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[ASK] Paint Problem

Monoxdifly

Well-known member
Aug 6, 2015
312
3 friends, Alan, Brian and Chester, paint a house. If Alan had to paint it on his own, it would take him one hour more than the time it would take for all three to paint it together. If Brian had to paint it on his own, it would take him five hours more than the time it would take for all three to paint it together, and Chester 8 hours more.
How much time would it take for Alan and Brian to paint it together?

If Alan had to paint it on his own, it would take him one hour more than the time it would take for all three to paint it together.
Would it be like this?
\(\displaystyle \frac1{A+1}=\frac1A+\frac1B+\frac1C\)?
Will it have something to do with quadratic equation?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
980
Actually, the solution I came up with involved both a cubic & quadratic equation. There may be some clever, short solution, but I was unable to reason one out.

let $t$ be the number of hours it takes for all three painters to complete the job.

$\dfrac{1}{t+1} + \dfrac{1}{t+5} + \dfrac{1}{t+8} = \dfrac{1}{t}$

working this equation yielded the cubic equation

$t^3+7t^2-20=0$

using the rational root theorem resulted in the factorization

$(t+2)(t^2+5t-10)=0$

solving the quadratic factor yielded $t=\dfrac{-5+\sqrt{65}}{2} \approx 1.53 \text{ hrs}$, the time required to complete the job with all three working.

for just A and B ...

$\dfrac{1}{t+1}+\dfrac{1}{t+5} = \dfrac{1}{x}$, where $x$ is the time to complete the job with just A and B working.

Being lazy, I used my calculator ... $x \approx 1.82 \text{ hrs}$

Just curious, where did you get this problem?
 

Monoxdifly

Well-known member
Aug 6, 2015
312
Just curious, where did you get this problem?
From another forum. I though it looked simple enough, so I saved it in case I needed it. Turned out I couldn't solve it, so I tried to retrace from which forum it was from, but Google search showed no result. Thanks for your help, though. :)
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
802
3 friends, Alan, Brian and Chester, paint a house. If Alan had to paint it on his own, it would take him one hour more than the time it would take for all three to paint it together. If Brian had to paint it on his own, it would take him five hours more than the time it would take for all three to paint it together, and Chester 8 hours more.
How much time would it take for Alan and Brian to paint it together?

If Alan had to paint it on his own, it would take him one hour more than the time it would take for all three to paint it together.
Would it be like this?
\(\displaystyle \frac1{A+1}=\frac1A+\frac1B+\frac1C\)?
Will it have something to do with quadratic equation?
Since you havn't said what "A", "B", and "C" represent it is impossible to say whether that equation is correct or not!

I will presume that "A" is the time, in hours, it would take Alan to do the job alone, that "B" is the time, in hours, it would take Brian to do the job alone, and that "C" is the time, in hours, it would take Chester to do the job alone. The rate at which each works will be 1 "job" divided by the time it takesto do job. Alan works at the rate of \(\displaystyle \frac{1}{A}\) job/hour, Brian works at the rate of \(\displaystyle \frac{1}{B}\) job/hour, and Chester works at the rate of \(\displaystyle \frac{1}{C}\) job/hour.

When people work together their rates add. Working together, their rate will be \(\displaystyle \frac{1}{A}+ \frac{1}{B}+ \frac{1}{C}=\frac{BC+AC+ AB}{ABC}\) job/hour and the time it would take for all three to do the job working together is
\(\displaystyle \frac{1}{\frac{BC+ AC+ AB}{ABC}}= \frac{ABC}{BC+ AC+ AB}\)

If Alan had to paint it on his own, it would take him one hour more than the time it would take for all three to paint it together.
So \(\displaystyle A= \frac{ABC}{BC+AC+AB}+ 1\).

If Brian had to paint it on his own, it would take him five hours more than the time it would take for all three to paint it together
So \(\displaystyle B= \frac{ABC}{BC+AC+AB}+ 5\).

and Chester 8 hours more.
So \(\displaystyle C= \frac{ABC}{BC+AC+AB}+ 8\).

Solve those three equations for A, B, and C.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
802
Since this has been here since January and I just can't leave it undone:

We have the three equations

$A= \frac{ABC}{AB+ AC+ BC}+ 1$

$B= \frac{ABC}{AB+ AC+ BC}+ 5$

$C= \frac{ABC}{AB+ AC+ BC}+ 8$



We can write those as

$A- 1= \frac{ABC}{AB+ AC+ BC}$

$B- 5= \frac{ABC}{AB+ AC+ BC}$ and

$C- 8= \frac{ABC}{AB+ AC+ BC}$



So A- 1= B- 5= C- 8

From A- 1= B- 5, A= B- 4.

From B- 5= C- 8, C= B+ 3.



So we can write $B- 5= \frac{ABC}{AB+ AC+ BC}$ as

$B- 5= \frac{B(B- 4)(B+3)}{B(B- 4)+ (B- 4)(B+ 3)+ B(B+ 3)}$.

$(B-5)B(B- 4)+ (B-5)(B+4)(B+3)+ (B-5)B(B+ 3)= B(B- 4)(B+3)$.



Multiply that out and each side will have a term of $B^3$ which will cancel leaving a quadratic equation to solve for B.