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I got kind of lost in this . . . so you have [(m liters)/(household)]/month and then i get lost

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up vote 3 down vote accepted

For all $n$ households taken together, the rate at which water is needed is $$\left(m\;\frac{\text{liters}}{\text{households}\times\text{months}}\right)\times\left(n\;\;\text{households}\right)=mn\;\frac{\text{liters}}{\text{months}}.$$ Thus, if $p$ liters is needed to last $x$ months, then $$p\;\;\text{liters}=\left(mn\;\frac{\text{liters}}{\text{months}}\right)\times(x\;\;\text{months}).$$ Solve for $x$.

This problem is an example of dimensional analysis (Wikipedia article).

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Since there are $n$ households, the village requires $mn$ litres of water per month. We have $p$ litres of water and $p$ litres contain $\dfrac{p}{mn}$ number of $mn$ litres which is the number of months it will last.

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Try this:

"$m$ litres will last $1$ household for $1$ month"


"$1$ litre will last $1$ household for $1/m$ months"


"$p$ litres will last $1$ household for $p/m$ months"

Can you continue this process?

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