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

9700 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 #