Alice and Bob do a job in 2 hours. Alice and Charlie can do the same job in 3 hours. Bob and Charlie can do the same job in 4 hours.

Question

how long would it take for Alice, Bob, and charlie to do the same job if they worked together?

Answer 1

PROBLEM

 

Alice and Bob do a job in 2 hours. Alice and Charlie can do the same job in 3 hours. Bob and Charlie can do the same job in 4 hours.
how long would it take for Alice, Bob, and charlie to do the same job if they worked together?

 

SOLUTION

 

This is a work problem; i.e., work done = rate × time

 

If you had two workers performing at different rates, to find their total rate you would look at the problem like this:

 

rate A + rate B = total rate

 

work A = rate A * time

 

work B = rate B * time

 

total work = work A + work B

 

total work = rate A * time + rate B * time

 

factoring out the time gives you

 

total work = time * (rate A + rate B)

 

Now, let's apply these concepts to the problem

 

Let A = Alice's rate (jobs/hr)

 

Let B = Bob's rate (jobs/hr)

 

Let C = Charlie's rate (jobs/hr)

 

So, if Alice and Bob can do one job in 2 hours, their equation would be:

 

2hrs•(A jobs/hr + B jobs/hr) = 1 job or just simply

 

2•(A + B) = 1 which becomes

 

2A + 2B = 1 [EQUATION 1]

 

For Alice and Charlie the equation would be:

 

3•(A + C) = 1 which becomes

3A + 3C = 1 [EQUATION 2]

 

For Bob and Charlie the equation would be:

 

4•(B + C) = 1 which becomes

4A + 4C = 1 [EQUATION 3]

 

Now you have a system of equations with three equations and three unknowns:

 

2A + 2B          = 1 [EQUATION 1]

 

3A          + 3C = 1 [EQUATION 2]

 

         4B + 4C = 1 [EQUATION 3]

 

Let's solve [EQUATION 1] for A and substitute that into [EQUATION 2]

 

2A + 2B = 1

 

        2A = 1 - 2B

 

          A = (1/2) - B [EQUATION 1]*

 

3A + 3C = 1

 

3[(1/2) - B] + 3C = 1

 

(3/2) - 3B + 3C = 1

 

         - 3B + 3C = 1 - (3/2)

 

         - 3B + 3C = (2/2) - (3/2)

 

         - 3B + 3C = - (1/2) [EQUATION 2]*

 

Now, combine [EQUATION 2]* and [EQUATION 3] and use the elimination method to solve for one of the variables

 

- 3B + 3C = - (1/2) [EQUATION 2]*

  4B + 4C =         1 [EQUATION 3]

 

 

4•[- 3B + 3C = - (1/2)] [EQUATION 2]*
3•[  4B + 4C =         1] [EQUATION 3]

 

- 12B + 12C = - 2 [EQUATION 2]*
  12B + 12C =    3 [EQUATION 3]

            24C =    1

                C = 1/24 jobs/hr

 

Substituting C into [EQUATION 3]

 

4B + 4C = 1 [EQUATION 3]

 

4B + 4•(1/24) = 1

 

4B + (4/24) = 1

 

4B + (1/6) = 1

 

            4B = 1 - (1/6)

 

            4B = 5/6

 

              B = 5/24 jobs/hr

 

Substituting B into [EQUATION 1]

 

2A + 2B = 1 [EQUATION 1]

 

2A + 2•(5/24) = 1

 

2A + 10/24 = 1

 

2A + 5/12 = 1

 

2A = 1 - 5/12

 

2A = 7/12

 

A = 7/24 jobs/hr

 

Working together their total rate would be:

 

(7/24) + (5/24) + (1/24) = 13/24 jobs/hr

 

Since work = rate × time

 

1 (job) = 13/24 (jobs/hr) × time (hrs)

 

Solving for time gives time = 1/[(13/24)] = 24/13 hrs = 1 11/13 hrs

 

 

Answer 2

This is a work problem...and they all work out the same way which is hard to explain, basically it is a rate problem.

The algebra is messy, but persevere and DO NOT try to clear fractions!

If you don't know why this is the way to work this problem, please send me a message and I will try to explain it further.

 

1/A + 1/B = 1/2

1/A + 1/C = 1/3

1/B + 1/C = 1/4

 

1/A = 1/2 - 1/B

1/A = 1/3 - 1/C

 

1/2- 1/B = 1/3 - 1/C

 

1/B - 1/C = 1/6

1/B + 1/C = 1/4

 

2/B = 5/12 => 1/B = 5/24

1/A + 5/24 = 1/2 => 1/A = 7/12

 

5/24 + 1/C = 1/4 => 1/C = 1/24

 

1/A + 1/B + 1/C = 1/x

 

14/24 + 5/24 + 1/24 = 1/x

20/24 = 1/x  => x = 6/5