Water is flowing out of an artesian spring at a rate of 8 cubic feet per minute. There are 7.5 gallons of water per cubic foot. How many minutes will it take for the water to fill up a 300 gallon tank?

As the water is flowing out of the artesian spring at a rate of `8` cubic feet per minute and each cubic foot has `7.5` gallons

the water is flowing out of the artesian spring at a rate of `8xx7.5=60` gallons per minute.

As the tank can fill up `300` gallon,

it should take `300/60=5` minutes to fill tank.

`color(blue)("Building the conversion")`

you can manipulate units of measurement the same way you do the numbers.

Let cubic feet per minute be `" "(ft^3)/min`

Let gallons of water per cubic foot be `" "g/(ft^3)`

So `" "(ft^3)/minxxg/(ft^3)" " ->" "(cancel(ft^3))/(cancel(ft^3))xxg/(min)`

`color(green)("From this observe that multiplying these together give you gallons")``color(green)("per minute."ul(" Associating this with the numbers gives:"))`

`=>8 (ft^3)/("min") xx 7.5 g/(ft^3)" " =" " (8xx7.5) g/("min")" "=" "60 g/("min")`

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`color(blue)("Determine the time for 300 gallons")`

Using ratio in fraction form

`(60g)/(1 "min") -=(300g)/("time")`

Turn the whole thing upside down

`(1 "min")/(60g)-=("time")/(300g)`

By example, using the same principle as `1/2` is the same as `2/4`
Multiply by 1 and you do not change the overall value. However, 1 comes in many forms.

`color(green)([(1 "min")/(60g)color(red)(xx1)]-=("time")/(300g)" "->" " [(1 "min")/(60g)color(red)(xx5/5)]-=("time")/(300g))`

`=(5" min")/(300 g)`

The time taken for 300 gallons is 5 minutes