We are given that the total number of employees in the organization is \(100x\), and each employee's wage is \(100y\). Let's analyze the situation step by step:
The manufacturing department employs 20% of the total workforce. Therefore, the number of employees in the manufacturing department is \(20x\). Additionally, the combined salary of all manufacturing staff members is one-sixth of the total salary received by all employees.
The combined salary of the manufacturing department is:
\(\frac{1}{6} \times 100x \times 100y = \frac{10000xy}{6}\)
The average salary in the manufacturing department is then:
\(\frac{\frac{10000xy}{6}}{20x} = \frac{500y}{6x}\)
The non-manufacturing department employs the remaining 80% of the workforce, meaning there are \(80x\) employees in the non-manufacturing department. The combined salary of the non-manufacturing department is:
\(\frac{50000xy}{6}\)
The average salary in the non-manufacturing department is:
\(\frac{\frac{50000xy}{6}}{80x} = \frac{500y}{24x}\)
We now calculate the ratio of the average salary in the manufacturing department to the average salary in the non-manufacturing department:
\(\frac{\frac{500y}{6x}}{\frac{500y}{24x}} = \frac{500y}{6x} \times \frac{24x}{500y} = \frac{24}{6} = 4:15\)
Thus, the ratio of the average salary in the manufacturing department to the average salary in the non-manufacturing department is 4:15.
Therefore, the correct choice is (B): 4:15.
When $10^{100}$ is divided by 7, the remainder is ?