Question 884541
Let x be the length of two candles. s1 and s2 rate of burning of A & B candles respectively.
For A, x= 4s1 => s1 = x/4
For B s2 = x/3
Let t be the time at which candles A is twice as much as candle B height
At time t, using the formula distance = speed * time
candle A:  {{{t=(x-y)/s1}}}  -> eq A
For candle B {{{t=(x-(y/2))/s2}}}
Equating above two we get
{{{(x-y)/s1=(2x-y)/2s2}}}
substitute for s1, s2
{{{(x-y)/(x/4) = (2*x-y)/(2*(x/3))}}}
=>{{{x=(5/2)*y}}}
Substituting for x,s1 intermns of y in eq A
{{{t = (x-y)/s1}}}
We get t = 2.4 hrs