Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in 3 hours while the other burns out in 4 hours. At what time (in PM) should both candles be lit so that at 9 PM one candle stub is twice the length of the other?

• 6:36 pm....let rate of burning of 1st and 2 nd candles be x and y respectively...den length of both candles is 3x and 4y...and as dese are equal we have 3x = 4y.....now let after t hour after burning the event described in the question takes place so after t hours stub of 1st candle=tx and length of other candle after t time= 4y-ty......and according to the question tx= 2*(4y-ty) =>tx= 8y-2ty....=>t(x+2y)=8y=>t=8y/(x+2y)...putting values from above equation...we get t=2.4 hour or 2 hours and 24 minutes so we lit both candles at 6:36 pm
• 11 years agoHelpfull: Yes(2) No(1)
• suppose length of each candle is 12 units
  and first candle burns 4 units in 1 hr and other candle burns 3 units in 1 hr, and suppose one candle stub is twice the length of the other after x hrs,then
  (12-3x)/(12-4x)=2
  x= 12/5 hrs = 2 hrs 24 mins
  
  so both candles should be lit at 6:36 PM.
• 11 years agoHelpfull: Yes(1) No(1)
• (x+1-1)^45+52
  binomial implies only -1^45 is the remainder from the first term that is -1
  so net remainder = -1+52=51
• 11 years agoHelpfull: Yes(0) No(2)