A farmer can plow the field in 8 days. After working for 3 days, his son joins him and together they plow the field in 3 more days. How many days will it require for the son to plow the field alone?

Apr 30, 2017

Solution 1 of 2
Go to solution 2 of 2 to see what it should look like.
12 days

$\textcolor{b r o w n}{\text{This is more a tutorial on how to handle units of measurement}}$

Explanation:

Method uses rates or work measured in units of 1 field per day

Let the rate of work per day for the father be ${W}_{f}$
Let the rate of work per day for the sun be ${W}_{s}$

Let the total work done by the father be ${T}_{f}$
Let the total work done by the son be ${T}_{s}$

Let the unit identifying counts of a field be $f$
Let the unit identifying counts of days be $d$

Example $2 d \to 2 \mathrm{da} y s \mathmr{and} \frac{1}{3} f \to \frac{1}{3} o f 1 f i e l d$
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$\textcolor{b l u e}{\text{Determine the rate of work for the father}}$

Modeling the number of days worked for completion of one field

${W}_{f} \times {T}_{f 1} = 1 f \text{ "->" } {W}_{f} \times 8 d = 1 f$

So the amount of work done in 1 day $\to {W}_{f} = \frac{1}{8} \frac{f}{d}$

Where $\frac{f}{d}$ represents 'fields per day'
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$\textcolor{b l u e}{\text{Determine the total amount of work done by the father }}$

Works for 3 days without the son $\text{ } \ldots \ldots \ldots . . \to 3 {\mathrm{dW}}_{f}$
Then works a further 3 days but with the son $\underline{\to 3 {\mathrm{dW}}_{f}} \leftarrow \text{ Add}$
Total work done by the father:$\text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .6 {\mathrm{dW}}_{f}$

Do not forget the $d$ is a unit of measurement. As is $f$
But ${W}_{f} = \frac{1}{8} \frac{f}{d}$ giving:

${T}_{f 2} = 6 {\mathrm{dW}}_{f} \text{ "=" "6cancel(d) xx 1/8 f/(cancel(d))" "=" } \frac{3}{4} f$
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$\textcolor{b l u e}{\text{Determine the total amount of work done by the son}}$

The father has done $\frac{3}{4} f$ so the son did:
$1 f - \frac{3}{4} f \text{ "=" } \frac{1}{4} f$
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$\textcolor{b l u e}{\text{Determine the work rate of the son}}$

The son worked for 3 days so we have the model:

$3 {\mathrm{dW}}_{s} = \frac{1}{4} f$
Do not forget the $d$ is a unit of measurement. As is $f$

$3 {\mathrm{dW}}_{s} = \frac{1}{4} f \text{ "->" }$

divide both sides by $3 d$

$\frac{3 d}{3 d} \times {W}_{s} = \frac{f}{4} \times \frac{1}{3 d} \text{ "->" } {W}_{s} = \frac{1}{12} \frac{f}{d}$
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$\textcolor{b l u e}{\text{Determine son's time for 1 field}}$

$1 f = {W}_{s} \times {T}_{s} \text{ } \to 1 f = \frac{1 f}{12 d} \times {T}_{s}$

Divide both sides by $\frac{1 f}{12 d}$ giving:

T_s=(12d)/(1cancel(f color(white)(.))) xx 1 cancel(fcolor(white)(.))

${T}_{s} = 12 d \text{ "->" "12" days}$

Apr 30, 2017

Solution 2 of 2
This more efficient calculated is done without explaining what to do with units and how they work.

12 days

Explanation:

Fathers rate of work: 1 field in 8 days $\implies \frac{1}{8}$ fields per day

When working with his son the father completed $6 \times \frac{1}{8} = \frac{3}{4}$ fields

So his son completed $1 - \frac{3}{4} = \frac{1}{4}$ fields in 3 days

Thus the sons rate of work is $\frac{1}{4} \div 3 = \frac{1}{12}$ fields per day

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Thus the number of days (d) it would take for the son to do one field is:

$d \to d \times \frac{1}{12} = 1 \text{ "=>" } d = 12$ days