Alex can complete a model rocket assembly in four hours less time than it takes Sean.
Working together, they can complete the assembly in one and a half hours. How long
would it take Alex to do the assembly working alone?
A long time.
To find out how long it would take Alex to complete the assembly working alone, we can follow these steps:
Step 1: Assign variables to the unknowns:
Let A be the time it takes Alex to complete the assembly alone.
Let S be the time it takes Sean to complete the assembly alone.
Step 2: Set up equations based on the given information:
From the problem statement, we know that Alex can complete the assembly in four hours less time than Sean, so we have the equation:
A = S - 4
Additionally, we are told that working together they can complete the assembly in one and a half hours, or 1.5 hours. This can be represented as:
1/A + 1/S = 1/1.5
Step 3: Substitute the value of A from the first equation into the second equation:
Substituting A = S - 4 into the equation 1/A + 1/S = 1/1.5, we get:
1/(S - 4) + 1/S = 1/1.5
Step 4: Solve the equation for S:
Multiply both sides of the equation by the least common multiple (LCM) of the denominators to eliminate fractions. In this case, the LCM is 1.5S(S - 4). We get:
1.5S + 1.5(S - 4) = S(S - 4)
Simplify the equation:
1.5S + 1.5S - 6 = S^2 - 4S
Combine like terms:
3S - 6 = S^2 - 4S
Rearrange the equation to set it to zero:
S^2 - 7S + 6 = 0
Step 5: Factor the quadratic equation:
(S - 1)(S - 6) = 0
So, S - 1 = 0 or S - 6 = 0
Solving for S:
If S - 1 = 0, then S = 1
If S - 6 = 0, then S = 6
Step 6: Determine the value of A:
Using the first equation A = S - 4, we can determine the value of A.
If S = 1, then A = 1 - 4 = -3 (extraneous solution)
If S = 6, then A = 6 - 4 = 2
Step 7: Check the answer:
To check if our solution is correct, we can substitute the values of A and S into the equation 1/A + 1/S = 1/1.5:
1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3
Since 2/3 is equal to 1/1.5, our solution is correct.
Therefore, it would take Alex 2 hours to complete the assembly working alone.