# How to find the elapsed time for one candle to be consumed when it is thicker by three units with respect of another?

The problem is as follows:

Assume that you have two candles. One is the triple of diameter of the other. Both of these candles are of the same quality and equal in length. Then they are lighted together at the same time. After hour they differ by 16 centimeters. Then after half an hour later, the length of one is triple the length of the other. How long does it take to burn out the thickest candle from the moment it was lit?

The choices given in my book are as follows:

$$\begin{array}{ll} 1.&\textrm{16 hours and 15 minutes}\\ 2.&\textrm{18 hours and 30 minutes}\\ 3.&\textrm{16 hours and 30 minutes}\\ 4.&\textrm{19 hours and 30 minutes}\\ \end{array}$$

I'm not sure how to solve this problem. The thing is that I don't know how to relate the thickness of the candles to make them work in an equation.

The only thing which it comes to my mind is that the one which is thicker will burn out at a lesser rate than the one which is less thicker.

I think one of the candles will have a rate of one third of the other based on their thickness but I don't know how to connect these ideas together. Can someone help me with the steps here?.

My book gives the hint of using the fact that after an hour both candles consume the same amount of volume, but I don't know how to prove this or if this was stated on purpose by my book. Will this help?.

It would help me a lot if the steps could explain what sort of interpretation should be done here.

After working out this problem a little bit further I arrived to this conclusion:

It seems that you can get some relationship between the descend lengths when you relate the volume of the candles hence it is given the radius for each.

Since it establishes that the diameter of the thicker candle is three times of the other this can be arranged as follows:

Let for the thicker candle:

$$\begin{matrix} &\textrm{candle 1 less thicker}&\textrm{candle 2 thickest one}\\ \textrm{diameter}&d&3d\\ \textrm{diameter in terms of radius}&\textrm{2r}&\textrm{6r}\\ \textrm{radius}&r&3r \end{matrix}$$

Thus to get the speeds for each candle it is only needed to get the volume consumed:

Letting:

Thinest one: $$v_{1}$$

Thickest one: $$v_{2}$$

Therefore: Assuming for both we wait for $$t$$ time units

$$v_{1}=\frac{\pi(r)^2\cdot a}{t}$$

$$v_{2}=\frac{\pi(3r)^2\cdot b}{t}$$

Then relating these consumed heights $$a$$ for the thinnest and $$b$$ for the thickest then we get:

$$\frac{v_{1}}{v_{2}}=\frac{1a}{9b}$$

I don't know how to prove this thing but it seems that the speed of the thinnest is a ninth of the thickest one but since in the latter equation it appears as two variables. This doesn't help me much to understand.

Does it exist a way to make my logic to work out?

Anyways: If we follow what it is indicated:

Assuming the length of the candles is $$l$$ then:

$$\left(l-\frac{a}{9}\right)-\left(l-a\right)=16$$

Solving this I'm getting $$a=18$$ centimeters.

Thus for the thickest candle it would be: (For one hour elapsed) it has descended.

$$\frac{18}{9}=2\,cm$$

Assuming the thickest candle is a ninth of the speed of the thinnest candle.

While for the other candle (thinnest) it would have descended $$18$$ centimeters.

Then as it indicates that the thickest candle will have a height which is three times the size of the thinnest candle after it has elapsed $$1+\frac{1}{2}$$ hour I'm getting:

For that amount of time:

The height of the thickest candle:

$$l-2\cdot\frac{3}{2}=3\left(l-18\cdot\frac{3}{2}\right)$$

This all gets the size of the candles to be $$39$$ centimeters.

As it requests that how long will it take for the thickest candle to burn out it will be:

$$39\textrm{cm}\cdot\frac{\textrm{1 hour}}{\textrm{2 cm}}=17\frac{1}{2}\,h$$

Which is equivalent to say $$19$$ hours and $$30$$ minutes or choice $$4$$.

Which corresponds to the answer according to my book.

