Two taps running together can fill
a tank in three and one thirteenth hours. If one tap takes three hours more
than the other to fill the tank, how much time will each tap take to fill the
We can picture this scenario of a
tank filling up with water thanks to the water coming out of two different taps. If we call these taps one and two,
respectively, we want to solve for how much time it will take for each one
individually to fill up the tank.
We’re given two pieces of
information about the scenario. The first is that in three and one
thirteenth hours, the taps working together fill the tank completely. And second, if we call 𝑡 sub one
the time it would take tap one to completely fill the tank by itself and 𝑡 sub two
that same time for the second tap, then we’re told that one of these taps — we’ll
call it 𝑡 sub two — takes three hours longer than the other tap 𝑡 sub one to fill
the tank. It’s these values 𝑡 sub two and 𝑡
sub one that we want to solve for based on this information.
To start solving for these values,
we can create a table with three different columns. The first column has the taps that
are in use filling the tank, the second has the time elapsed, and the third shows
the level of the tank’s fullness.
If both taps one and two are in use
for a time of three and one thirteenth hours, we can say that the tank level is full
and we’ll represent that with one. If we only let tap one be in use
and let that tap run for three and one thirteenth hours, then we can say that the
tank level will be equal to three and one thirteenth hours divided by 𝑡 one, which
is the time it takes tap one all by itself to fill the tank.
In other words, the ratio of these
two times shows the level of the tank’s fullness after three and one thirteenth
hours have elapsed with the tank being filled just from tap one. In a similar way, if we only turn
on tap two and let that run for three and one thirteenth hours, then the tank’s
fullness can be represented by three and one thirteenth hours divided by 𝑡 one plus
three hours. Recall that’s equal to 𝑡 two.
Now, we can combine the last two
rows of our table, representing the effects of taps one and two individually, and
relate them to the effects of taps one and two at the same time, the first row. We can say that the tank level due
to just tap two plus the tank level due to just tap one is equal to the tank level
due to both taps combined.
We now have an equation that we can
solve for 𝑡 one, the time it takes for the tank to fill just with tap one in
operation. We can now consider how to
rearrange this mathematical relationship to solve for 𝑡 one.
As a first step, we can convert
three and one thirteenth into a simplified fraction. Three and one thirteenth is equal
to 40 thirteenth. And if we multiply both sides of
the equation by 13 over 40, our equation simplifies to one over 𝑡 one plus one over
𝑡 one plus three is equal to 13 fortieths.
Then, if we multiply both sides of
the equation by 𝑡 one times the quantity 𝑡 one plus three, our left-hand side
simplifies to 𝑡 one plus three plus 𝑡 one or two 𝑡 one plus three, which is equal
to 13 fortieths times 𝑡 one times the quantity 𝑡 one plus three.
Now, looking at the right-hand
side, let’s multiply out our factor of 𝑡 one and ultimately reach a quadratic
equation in 𝑡 one. After multiplying through by 𝑡
one, we can then multiply both sides of our equation by forty, cancelling out that
factor on the right-hand side. This gives us 80 times 𝑡 one plus
120 on the left being equal to 13 𝑡 one squared plus 39 𝑡 one on the right. If we then subtract 80 𝑡 one and
120 from both sides of the equation, we now arrive at a quadratic equation for 𝑡
The next question is what’s the
best way to solve for 𝑡 one given this relationship. We could use the quadratic equation
identifying the factors of 𝑎, 𝑏, and 𝑐. But there’s another way to solve
for 𝑡 one that involves less calculation. Let’s consider this middle term in
the expression negative 41 𝑡 one.
We want to know if we can write
this term in such a way that by splitting it up, we can create two terms from this
one, which will share factors with 13 and 120. What if we split negative 41 𝑡 one
into negative 65 𝑡 one plus 24 𝑡 one? In this case, our value of 65 and
13 are related by an integer multiple as are the values of 24 and 120. So let’s see how this works.
Now that we’ve split our term of
negative 41 𝑡 one into negative 65 𝑡 one plus 24 𝑡 one, let’s consider the first
two terms on the right-hand side of this expression as well as the second two
Considering the first two terms,
it’s possible to express them as 13 𝑡 one times the quantity 𝑡 one minus five. And regarding the second two terms,
it’s possible to express them as 24 times the quantity 𝑡 one minus five. Based on this, we see that 𝑡 one
minus five can be factored out of this expression. Therefore, we can write that zero
is equal to the quantity 𝑡 one minus five multiplied by the quantity 13 𝑡 one plus
Looking now at these 𝑡 one terms,
we wonder for what value of 𝑡 one will this overall expression be equal to
zero? In the first term, we see that 𝑡
one equaling five will lead to five minus five or zero. And in the second case, we see that
𝑡 one being equal to negative 24 over 13 will lead to that second term being zero
and therefore the overall expression being zero.
At this point, we recall that 𝑡
one is the time it takes for faucet one to fill the tank all by itself. This time value must be positive,
which means our potential solution negative 24 over thirteenth being a negative
value can’t apply. The only root of this equation that
makes physical sense is for 𝑡 one to be equal to five; that is, five hours.
We can say then that 𝑡 one is
equal to five hours. And when we recall that 𝑡 two is
equal to 𝑡 one plus three hours, that means 𝑡 two must be eight hours. These are the times it takes for
the individual taps to fill the tank.