Mind too small for quadratic equation problem

Question

I was struggling for a long time with the following question, eventually unable to find it. I'm really frustrated with it because the answer is not difficult per se but to me requires a complete shift in approaching the problem, and I am just unable to make such a leap by myself. So, is it a hard question or am I just stupid?

Here goes: Two pipes together can fill a reservoir in 6 hr 40 min. Find the time each alone will take to fill the reservoir if one of the pipes can fill it in 3 hr less time than the other.

Edit: The solution is in my textbook so not quite asking for that. Once you formulate the quadratic equation it's straightforward but for me this step really takes a leap of the imagination to find it and I wonder if I'm alone in this. Can't really post the answer because then it'll be obvious.

So basically I'm asking if anyone can formulate the equation and tell me if they found it difficult.

Thanks for the willingness to answer already!

Answer 1

Suppose that one pipe takes $t$ hours to fill the reservoir, and the other pipe takes $t-3$ hours.

Then in $t(t-3)$ hours, pipe 1 has filled $t-3$ reservoirs, and pipe 2 has filled $t$ reservoirs. Together, they fill $2t - 3$ reservoirs in $t(t-3)$ hours, or $1$ reservoir in $\frac{t(t-3)}{2t-3}$ hours.

Converting $6$ hours $40$ minutes into hours, we get:

$$\frac{t(t-3)}{2t-3} = \frac{20}{3}$$ $$\Rightarrow 3t(t-3) = 20(2t-3)$$ $$\Rightarrow 3t^2 - 9t = 40t - 60 \implies 3t^2 - 49t + 60 = 0$$ $$\Rightarrow (3t - 4)(t - 15) = 0 \implies t = 15$$

so working alone, one pipe fills it in $15$ hours, and the other fills it in $12$.

The smaller solution is discarded because firstly, it would make $t-3$ negative (which does not make sense), and working alone should result in a longer time than working together.

As a sanity check, you can repeat this process with $t = 15$ (or using any other method), and doing this gives you $\frac{20}{3}$ hours to fill up the reservoir.

Answer 2

We are looking at the percentage of the reservoir that is full as a function of time, call the function $p_i(t)$, $0 \leq p_i(t) \leq 100$, where $i=1,2$ standing for pipe 1 or pipe 2. Note that regardless of the combination of pipes, the reservoir starts empty, so $p_1(0)=p_2(0)=0$.

One of the pipes fills the reservoir to $100$ after a certain period of time, call it $T$: $$ p_1(T) = 100 .$$ The other pipe takes 3 hours less: $$ p_2(T-3) = 100 .$$ So we can equate the two equations to see $$ p_1(T) = p_2(T-3) . \quad (*)$$

Now assuming that in each pipe the water flows at a linear rate, we have that for any point in time $0 \leq t \leq T$, $$ p_1 (t) = \left( \frac{p_1(T) - p_1(0) }{T-0} \right) (t-0) - p_1(0) = \frac{100t}{T} $$ and similarly, in combination with $(*)$, $$ p_2 (t) = \frac{100t}{T-3}. $$

Finally we use the information that both pipes together fill the reservoir in $t=6+(4/6)=40/6=20/3$ hours. So $$ 100 = p_1 (20/3) + p_2 (20/3) = (2000/3) \left( \frac{1}{T} + \frac{1}{T-3} \right) $$ Now solving for $T$ gives you your quadratic and your answer

Answer 3

Thanks for the comments and answers. The answer my book gives wasn't given, but this is the most straightforward:

Let $x$ = time (hours) required by smaller pipe, $x-3$ = time required by larger pipe. Then $\frac{1}{x}$ = part filled in 1 hr by smaller pipe and $\frac{1}{x-3}$ = part filled in 1 hr by larger pipe.

Since the two pipes together fill $\frac{1}{20/3} = \frac{3}{20}$ of the reservoir in $1$ hr, $\frac{1}{x} + \frac{1}{x-3} = \frac{3}{20}$.

This is the quadratic equation that you can then solve easily to get the answer.

So, the whole idea is that you take think in parts per hour instead of hours per reservoir. Couldn't imagine myself thinking of that.

Answer 4

Here is the logic:-

Suppose that pipe-1 can fill the tank in x hours.

Then, in 1 hour's time, ($\dfrac {1}{x}$) of tank have been filled.

Similarly, for pipe-2, in 1 hour's time, ($\dfrac {1}{x - 3}$) of the tank have been filled.

If they worked together, in 1 hour's time, $(\dfrac {1}{x} + \dfrac {1}{x - 3})$ of the tank have been filled.

Then, $(\dfrac {1}{x} + \dfrac {1}{x - 3}) = ...$