Again for all of this to work I assumed these two things:

The speed of the thickest candle is a ninth of the speed of the thinnest candle. And when the thickest candle descends $$b$$ this $$b=\frac{1a}{9}$$

But I got tangled with the equations as I don't know exactly how to justify these

Therefore It would really help me a lot if someone could help me with this part because this makes me confused.

• yes, it is reasonable to assume that the volume burnout should be (approx.) the same. Commented Apr 2, 2021 at 22:38
• The "after an hour" datum seems to be a red herring -- it's the only given that that contains a length in absolute units, so there's nothing to combine it with. (Or perhaps the problem has a part 2 that asks how tall the candles were initially?) Commented Apr 2, 2021 at 22:43
• @Troposphere Ironically, you also need the "after an hour" phrase to find the matching interpretation of after half an hour later. Commented Apr 3, 2021 at 0:02
• One suspect aspect to this problem: the shortest any candle could be here and be consistent with an answer is (16hrs)(16cm/hr)=250cm=2.5m That is a very long candle. I would be suspect of a question with such a ridiculous answer (even though the longest candle ever is 25m, yes I just looked this up, 2.5m for an answer here is still an unreasonable candle). Commented Apr 3, 2021 at 1:44

Alternative approach

First of all, I investigated the two interpretations of (after half an hour) and (after half an hour later) and came to the same conclusion that David Stork did: only the second interpretation yields one of the offered choices.

I then found that the first piece of information, is indeed a red herring.

Suppose that the candles burn at $$\displaystyle ~~\frac{2k ~\text{cubic cm}}{\text{hour}}.$$

In the following discussion, assume that all units of length are (cm) or (cubic cm), and that all lengths of time are in hours.

Further suppose that the initial volume of the two candles is $$\displaystyle \pi H (3r)^2, ~~\text{and} ~~\pi H (r)^2.$$

You need to compute $$\displaystyle \frac{\pi H (3r)^2}{2k}.$$

The only equation that you need to consider is that after 1.5 hours, the thicker candle now has a height 3 times the thinner candle.

After 1.5 hours, the thicker candle will have its volume changed from
$$(9 \pi H r^2) ~~\text{to}~~ (9 \pi H r^2 - 3k).\tag1$$

You calculate the new height as

$$\frac{\text{new volume}}{9 \pi r^2}.\tag2$$

So, all that you have to do, is

• similarly calculate the new height of the thinner candle
• derive the equation that represents that the thicker candle is now three times as tall as the thinner candle
• use this equation to derive the value of the fraction $$\displaystyle \frac{\pi H (3r)^2}{2k}.$$

Responding to the question/comment of Chris Steinbeck Bell:

Since you have shown significant work in your query, I am allowed to present a full solution. Normal mathSE protocol is that I dissect your work, rather than simply taking my work to completion. In this case, it is far easier for me to ignore your work, and simply complete my analysis.

I ask that you study my analysis, and then continue to pose comment/questions, as you need to, following my answer.

I assumed that both candles had the shape of a cylinder, so

$$\text{Volume} ~=~ \pi(\text{radius})^2(\text{height}).$$

I assumed that since both candles had the same quality, they both diminished at the same rate, in (cubic cm) per hour.

I arbitrarily decided to assign a fixed constant of $$2k$$ to this rate. This means that $$k$$ is a fixed unknown value.

Further since the candle is diminishing at $$2k$$ per hour, the number of hours that the candle will last is

$$\frac{\text{original volume}}{2k}.$$

This is the value that must be computed. With the assigned constants, this will be

$$\frac{\pi H (3r)^2}{2k}.\tag3$$

Using equations (1) and (2) from the first part of my answer, the new height (i.e. length) of the thick candle, after $$1.5$$ hours is

$$\frac{9 \pi H r^2 - 3k}{9\pi r^2} = H - \frac{k}{3\pi r^2}.\tag4$$

The next step is to similarly calculate the new height of the thin candle, after $$1.5$$ hours.

After 1.5 hours, the thinner candle will have its volume changed from

$$\displaystyle (\pi H r^2) ~~\text{to}~~ (\pi H r^2 - 3k).$$

Therefore, the new height of the thinner candle will be

$$\frac{\pi H r^2 - 3k}{\pi r^2} = H - \frac{3k}{\pi r^2}.\tag5$$

You are given that after $$(1.5)$$ hours, the thick candle will have a height that is $$3$$ times the height of the thin candle.

