# Len can complete a task in 4 hours less than Ron. On the other hand if they both work together on the task it is completed in 4 hours. How long would it take for each of them to complete the task on their own?

## This is more a tutorial on mathematical approach. The solution takes a bit of work so it is split into parts.

Dec 29, 2017

$\textcolor{red}{\text{Solution part 1}}$

#### Explanation:

The general approach is first to define the given key information in formats that may be manipulated. Then to eliminate what is not needed. Use what is left through some format of comparison to determine the target values.

There are a lot of variables so we need to reduce them by substitution if we can.

$\textcolor{b l u e}{\text{Defining the key points}}$

Let the total amount of work needed for the task be $W$
Let the work rate of Ron be ${w}_{r}$
Let the time Ron would need to complete all the task be ${t}_{r}$

Let the work rate of Len be ${w}_{L}$
Let the time Len would need to complete all the task be ${t}_{L}$

Then we have:

${w}_{r} {t}_{r} = W \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$
${w}_{L} {t}_{L} = W \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

From the question we also have:

${t}_{L} = {t}_{r} - 4 \text{ } \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(3\right)$

Working together for 4 hours we have:

$4 {w}_{r} + 4 {w}_{L} = W \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(4\right)$
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$\textcolor{b l u e}{\text{Looking for usable connections}}$

Using $E q n \left(1\right) \mathmr{and} E q n \left(2\right)$ noting that $W$ is a common value we can start to experiment to see if we can eliminate one or more of the unknowns. There are too many.

Lets express work rates in terms of $W$ forming a link

Eqn(1)-> w_rt_r=W color(white)("d")=>color(white)("d")w_r=W/t_r" "....Equation(1_a)

Eqn(2)->w_Lt_L=W color(white)("d")=>color(white)("d")w_L=W/t_L" ".....Equation(2_a)

Ok, lets see if we can 'get rid' of one more. We now that from $E q n \left(3\right) \textcolor{w h i t e}{\text{d}} {t}_{L} = {t}_{r} - 4$ so we can do another substitution in $E q n \left({2}_{a}\right)$ giving:

Eqn(2_a)->w_L=W/t_L color(white)("d")=>color(white)("d")w_L=W/(t_r-4)" ".....Equation(2_b)

Now we can substitute into $E q n \left(4\right)$ and see what we get.
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$\textcolor{m a \ge n t a}{\text{See solution part 2}}$

Dec 29, 2017

$\textcolor{m a \ge n t a}{\text{Solution part 2}}$

#### Explanation:

Continued from solution part 1

Substitute in $E q n \left(4\right)$ using $E q n \left({1}_{a}\right) \mathmr{and} E q n \left({2}_{b}\right)$

color(green)([4color(red)(w_r)]+[4color(red)(w_L)]=Wcolor(white)("d")->color(white)("d")[4color(red)(xxW/t_r)]+[4color(red)(xxW/(t_r-4))]=W

color(white)("dddddddddddddddd")color(green)(->color(white)("ddd")(4W)/(t_r)color(white)("dd")+color(white)("dd")(4W)/(t_r-4)color(white)("ddd")=W)

As there are $W ' s$ on both sides (in everything) we can 'get rid of them. Divide both sides by $W$

color(white)("dddddddddddddddd")color(green)(->color(white)("ddd")4/(t_r)color(white)("dd")+color(white)("dd")4/(t_r-4)color(white)("ddd")=1)

We now need to make the denominators all the same and we $\underline{\text{'force'}}$ them to be so.

Notice that there is only a ${t}_{r}$ as the denominator on the left fraction. So we need a ${t}_{r}$ that we can factor in the right hand denominator but in such a way that is just another way of writing ${t}_{r} - 4$. Note that ${t}_{r} \left(1 - \frac{4}{t} _ r\right)$ is such a thing. Multiply it out and you get ${t}_{r} - 4$. So we write:

color(white)("dddddddddddddddddd")color(green)(->color(white)("dd")4/t_rcolor(white)("d")+color(white)("d")4/(t_r(1-4/t_r))color(white)("d")=1)

Now we need to change $\frac{4}{t} _ r$ to have the same denominator as the right fraction. Multiply by 1 but in the form $\frac{1 - \frac{4}{t} _ r}{1 - \frac{4}{t} _ r}$

color(white)("dddddddddddddd")color(green)(->color(white)("dd")(4(1-4/t_r))/(t_r(1-4/t_r))color(white)("d")+color(white)("d")4/(t_r(1-4/t_r))color(white)("d")=1)

color(white)("dddddddddddddd")color(green)(->color(white)("ddddddd")(4(1-4/t_r)+4)/(t_r(1-4/t_r))color(white)("dddddd")=1)

$\textcolor{w h i t e}{\text{ddddddddddddddd")->color(white)("dddddd}} 4 \left(1 - \frac{4}{t} _ r\right) + 4 = {t}_{r} \left(1 - \frac{4}{t} _ r\right)$

$\textcolor{w h i t e}{\text{ddddddddddddddd")->color(white)("dddddddd")4-16/t_rcolor(white)("d}} + 4 = {t}_{r} - 4$

$\textcolor{w h i t e}{\text{ddddddddddddddd")->color(white)("ddddddddd}} 0 = {t}_{r} + \frac{16}{t} _ r - 12$

We need to 'get rid' of the denominator ${t}_{r}$ so multiply both sides by ${t}_{r}$

$\textcolor{w h i t e}{\text{ddddddddddddddd")->color(white)("ddddddddd}} 0 = {\left({t}_{r}\right)}^{2} + 16 - 12 {t}_{r}$
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$\textcolor{m a \ge n t a}{\text{See part 3}}$

Dec 29, 2017

$\textcolor{red}{\text{Solution Part 3}}$

${t}_{r} = 6 + 2 \sqrt{5}$

${t}_{L} = {t}_{r} - 4 = 2 + 2 \sqrt{5}$

#### Explanation:

In part 2 we ended up with:

$0 = {\left({t}_{r}\right)}^{2} + 16 - 12 {t}_{r}$

$0 = {\left({t}_{r}\right)}^{2} - 12 {t}_{r} + 16$

Completing the square

$0 = {\left({t}_{r} - 6\right)}^{2} + k + 16$ where ${\left(- 6\right)}^{2} + k = 0 \implies k = - 32$

$0 = {\left({t}_{r} - 6\right)}^{2} - 32 + 16$

$0 = {\left({t}_{r} - 6\right)}^{2} - 20$

${t}_{r} = 6 \pm 2 \sqrt{5}$ Note that $6 - 2 \sqrt{5}$ does not work so we have:

${t}_{r} = 6 + 2 \sqrt{5}$

Thus ${t}_{L} = {t}_{r} - 4 = 2 + 2 \sqrt{5}$