Candle problem about burn rates

Question

Two candles, one of which is two centimeters longer than the other, are lit. The longer and thinner one is lit at noon and the shorter but fatter one is lit 15 minutes later. Each candle burns at a steady rate, and by 4 PM both are the same length. The thinner one burns out at 8 PM and the fatter one burns out one hour later. Let x and y be the lengths of the two candles. Find x + y.

(A) 15 (B) 30 (C) 40 (D) 45 (E) 50

I tried setting up equations using the length, width, and burn rate of the candles, but it seems too complicated. Is there a better way?

score 3 · Accepted Answer · answered Jan 11 '20 at 20:09

3

At 4pm the candles are the same length:

Lets say $z =$ height at 4:00

At 8 pm the faster burning candle burns out at 9 pm the slower burning candle burns out.

The fast candle burns $\frac {z}{4}$ inches per hour. The slow candle burns $\frac {z}{5}$ inches per hour.

The faster burning candle burns for 8 hours total, the slow candle burns for 8:45 = $\frac {35}{4}$ hours

$x = 8\frac {z}{4} = 2z\\ y = \frac {35}{4} \frac {z}{5} = \frac {7}{4} z\\ x - y = \frac 14 z = 2\\ z = 8\\ x = 16\\ y = 14\\ x + y = 16 + 14 = 30$

answered Jan 11 '20 at 20:09

user317176

11,017

It would be good to define $x$ as the length of the longer candle and $y$ as the length of the shorter. – Ross Millikan Jan 11 '20 at 20:17
@RossMillikan I thought about that, but then didn't as they seemed to be defined in the original problem. – user317176 Jan 13 '20 at 22:25

score 1 · Answer 2 · answered Jan 11 '20 at 19:53

I got $30$ as the answer.

Solution. Let the candles be labelled as $A$ and $B$ the height of the shorter candle be $x$. The time for which each candle burns is proportional to its height, so $t\propto h$. For candle $A$, $$\int_0^4\mathrm{d}t_1=k_1\int_{x+2}^h\mathrm{d}h_1$$ $$\implies k_1 = \frac{4}{h-x-2}.$$ For candle $B$, $$\int_{0.25}^4\mathrm{d}t_2=k_2\int_x^h\mathrm{d}h_2$$ $$\implies k_2 = \frac{3.75}{h-x}.$$ Again, we have for candle $A$ $$\int_0^8\mathrm{d}t_1=k_1\int_{x+2}^0\mathrm{d}h_1$$ $$\implies 8 = -k_1(x+2)$$ and for candle $B$ $$\int_{0.25}^9\mathrm{d}t_2 = k_2\int_x^0\mathrm{d}h_2$$ $$\implies 8.75 = k_2(-x).$$ Substituting for $k_2$ in the above equation we get $$8.75 = \frac{3.75}{h-x}(-x)\implies h = \frac{5x}{8.75}$$ and we put this value of $h$ in the first equation to get $k_1 = \frac{4}{\frac{5x}{8.75}-x-2}$. Put this value of $k_1$ in the third equation to solve for $x$: $$8 = -\left(\frac{4}{\frac{5x}{8.75}-x-2}\right)(x+2)$$ from which we readily obtain $x=14$. Thus $y=16$ and $x+y=30$.

Candle problem about burn rates

2 Answers2