \(30\). Let's assume the total work is \(30\) units.
2. Efficiency of A = \(30 \text{ units} / 15 \text{ days} = 2 \text{ units/day}\).
3. Efficiency of B = \(30 \text{ units} / 10 \text{ days} = 3 \text{ units/day}\).
4. Total Combined Efficiency = \(2 + 3 = 5 \text{ units/day}\).
5. Total Days = \(\text{Total Work} / \text{Combined Efficiency} = \frac{30}{5} = 6 \text{ days}\).
Both methods yield the same logical result.
Step 4: Final Answer:
They will complete the work together in 6 days.