Using equations (4) and (5) above, this implies that

$$\left[H - \frac{k}{3\pi r^2}\right] ~=~ 3 \times \left[H - \frac{3k}{\pi r^2}\right].\tag6$$

Because the math is so convoluted here, I am going to introduce a new variable

$$A = \frac{k}{\pi r^2}.\tag7$$

This will greatly facilitate the algebra. Now, equation (6) above becomes:

$$\left[H - \frac{A}{3}\right] ~=~ 3 \times [H - 3A] \implies [3H - A] ~=~ 9[H - 3A] \implies$$

$$26A = 6H \implies 39A = 9H \implies (19.5) = \frac{9H}{2A}.\tag8$$

The final expression in equation (8) seems weird at first glance. However, if you use equation (7) above to compare this expression to expression (3) above, you see that

$$\frac{9H}{2A} = \frac{9H\pi r^2}{2k}$$

which is the exact fraction that you are seeking. Therefore, the thick candle will take $$(19.5)$$ hours to burn.

• I have edited my question. Can you take a look in the new questions that have arised from attempting to solve it on my own?. By the way, from where does it come the $2k$ in the denominator?. I'm sorry I forgot to include the clue that in the problem indicated that the elapsed time was half an hour later. Again, can you please take a look into the edited question? Commented Apr 3, 2021 at 1:14

This is a terrible question. You need to assume something about how diameter tells you something about burn rate. I think you are supposed to assume that the volume consumed per unit time is the same for both candles and that the candles burn evenly in length from the starting end. If you have ever watched a candle this is nonsense, but that is what you have.

Given that assumption, the thick one burns at $$\frac 19$$ the rate of the thin one, because that is the ratio of the areas. Let $$L$$ be the starting length and $$V$$ the rate of burning of the thick one per hour. The thin one then burns at $$9V$$. The length of the thick one is $$L-Vt$$ if $$t$$ is measured in hours. We are given that $$L-V\frac 32=3\left(L-9V\frac 32\right)$$ So $$L-V\frac 32=3\left(L-9V\frac 32\right)\\ L-\frac 32V=3L-\frac{81}2V\\ 39V=2L\\\frac LV=19.5$$ and the expected answer is $$19$$ hours $$30$$ minutes.

• I think a simple approach (like this one) is good for this problem. I would just add that in the last step, $39/2 = 19.5$. Commented Apr 3, 2021 at 15:15
• @DavidK: Oops. Thanks. Fixed Commented Apr 3, 2021 at 18:12

$$L_1(t) = L - m_1 t$$

$$L_2(t) = L - m_2 t$$

where by physics all we know is that $$m_1 > m_2$$.

$$L_1(1) = L_2(1) - 16$$

$$L_1(1.5) = L_2(1.5)/3$$

(The problem statement is ambiguous: You write "after half an hour." I'm assuming this means "half an hour later", i.e., 1.5 hours from the start.)

Solve these and compute $$T = L/m_2$$.

• I assume that you mean that $m$ is the slope. It would help a lot if you could indicate who are the slopes here and why? The slope is the speed. I recall that from physics. But here it is not very intuitive for me. Can you please state these in your answer so I can understand?. I've edited my question. Can you please take a look into my doubts from my attempt? i'll appreciate if you could do that. I don't know how to use the equations you presented. Should I equate both, or divide them?. What?. Please help. Commented Apr 3, 2021 at 1:26
• Substituting the formulas for $L_1$ and $L_2$ in the last two equations, we end up with two equations with three unknowns: $L,m_1,m_2.$ We can derive $m_2 = \frac23L-24$ from this, but that doesn't give us $L/m_2.$ I think we need one more equation, and the hint that "both candles consume the same amount of volume" is meant to tell us that $m_1 = 9m_2.$ And then it turns out the first equation is actually unnecessary. I wonder if the book's author realized that. Commented Apr 3, 2021 at 15:05