# You have two candles of equal length. Candle A takes six hours to burn, and candle B takes three hours to burn. If you light them at the same time, how long will it be before candle A is twice as long as Candle B? Both candles burn st a constant rate.

Jul 13, 2018

Two hours

#### Explanation:

Start by using letters to represent the unknown quantities,

Let burn time = $t$
Let initial length $= L$
Let length of candle A = $x$ and length of candle B = $y$

Writing equations for what we know about them:

What we are told:
At the start (when $t = 0$), $x = y = L$

At $t = 6$, $x = 0$
so burn rate of candle A
= $L$ per 6 hours $= \frac{L}{6 h o u r s} = \frac{L}{6} p e r h o u r$

At $t = 3$, $y = 0$
so burn rate of candle B = $\frac{L}{3} p e r h o u r$

Write eqns for $x$ and $y$ using what we know.
e.g. $x = L - \text{burn rate} \cdot t$

$x = L - \frac{L}{6} \cdot t$ .............(1)
Check that at $t = 0$, $x = L$ and at $t = 6$, $x = 0$. Yes we do!

$y = L - \frac{L}{3} \cdot t$ ..............(2)

Think about what we are asked for: Value of $t$ when $x = 2 y$

Using eqns (1) and (2) above if $x = 2 y$ then

$L - \frac{L}{6} \cdot t = 2 \left(L - \frac{L}{3} \cdot t\right)$

expand and simplify this

$L - \frac{L}{6} \cdot t = 2 L - 2 \frac{L}{3} \cdot t$

$\cancel{L} - \cancel{L} - \frac{L}{6} \cdot t + 2 \frac{L}{3} \cdot t = 2 L - L - \cancel{2 \frac{L}{3} \cdot t} + \cancel{2 \frac{L}{3} \cdot t}$

$- \frac{L}{6} \cdot t + 2 \frac{L}{3} \cdot t = 2 L - L$ ....... but $\frac{L}{3} = 2 \frac{L}{6}$

$- \frac{L}{6} \cdot t + 2 \left(2 \frac{L}{6}\right) \cdot t = L$

$- \frac{L}{6} \cdot t + 4 \frac{L}{6} \cdot t = L$

$\frac{3 L}{6} \cdot t = L$

$\frac{\cancel{3 L}}{\cancel{6}} \cdot t \cdot \frac{\cancel{6}}{\cancel{3 L}} = \cancel{L} \cdot \frac{6}{3 \cancel{L}}$

$t = \frac{6}{3} = 2$