6.	Ajay, Maya, and Mona can finish painting the fence in 2 hours. If Ajay does the job alone, he can finish it in 5 hours. If Maya does the job alone, she can finish it in 6 hours. How long will it take for Mona to finish the job alone?

Let's denote the following:
- \( A \) as Ajay's rate of work (fence per hour),
- \( M_1 \) as Maya's rate of work (fence per hour),
- \( M_2 \) as Mona's rate of work (fence per hour).

Given:
- Ajay, Maya, and Mona together can finish painting the fence in 2 hours. So their combined rate is \( \frac{1}{2} \) fence per hour.
- Ajay alone can finish the job in 5 hours, so his rate \( A = \frac{1}{5} \) fence per hour.
- Maya alone can finish the job in 6 hours, so her rate \( M_1 = \frac{1}{6} \) fence per hour.

We need to find Mona's rate \( M_2 \), which will give us the time it takes for her to finish the job alone.

### Step 1: Set up the equation
The combined rate of work for all three together is:
\[
A + M_1 + M_2 = \frac{1}{2}
\]

Substitute the known values:
\[
\frac{1}{5} + \frac{1}{6} + M_2 = \frac{1}{2}
\]

### Step 2: Solve for \( M_2 \)
First, find the sum of \( \frac{1}{5} \) and \( \frac{1}{6} \):

\[
\frac{1}{5} + \frac{1}{6} = \frac{6 + 5}{30} = \frac{11}{30}
\]

Now substitute back into the equation:

\[
\frac{11}{30} + M_2 = \frac{1}{2}
\]

Subtract \( \frac{11}{30} \) from both sides:

\[
M_2 = \frac{1}{2} - \frac{11}{30}
\]

Find a common denominator to subtract:

\[
\frac{1}{2} = \frac{15}{30}
\]

So,

\[
M_2 = \frac{15}{30} - \frac{11}{30} = \frac{4}{30} = \frac{2}{15}
\]

### Step 3: Find the time for Mona to finish the job alone
Since \( M_2 = \frac{2}{15} \) fence per hour, the time it takes for Mona to finish the job alone

Learn how to calculate Mona's rate of painting from a problem involving Ajay, Maya, and Mona painting a fence together in 2 hours. Ajay and Maya's individual rates are provided, allowing for Mona's rate to be determined through solving equations with fractions. Suitable for students in Grades 7-9.

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