Two motor mechanics, Taju and Siraj, working together can overhaul a scooter in 6 hours. Raju alone can do the job in 15 hours. In how manu hours can siraj alone do it?

Answer

To find out how many hours Siraj alone can do the job of overhauling a scooter, we can follow these steps:

### Step 1: Determine Raju's Work Efficiency
Raju can overhaul a scooter in 15 hours. Therefore, his work efficiency is:
\[ \text{Raju's efficiency} = \frac{1}{15} \text{ (work done in 1 hour)} \]

### Step 2: Determine Combined Work Efficiency of Raju and Siraj
When Raju and Siraj work together, they can complete the job in 6 hours. Thus, their combined work efficiency is:
\[ \text{Raju + Siraj's efficiency} = \frac{1}{6} \text{ (work done in 1 hour)} \]

### Step 3: Calculate Siraj's Work Efficiency
To find Siraj's efficiency, we subtract Raju's efficiency from their combined efficiency:
\[ \text{Siraj's efficiency} = \text{(Raju + Siraj's efficiency)} - \text{(Raju's efficiency)} \]
\[ \text{Siraj's efficiency} = \frac{1}{6} - \frac{1}{15} \]

### Step 4: Find the LCM of the Denominators
To subtract the fractions, we need to find the least common multiple (LCM) of 6 and 15. The LCM of 6 and 15 is 30.

### Step 5: Convert Fractions to a Common Denominator
Now we convert the fractions:
\[ \frac{1}{6} = \frac{5}{30} \quad \text{(since } 6 \times 5 = 30\text{)} \]
\[ \frac{1}{15} = \frac{2}{30} \quad \text{(since } 15 \times 2 = 30\text{)} \]

### Step 6: Perform the Subtraction
Now we can perform the subtraction:
\[ \text{Siraj's efficiency} = \frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10} \]

### Step 7: Calculate the Time Taken by Siraj Alone
Since Siraj's efficiency is \(\frac{1}{10}\), it means he can complete the job in 10 hours. Therefore:
\[ \text{Time taken by Siraj alone} = 10 \text{ hours} \]

### Final Answer
Siraj alone can overhaul the scooter in **10 hours**.

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