# Word problem

1. Dec 13, 2011

### mindauggas

1. The problem statement, all variables and given/known data

Machine A can do a job, working alone, in 4 hours less than machine B. Working together, they can complete
the job in 5 hours. How long would it take each machine, working alone, to complete the job?

2. Relevant equations

Ans. Machine A: 8.4 hours; machine B: 12.4 hours, approximately

3. The attempt at a solution

Don't know how to reason this one through ...

2. Dec 13, 2011

### Staff: Mentor

Let the time that machine B takes to carry out the job be b hours.

Using the information we are given, we can write the time that machine A takes to carry out the job as the expression ..... ?

3. Dec 13, 2011

### mindauggas

a=b-4

And then ... ?

4. Dec 13, 2011

### Ray Vickson

In problems of this type you need to ASSUME something about how two machines work together. Clearly, the work times themselves don't add, so what does happen?

If we assume the RATES add (so that rate(A+B) = rate(A) + rate(B)) then we get two equations for the rates. If Ra = rate of A and Rb = rate of B (meaning the number of jobs per hour these machines can complete), time(A) = 1/Ra, time(B) = 1/Rb, and time(A+B) = 1/(Ra+Rb). You can take it from there.

RGV

5. Dec 13, 2011

### mindauggas

So in this particular case i have

time(A) = 1/Ra = B-4

time(B) = 1/Rb = B

time(A+B) = 1/(Ra+Rb) = $\frac{1}{(\frac{1}{(B-4)})+(\frac{1}{B})}$

But is it really the case???

6. Dec 13, 2011

### Ray Vickson

If the addition-of-rates assumption holds, then YES, that is the case; if the addition-of-rates assumption does not hold, then NO, that is not the case. The question itself (at least as you stated it here) leaves undetermined the manner in which two machines work together; the additive-rates method applies in some real-world situations (at least approximately), but not in some others.

Instead of further agonizing about the problem, I suggest you just go ahead and solve it to see what you get.

RGV

7. Dec 14, 2011

### mindauggas

I get the quadratic B$^{2}$-14B+20=0

B$_{1}$=12,4 (so we get the correct answer)

B$_{2}$=1,6 (i need to discard it, because the other assumption is A=B-4 and we can't have a negative work done (at least in this situation)).

Thanks RGV/