Ravi can build a wall in the same time in which Mahesh and Suresh together do it. If Ravi and Mahesh together could do it in 10 days and Suresh alone in 15 days. In what time Mahesh alone do it?

Options

A. 50 days
B. 60 days
C. 59 days
D. None of these

Answer

To solve the problem step by step, we will denote the work done by Ravi, Mahesh, and Suresh as R, M, and S respectively.

### Step 1: Determine the work done by Ravi and Mahesh together
Ravi and Mahesh together can complete the work in 10 days. Therefore, the work done by them in one day is:
\[ \text{Work done by Ravi and Mahesh in one day} = \frac{1}{10} \]

### Step 2: Determine the work done by Suresh alone
Suresh can complete the work alone in 15 days. Therefore, the work done by Suresh in one day is:
\[ \text{Work done by Suresh in one day} = \frac{1}{15} \]

### Step 3: Establish the relationship between Ravi, Mahesh, and Suresh
According to the problem, Ravi can build the wall in the same time that Mahesh and Suresh can together do it. This means:
\[ R = M + S \]

### Step 4: Calculate the total work done by all three together
If we denote the work done by Ravi in one day as R, we can express the total work done by Ravi, Mahesh, and Suresh together in one day as:
\[ R + M + S = \frac{1}{10} + \frac{1}{15} \]

### Step 5: Find a common denominator and add the fractions
The least common multiple of 10 and 15 is 30. Thus, we can rewrite the fractions:
\[ \frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
Adding these gives:
\[ R + M + S = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \]

### Step 6: Substitute R in terms of M and S
Since \( R = M + S \), we can substitute this into the equation:
\[ (M + S) + M + S = \frac{1}{6} \]
This simplifies to:
\[ 2M + S = \frac{1}{6} \]

### Step 7: Substitute S in terms of M
We know from earlier that S = \(\frac{1}{15}\). Substituting this into the equation gives:
\[ 2M + \frac{1}{15} = \frac{1}{6} \]

### Step 8: Solve for M
To solve for M, we first find a common denominator for \(\frac{1}{6}\) and \(\frac{1}{15}\). The least common multiple of 6 and 15 is 30:
\[ \frac{1}{6} = \frac{5}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
Now substituting back into the equation:
\[ 2M + \frac{2}{30} = \frac{5}{30} \]
Subtract \(\frac{2}{30}\) from both sides:
\[ 2M = \frac{5}{30} - \frac{2}{30} = \frac{3}{30} = \frac{1}{10} \]
Dividing both sides by 2 gives:
\[ M = \frac{1}{20} \]

### Step 9: Find the time taken by Mahesh to complete the work alone
If Mahesh can do \(\frac{1}{20}\) of the work in one day, then the time taken by Mahesh to complete the whole work alone is:
\[ \text{Time taken by Mahesh} = \frac{1}{\frac{1}{20}} = 20 \text{ days} \]

### Conclusion
Thus, Mahesh alone can complete the work in **20 days**.

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