Algebra

Melody: I just noticed what the problem is with the discrepancy bwtween my answer and Rom's. His equations are just fine.The mistake he made was in substitiution of r1=1/20 into equation 1. He simply made a mistake of getting an answer of 1/10 or 10 days for the second man. From his equation, you get r2=1/30 , or 30 days, NOT 1/10. So his answer is the same as mine. Resolved!!.

Two workers, if they were working together, could finish a certain job in 12 days. If one of the workers does the first half of the job and then the other one does the second half, the job will take them 25 days. How long would it take each worker to do the entire job by himself?

The key to this is that the rates they work at add.

\(\text{We get 3 equations.}\\ \\ (r_1+r_2)12=1 \\ \\ r_1 t = r_2(25-t) =\dfrac 1 2\\ \\ r_1 t + r_2 (25-t)=1\)

you can pound through these equations to obtain

\(r_1=\dfrac 1 {20},~~r_2=\dfrac{1}{10}\\ \\ \text{And thus worker 1 would finish in }\dfrac 1{r_1} = 20 \text{ days.}\\ \\ \text{and worker 2 would finish in }\dfrac{1}{r_2} = 10 \text{ days.}\) 

Let first worker=A
Let second worker=B
1/A + 1/B=1/12
A/2 + B/2=25, solve for A, B
A=20 days working alone
B=30 days working alone

Sorry Rom: I get different answer. Remember that 1/12 of the work is done in 1 day. Therefore, working together, they would finish the job in: 1/20+1/30=1/12. And each man working alone, would finish 1/2 of the 2nd. job in: 20/2=10 days for worker A. And 30/2=15 days for worker B. So that: 10 + 15 =25 days. At least that is my take on it!!.

I am going to try and adjudicate.    :)

  Two workers, if they were working together, could finish a certain job in 12 days. If one of the workers does the first half of the job and then the other one does the second half, the job will take them 25 days. How long would it take each worker to do the entire job by himself?

Rom says     the workers will take 10days and 20 days to finish 1 job individually

Then if they did half each that would be 5 days and 10 days = 15 days to finish

That doesn't seem right ://

Guest says 20 days and 30 days then that would be 10+15=25 if they do half each.

Mmm if they work together then. Obviously the faster one does more of the job.

\((\frac{1\;job}{20\;days}+\frac{1\;job}{30\;days})*12days\\ =(\frac{1*3\;job}{20*3\;days}+\frac{1*2\;job}{30*2\;days})*12days\\ =(\frac{3\;job}{60\;days}+\frac{2\;job}{60\;days})*12days\\ =(\frac{5\;job}{60\;days})*12days\\ days\;\;cancel\\ =(\frac{5\;job}{60})*12\\ =(\frac{5\;job}{5})*1\\ =1 \;job\)

That is what I need to see.

I am going with our guests answer - sorry Rom.

I think you both did well.  I am going to think about this one some more.

Yes that is what I thought was most likely but I still have not spent the time on this question that I would like too.

Questions like this are always tricky.  :)

I just wanted to come back and play with this question. 

I am just repeating what Rom has done, in the hope that I can do it myself next time.  Thanks Rom.

Two workers, if they were working together, could finish a certain job in 12 days. If one of the workers does the first half of the job and then the other one does the second half, the job will take them 25 days. How long would it take each worker to do the entire job by himself?

\(\mbox{Let worker 1  do }\quad   r_1 \;\frac{jobs}{day} \quad \mbox{and take t days to do half a job.}\\ \mbox{Let worker 2  do }\quad   r_2 \;\frac{jobs}{day} \quad \mbox{and take (25- t) days to do half a job.}\)

We have 3 unknowns and we have 4 equations

Since it takes 12 days for them to paint the house together

\(r_1+r_2=\frac{1}{12}\;\frac{job}{day} \qquad[1]\\ \frac{r_1t}{2}+\frac{r_2(25-t)}{2}=1\;job \quad[2]\\ r_1t=\frac{1}{2}\;\;job\qquad[3]\\ r_2(25-t)=\frac{1}{2}\;\;job\qquad[4]\\~\\~\\ r_1=\frac{1}{2t}\;\;job\qquad[3b]\\ r_2=\frac{1}{2(25-t)}\;\;job\qquad[4b]\\ \mbox{Sub [3b] and [4b] into [1] }\\ \frac{1}{2t}+\frac{1}{2(25-t)}=\frac{1}{12}\\ \frac{1}{t}+\frac{1}{(25-t)}=\frac{1}{6}\\ \frac{6(25-t)}{6t(25-t)}+\frac{6t}{6t(25-t)}=\frac{t(25-t)}{6t(25-t)}\\ 6(25-t)+6t=t(25-t)\\ 150=t(25-t)\\ t^2-25t+150=0\\ (t-15)(t-10)=0\\ t=15\qquad or \qquad t=10 \)

So one takes 15 days to do 1/2 a job and

the other takes 10 days to do half a job.

Hence, it would take them 30 days and 20 days respectively to do one whole job.

And I did not even need to use equation 2