Surya works 3 times as fast as Ramya and is able to complete a piece of work in 40 days less than the number of days taken by Ramya. Find the time in which they can complete the work together.
15 days
Solving Work and Time Problems: Surya and Ramya
This question involves the concept of work and time, specifically dealing with the relationship between the speed (or efficiency) of workers and the time they take to complete a task. We are given information about the relative speeds of Surya and Ramya and the difference in the time they take individually. We need to find the time they take to complete the work when working together.
Understanding the Relationship between Speed and Time
Speed or efficiency is inversely proportional to the time taken to complete a fixed amount of work. This means if someone works faster, they take less time. If someone works N times as fast as another person, they will take \( \frac{1}{N} \) times the time taken by the other person.
Step-by-Step Solution to Find Individual Times
Let's denote:
- Time taken by Ramya to complete the work alone as \( R \) days.
- Time taken by Surya to complete the work alone as \( S \) days.
According to the question:
- Surya works 3 times as fast as Ramya. This means Surya's speed is 3 times Ramya's speed.
- The time taken is inversely proportional to speed. So, Surya will take \( \frac{1}{3} \) of the time Ramya takes.
- Therefore, \( S = \frac{R}{3} \).
Also, we are given that Surya is able to complete the work in 40 days less than Ramya.
- This means the difference between Ramya's time and Surya's time is 40 days.
- So, \( R - S = 40 \).
Now we have a system of two equations:
- \( S = \frac{R}{3} \)
- \( R - S = 40 \)
We can substitute the first equation into the second equation:
\( R - \frac{R}{3} = 40 \)
To solve for \( R \), find a common denominator (which is 3):
\( \frac{3R - R}{3} = 40 \)
\( \frac{2R}{3} = 40 \)
Multiply both sides by 3:
\( 2R = 40 \times 3 \)
\( 2R = 120 \)
Divide by 2:
\( R = \frac{120}{2} \)
\( R = 60 \)
So, Ramya takes 60 days to complete the work alone.
Now, find Surya's time \( S \):
\( S = \frac{R}{3} = \frac{60}{3} = 20 \)
So, Surya takes 20 days to complete the work alone. Let's check the difference: \( R - S = 60 - 20 = 40 \) days, which matches the information given in the question.
Calculating Work Rate
The work rate is the amount of work done per unit of time (in this case, per day). If a person takes \( T \) days to complete a work, their daily work rate is \( \frac{1}{T} \) of the total work.
- Ramya's daily work rate = \( \frac{1}{R} = \frac{1}{60} \) of the work per day.
- Surya's daily work rate = \( \frac{1}{S} = \frac{1}{20} \) of the work per day.
Calculating Combined Work Rate and Time Together
When Surya and Ramya work together, their work rates add up. Combined daily work rate = Ramya's daily rate + Surya's daily rate
Combined daily work rate = \( \frac{1}{60} + \frac{1}{20} \)
To add these fractions, find a common denominator, which is 60.
\( \frac{1}{60} + \frac{1 \times 3}{20 \times 3} = \frac{1}{60} + \frac{3}{60} = \frac{1 + 3}{60} = \frac{4}{60} \)
Simplify the fraction:
\( \frac{4}{60} = \frac{1}{15} \)
Their combined daily work rate is \( \frac{1}{15} \) of the work per day.
The time taken to complete the work together is the reciprocal of their combined daily work rate.
Time taken together = \( \frac{1}{\text{Combined daily work rate}} \)
Time taken together = \( \frac{1}{\frac{1}{15}} = 15 \)
So, Surya and Ramya can complete the work together in 15 days.
Summary of Results
| Worker | Speed (relative to Ramya) | Time Taken (days) | Daily Work Rate |
|---|---|---|---|
| Ramya | 1x | 60 | \( \frac{1}{60} \) |
| Surya | 3x | 20 | \( \frac{1}{20} \) |
| Together | - | 15 | \( \frac{1}{15} \) |
The time they take to complete the work together is 15 days.
Revision Table: Key Concepts in Work and Time
| Concept | Description | Formula |
|---|---|---|
| Work Rate | Amount of work done per unit time. | Work Rate = \( \frac{\text{Total Work}}{\text{Time Taken}} \) |
| Time and Work Rate | Inversely proportional for constant work. Faster rate means less time. | \( \text{Time} \propto \frac{1}{\text{Work Rate}} \) |
| Individual Work Rate | If a person finishes work in T days, their daily rate is \( \frac{1}{T} \). | Daily Rate = \( \frac{1}{\text{Days Taken}} \) |
| Combined Work Rate | Sum of individual work rates when working together. | Combined Rate = Rate1 + Rate2 + ... |
| Time Taken Together | Reciprocal of the combined work rate. | Time Together = \( \frac{1}{\text{Combined Rate}} \) |
Additional Information: Solving Work and Time Problems
Work and time problems are common in aptitude tests. They often involve calculating the time taken by individuals or groups to complete a task, their efficiency, or the amount of work done in a specific time. Here are some points to remember:
- Assume the total work is 1 unit, or any convenient number like the LCM of the times given. Assuming 1 unit is simple when using the daily rate method.
- Efficiency is directly proportional to the amount of work done in a given time, and inversely proportional to the time taken to complete the work.
- If A does a work in 'a' days and B does the same work in 'b' days, their combined daily work rate is \( \frac{1}{a} + \frac{1}{b} \). The time taken together is \( \frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{ab}{a+b} \) days. This formula is useful for two people working together.
- Problems can involve pipes filling or emptying tanks, which are similar to work and time problems where the tank capacity is the 'work'.
Always read the question carefully to identify whether workers are working together, separately, or if their efficiency changes over time.
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