Question 87074
Each candle has a rate of burn, and what is being assumed in
the problem (and of course is debateable) is that the rate of
burn for each is constant and may be different for each candle.
The rate of burn is how far it burns down in a given time period.
First look at when each is lit and when it burns out.
Call the 1st one candle A and the 2nd one candle B.
The rate of burn for A is {{{r[A]}}}
The rate of burn for B is {{{r[B]}}}
Call {{{h}}} the starting height of A in cm
The {{{h-1}}} is the starting height of B in cm
From 5:30 to 11:30 for A is 6 hours
From 7:00 to 11:00 for B is 4 hours
{{{r[A] = h/6}}}
{{{r[B] = (h-1)/4}}}
Now look at the information relating to the candles being  the same
height at 9:30. Candle A will burn down the same amount as B plus an
extra cm.
So, in words, (A's rate)(A's time) = (B's rate)(B's time) + 1 cm
A's time is 4 hrs, and B's time is 2.5 hours
{{{4(h/6) = 2.5((h-1)/4) + 1}}}
multiple both sides by LCD which is 12
{{{ 8h = 3*2.5(h-1) + 12}}}
{{{8h = (15/2)*(h-1) + 12}}}
multiply both sides by 2
{{{16h = 15h - 15 + 24}}}
{{{h = 9}}}
So, the height of the longest candle is 9 cm
Check this by plugging back into equations
What are the rates of A and B?
{{{r[A] = h/6}}}
{{{r[A] = 9/6}}}
{{{r[A] = 3/2}}}
{{{r[B] = (h-1)/4}}}
{{{r[B] = 8/4}}}
{{{r[B] = 2}}}
{{{4(h/6) = 2.5((h-1)/4) + 1}}}
{{{4(3/2) = 2.5*2 + 1}}}
{{{6 = 5 + 1}}}
{{{6 = 6}}}
How about the rate*time for each to burn all the way down?
{{{6(h/6) = 4((h-1)/4) + 1}}}
h = 9
{{{9 = 4*2 + 1}}}
{{{9 = 8 + 1}}}
{{{9 = 9}}}
OK