Question

A fountain has two drainage valves. With only the first valve open, the fountain completely drains in 4 hours. With only the second valve open, the fountain completely drains in 5.25 hours. About how many hours will the fountain take to completely drain with both valves open?

Answer 1

Answer: it will take about 2.32 hours for the fountain to completely drain with both valves open

Step-by-step explanation:

The fountain has two drainage valves, First valve and second valve.

With only the first valve open, the fountain completely drains in 4 hours. This means that the rate at which the first valve drains is at a rate of 1/4 per hour

With only the second valve open, the fountain completely drains in 5.25 hours. This means that the rate at which the second valve drains is at a rate of 1/5.5 per hour

If both valves are opened, the fountain will drain faster but both valves will still be draining at their individual unit rates. They will be draining simultaneously. It means that their individual unit rates are additive. .Assuming it takes t hours to drain completely if both valves are open. Then their combined unit rate will be 1/t per hour. Therefore

1/4 + 1/5.5 = 1/t

9.5/22 = 1/t

t = 22/9.5 = 2.32 hours

Answer 2

You can use variables to model the situation and convert the description to mathematical expression.

The time that the fountain will take to get drained completely with both valves open is approximately 2.27 hours.

How to form mathematical expression from the given description?

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on  methods can be used to convert description to mathematical expressions.

Using above methodology to get to the solution

Let the fountain contains V amount of water

Let the first valve emits x liter of water per hour

Let the second valve emits y liter of water per hour

Then, from the given description, we have:

With only the first valve open, the fountain completely drains in 4 hours

or

$x \times 4 = V\\\\x = \dfrac{V}{4}$ (it is since V volume of water is drained when we let x liter of water drained for 4 hours, thus adding x 4 times which is equivalent of x times 4)

Similarly,

With only the second valve open, the fountain completely drains in 5.25 hours
or

$y \times 5.25 = V\\y = \dfrac{V}{5.25}$

Let after h hours, the fountain gets drained, then,

$x \times h + y \times h = V\\\\\dfrac{V}{4}h + \dfrac{V}{5.25}\times h = V\\\\\text{Multiplying both the sides with} \: \dfrac{4 \times 5.25}{V}\\\\5.25 \timses h + 4 h = 4 \times 5.25\\9.25h = 21\\\\h = \dfrac{21}{9.25} \approx 2.27$

Thus,

The time that the fountain will take to get drained completely with both valves open is approximately 2.27 hours.

Learn more about linear equations representing situations here:

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