Question

A plumber and his assistant work together to replace the pipes in an old house. The plumber charges $40 an hour for his own labor and $15 an hour for his assistant's labor. The plumber works twice as long as his assistant on this job, and the labor charge on the final bill is $3325. How long did the plumber and his assistant work on this job

Answer 1

Answer:

The plumber worked 70 hours and his assistant 35 hours.

Explanation:

To solve this problem, we will use the substitution method. Step by step explanation:

1. Defining the variables:

• P: For plumber
• A: For assistant

2. Defining the equations:

1. P = 2A (As the plumber works exactly double what the assistant works)
2. $40P + $15A = $3325

3. Replacing P in equation 2:

• $40(2A) + 15A = 3325\\80A + 15A = 3325\\95A = 3325\\A = \frac{3325}{95} \\A = 35$

4. Replacing A in equation 1 to find the value of P:

• $P = 2(35)\\P = 70$

This means the plumber worked 70 hours and his assistant 35 hours for a final bill of $3325.