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.

Question

10021 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T04:02:05

Two hours

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 $= L/(6hours) = L/6 per hour$

At $t=3$ , $y = 0$
so burn rate of candle B = $L/3 per hour$

Write eqns for $x$ and $y$ using what we know.
e.g. $x = L - "burn rate" * t$

$x = L - L/6*t$ .............(1)
Check that at $t = 0$ , $x=L$ and at $t = 6$ , $x=0$ . Yes we do!

$y = L - L/3*t$ ..............(2)

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

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

$L - L/6*t =2(L - L/3*t)$

expand and simplify this

$L - L/6*t =2L - 2L/3*t$

$cancelL -cancel L-L/6*t+2L/3*t =2L - L -cancel(2L/3*t)+cancel(2L/3*t)$

$-L/6*t+2L/3*t =2L - L$ ....... but $L/3 = 2L/6$

$-L/6*t+2(2L/6)*t =L$

$-L/6*t+4L/6*t =L$

$(3L)/6*t =L$

$cancel(3L)/cancel6*t *cancel6/cancel(3L)=cancelL*6/(3cancelL)$

$t = 6/3 = 2$