A and B together can complete a certain work in 20 days whereas B and C together can complete it in 24 days. If A is twice as good a workman as C, then in what time will B alone do 40% of the same work?
12 days
Understanding the Time and Work Problem
This problem involves understanding the concept of time and work, specifically how individuals working together complete a task and how their efficiencies relate to each other.
We are given information about the time taken by pairs of workers (A and B, B and C) to complete a certain work. We are also given a relationship between the efficiencies of A and C. We need to find the time taken by B alone to complete 40% of the same work.
Setting up the Problem: Efficiency and Total Work
Let's assume the total work to be a certain number of units. A common method is to take the Least Common Multiple (LCM) of the days given. The days given are 20 days (for A and B) and 24 days (for B and C).
LCM(20, 24) = 120 units.
Let the efficiency of A, B, and C (i.e., the amount of work they do per day) be \(a\), \(b\), and \(c\) units/day, respectively.
Formulating Equations based on Given Information
1. A and B together complete the work in 20 days:
Total work = (Combined efficiency) \(\times\) Time taken
\(120 = (a + b) \times 20\)
Dividing both sides by 20, we get the combined efficiency of A and B:
\(a + b = \frac{120}{20} = 6\) units/day. (Equation 1)
2. B and C together complete the work in 24 days:
Total work = (Combined efficiency) \(\times\) Time taken
\(120 = (b + c) \times 24\)
Dividing both sides by 24, we get the combined efficiency of B and C:
\(b + c = \frac{120}{24} = 5\) units/day. (Equation 2)
3. A is twice as good a workman as C:
This means the efficiency of A is twice the efficiency of C.
\(a = 2c\) units/day. (Equation 3)
Solving the System of Equations
We have a system of three linear equations with three variables (\(a\), \(b\), \(c\)):
1. \(a + b = 6\)
2. \(b + c = 5\)
3. \(a = 2c\)
Substitute Equation 3 into Equation 1:
\(2c + b = 6\) (Equation 4)
Now we have two equations (Equation 2 and Equation 4) involving only \(b\) and \(c\):
Equation 4: \(b + 2c = 6\)
Equation 2: \(b + c = 5\)
Subtract Equation 2 from Equation 4:
\((b + 2c) - (b + c) = 6 - 5\)
\(b + 2c - b - c = 1\)
\(c = 1\) unit/day.
Now that we have the value of \(c\), we can find \(a\) using Equation 3:
\(a = 2c = 2 \times 1 = 2\) units/day.
Finally, we can find \(b\) using Equation 2 (or Equation 1):
Using Equation 2: \(b + c = 5\)
\(b + 1 = 5\)
\(b = 5 - 1 = 4\) units/day.
So, the efficiencies are: A does 2 units/day, B does 4 units/day, and C does 1 unit/day.
Calculating Time for B Alone to do 40% Work
The total work is 120 units.
We need to find the time B alone takes to complete 40% of the work.
Amount of work B needs to complete = 40% of 120 units
Work = \(0.40 \times 120 = \frac{40}{100} \times 120 = \frac{2}{5} \times 120 = 2 \times 24 = 48\) units.
The efficiency of B is 4 units/day.
Time taken by B alone to complete 48 units of work = \(\frac{\text{Amount of Work}}{\text{Efficiency of B}}\)
Time = \(\frac{48}{4} = 12\) days.
Conclusion
B alone will take 12 days to complete 40% of the same work.
| Workers | Combined Time (days) | Combined Efficiency (units/day) |
|---|---|---|
| A + B | 20 | \(\frac{120}{20} = 6\) |
| B + C | 24 | \(\frac{120}{24} = 5\) |
| Worker | Efficiency (units/day) |
|---|---|
| A | 2 |
| B | 4 |
| C | 1 |
40% of total work = \(0.40 \times 120 = 48\) units.
Time for B to do 40% work = \(\frac{48 \text{ units}}{4 \text{ units/day}} = 12 \text{ days}\).
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Time and Work Relationship | Work = Efficiency \(\times\) Time |
| Efficiency | The amount of work done per unit of time. Higher efficiency means less time taken to complete work. |
| Combined Efficiency | When multiple people work together, their efficiencies add up. |
| Assuming Total Work | Taking LCM of the given time periods simplifies calculations by making the total work an integer. |
Additional Information: Variations in Time and Work Problems
Time and work problems can have several variations, including:
- Problems involving different efficiencies of workers (like A is twice as good as C).
- Problems where workers join or leave the work at different times.
- Problems involving wages earned for the work done, which are usually proportional to the work done or efficiency.
- Problems involving pipes and cisterns, which are similar to time and work problems where pipes fill or empty a tank (work) over time.
Solving these problems often involves calculating individual or combined efficiencies and then using the relationship Work = Efficiency \(\times\) Time to find the unknown quantity.
Important Questions from Work Efficiency
14 men can complete a work in 15 days. If 21 men are employed, then in how many days will they complete the same work?
A can do a certain work in 15 days, while B can do the same work in 21 days. If they work together, then in how many days will the same work be completed?
To do a certain work, A and B work on alternate days with B beginning the work on the first day. A alone can complete the same work in 24 days. If the work gets completed in \(11 \frac{1}{3}\) days, then B alone can complete \(\rm \frac{7}{9}^{th}\) part of the original work in:
Two men and 7 women can complete a work in 28 days whereas 6 men and 16 women can do the same work in 11 days. In how many days can 7 men complete the same work?
Five men and 2 boys can do in 30 days as much work as 7 men and 10 boys can do in 15 days. How many boys should join 40 men to do the same work in 4 days?
