Ajay and Babu can finish a task in 6 days working together. If Ajay works at one-third of his normal efficiency and Babu works at twice his normal efficiency, the time required to finish the same task doesn’t change. How many days will Ajay alone take to finish a different task, if Babu alone can finish it in 18 days?

To solve this problem, let's break it down step by step.

### Step 1: Define the efficiencies of Ajay and Babu

Let the normal efficiency of Ajay be \( A \) and that of Babu be \( B \). This means:
- Ajay completes \( A \) fraction of the task in one day.
- Babu completes \( B \) fraction of the task in one day.

### Step 2: Calculate their combined efficiency when they work together

Ajay and Babu together can finish the task in 6 days, so their combined efficiency is:
\[
A + B = \frac{1}{6}
\]

### Step 3: Modify their efficiencies as per the new conditions

Now, Ajay works at one-third of his normal efficiency, and Babu works at twice his normal efficiency. Their modified efficiencies will be:
- Ajay's modified efficiency = \( \frac{A}{3} \)
- Babu's modified efficiency = \( 2B \)

Even with these changes, they still take 6 days to finish the task. So, their combined efficiency in this case is:
\[
\frac{A}{3} + 2B = \frac{1}{6}
\]

### Step 4: Solve the equations

We now have two equations:
1. \( A + B = \frac{1}{6} \)
2. \( \frac{A}{3} + 2B = \frac{1}{6} \)

Let's multiply the second equation by 3 to eliminate the fraction:
\[
A + 6B = \frac{1}{2}
\]

Now, subtract the first equation from this result:
\[
(A + 6B) - (A + B) = \frac{1}{2} - \frac{1}{6}
\]
\[
5B = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}
\]
\[
B = \frac{1}{15}
\]

Now, substitute \( B = \frac{1}{15} \) back into the first equation:
\[
A + \frac{1}{15} = \frac{1}{6}
\]
\[
A = \frac{1}{6} - \frac{1}{15}
\]

To solve for \( A \), find the common denominator (which is 30):
\[
A = \frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10}
\]

### Step 5: Determine the time Ajay takes to finish the different task alone

Ajay's efficiency is \( A = \frac{1}{10} \), which means Ajay can finish \( \frac{1}{10} \) of the task in one day.

So, Ajay alone will take:
\[
\text{Time taken by Ajay} = \frac{1}{\frac{1}{10}} = 10 \text{ days}
\]

Thus, Ajay alone will take **10 days** to finish a different task if Babu alone can finish it in 18 days.

Would you like any further details or have any questions?

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Here are some related questions you might find interesting:
1. What happens if Ajay works at half his efficiency and Babu at three times his efficiency?
2. How would the time change if Babu's efficiency was reduced by half instead of doubled?
3. If Ajay and Babu switch their normal efficiencies, how many days will they now take together to finish the task?
4. How would the result change if Ajay could finish the task alone in 12 days?
5. What would be the effect on the result if both Ajay and Babu’s efficiencies were doubled?

**Tip:** Always check the units when dealing with work and time problems to ensure accuracy.

Explore a math problem involving work efficiency where Ajay and Babu complete tasks at altered efficiencies. Learn how to solve the problem step-by-step, determining Ajay's time to finish a task alone. Suitable for high school students studying work and time problems.

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