X, Y and Z can complete a piece of work individually in 6 hours, 8 hours and 8 hours respectively. However, only one person at a time can work in each hour and nobody can work for two consecutive hours. All are engaged to finish the work. What is the minimum amount of time that they will take to finish the work.
6 hours 45 minutes
The question asks for the minimum time required for three individuals, X, Y, and Z, to complete a piece of work. We are given their individual times to complete the work, and specific constraints on how they can work together:
Our goal is to find the most efficient working pattern that satisfies these rules and determines the shortest possible time to finish the entire work.
First, let's determine how much work each person can complete in one hour. This is their work rate.
The rates are: X (\(\frac{1}{6}\)), Y (\(\frac{1}{8}\)), Z (\(\frac{1}{8}\)). X is the fastest worker.
To minimize the total time, we need to maximize the amount of work done in each hour, given the constraints. The constraint that no one can work for two consecutive hours means we must alternate workers every hour.
Since X is the fastest worker, we should try to use X as frequently as possible. Under the alternating rule, X can work every other hour. In the hours X is not working, Y and Z must take turns (since all are engaged and must alternate with X and each other). A pattern like X, Y, X, Z, X, Y, X, Z... seems efficient because it uses X every second hour and alternates between Y and Z in the hours X is resting. This pattern satisfies all conditions:
Let's calculate the work done hour by hour following the pattern X, Y, X, Z, X, Y, X, Z...:
| Hour | Worker | Work Done in Hour | Total Work Done | Remaining Work |
|---|---|---|---|---|
| 1 | X | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(1 - \frac{1}{6} = \frac{5}{6}\) |
| 2 | Y | \(\frac{1}{8}\) | \(\frac{1}{6} + \frac{1}{8} = \frac{4+3}{24} = \frac{7}{24}\) | \(1 - \frac{7}{24} = \frac{17}{24}\) |
| 3 | X | \(\frac{1}{6}\) | \(\frac{7}{24} + \frac{1}{6} = \frac{7+4}{24} = \frac{11}{24}\) | \(1 - \frac{11}{24} = \frac{13}{24}\) |
| 4 | Z | \(\frac{1}{8}\) | \(\frac{11}{24} + \frac{1}{8} = \frac{11+3}{24} = \frac{14}{24} = \frac{7}{12}\) | \(1 - \frac{7}{12} = \frac{5}{12}\) |
| 5 | X | \(\frac{1}{6}\) | \(\frac{7}{12} + \frac{1}{6} = \frac{7+2}{12} = \frac{9}{12} = \frac{3}{4}\) | \(1 - \frac{3}{4} = \frac{1}{4}\) |
| 6 | Y | \(\frac{1}{8}\) | \(\frac{3}{4} + \frac{1}{8} = \frac{6+1}{8} = \frac{7}{8}\) | \(1 - \frac{7}{8} = \frac{1}{8}\) |
After 6 hours, \(\frac{7}{8}\) of the work is completed, and \(\frac{1}{8}\) work remains. The pattern dictates that X works in the 7th hour.
Remaining work = \(\frac{1}{8}\).
X's work rate = \(\frac{1}{6}\) work per hour.
Time needed by X to complete the remaining \(\frac{1}{8}\) work = \(\frac{\text{Remaining Work}}{\text{X's Rate}} = \frac{\frac{1}{8}}{\frac{1}{6}}\) hours.
Calculating the time: \(\frac{1}{8} \times \frac{6}{1} = \frac{6}{8} = \frac{3}{4}\) hours.
Convert the fraction of an hour to minutes: \(\frac{3}{4} \times 60\) minutes = 45 minutes.
So, the work finishes after 6 full hours and an additional 45 minutes into the 7th hour.
The total minimum time taken to finish the work following the optimal pattern is 6 hours and 45 minutes.
| Concept | Description | Formula |
|---|---|---|
| Work Rate | The amount of work done by a person per unit of time. | Rate = \(\frac{1}{\text{Time Taken}}\) |
| Total Work | Usually considered as 1 unit. | Work = Rate \(\times\) Time |
| Time Taken | Total time to complete the work. | Time = \(\frac{\text{Total Work}}{\text{Combined Rate}}\) (for simultaneous work) or calculated based on individual contributions over time. |
| Alternating Work | Workers take turns, often calculated based on cycles of work done by a group. | Calculate work done in one cycle, then find number of cycles and remaining work. |
In work and time problems with constraints like alternating work or non-consecutive hours, the most efficient strategy typically involves having the most efficient worker (the one with the highest rate) work for the largest possible share of the total time, while still adhering to the rules. By having X work every other hour and alternating the less efficient workers Y and Z in the gaps, we ensure X's higher rate contributes significantly to completing the work quickly. The 'all engaged' rule confirms that no worker can be left out entirely from the process.
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