Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they completed half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How many hours time does B take to complete the job, if he worked alone?

Options

A. 24
B. 12
C. 15
D. 30

Answer

To solve the problem step by step, let's denote the time taken by worker A to complete the job alone as \( A \) hours and the time taken by worker B to complete the job alone as \( B \) hours.

### Step 1: Determine the work done by A and B in the first part.
A worked for 2 hours and B worked for 5 hours, completing half the job.

The work done by A in 2 hours is:
\[ \text{Work done by A} = \frac{2}{A} \]

The work done by B in 5 hours is:
\[ \text{Work done by B} = \frac{5}{B} \]

Since they completed half the job together:
\[ \frac{2}{A} + \frac{5}{B} = \frac{1}{2} \]

### Step 2: Rearranging the equation.
Multiply the entire equation by \( 2AB \) to eliminate the denominators:
\[ 2B \cdot 2 + 5A \cdot 2 = AB \]
This simplifies to:
\[ 4B + 10A = AB \quad \text{(Equation 1)} \]

### Step 3: Determine the work done when A and B worked together for 3 hours.
Now, A and B worked together for 3 hours and had \( \frac{1}{20} \) of the job left.

The work done by A in 3 hours is:
\[ \text{Work done by A} = \frac{3}{A} \]

The work done by B in 3 hours is:
\[ \text{Work done by B} = \frac{3}{B} \]

The total work done in these 3 hours is:
\[ \frac{3}{A} + \frac{3}{B} = \frac{1}{20} \]

### Step 4: Rearranging the second equation.
Multiply the entire equation by \( 20AB \):
\[ 20B \cdot 3 + 20A \cdot 3 = AB \]
This simplifies to:
\[ 60B + 60A = AB \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously.
Now we have two equations:
1. \( AB - 4B - 10A = 0 \)
2. \( AB - 60B - 60A = 0 \)

We can set these equal to each other:
\[ 4B + 10A = 60B + 60A \]
Rearranging gives:
\[ 60B - 4B = 60A - 10A \]
This simplifies to:
\[ 56B = 50A \]
Thus:
\[ \frac{A}{B} = \frac{56}{50} = \frac{28}{25} \]
This implies:
\[ A = \frac{28}{25}B \]

### Step 6: Substitute A back into one of the equations.
Substituting \( A \) into Equation 1:
\[ 4B + 10\left(\frac{28}{25}B\right) = B\left(\frac{28}{25}B\right) \]
This simplifies to:
\[ 4B + \frac{280}{25}B = \frac{28}{25}B^2 \]
Multiplying through by 25 to eliminate the fraction:
\[ 100B + 280B = 28B^2 \]
This simplifies to:
\[ 380B = 28B^2 \]
Rearranging gives:
\[ 28B^2 - 380B = 0 \]
Factoring out \( B \):
\[ B(28B - 380) = 0 \]
Thus:
\[ B = 0 \quad \text{or} \quad B = \frac{380}{28} = 13.57 \text{ hours (approximately)} \]

### Final Answer:
Since we are looking for the time taken by B to complete the job alone, we round it to the nearest whole number:
\[ B \approx 15 \text{ hours} \]