Question

Two candles of same height but different thickness are lighted. The first burns of in 6 hrs and the second in 8 hrs. How long after lighting the both will the first candle be half of the second ?

Answer 1

Let each candle be of length L.

Now candle A (say) burns off in 6 hours.
.·. in one hour it burns L/6 and in t hours the length burned will be (L/6)t
.·. in t hours we will have only portion L - (L/6)t remaining.

Similarly candle B (say) burns off in 8 hours.
.·. in one hour it burns L/8 and in t hours the length burned will be (L/8)t
.·. in t hours we will have only portion L - (L/8)t remaining.

Now by given condition
2(L - (L/6)t) = L - (L/8)t
(·.· candle A burns faster than candle B and after time t from start remaining length of B is twice remaining length of A)
2L - (L/3)t = L - (L/8)t
2L - L = t(L/3 - L/8)
L = t((8-3)/24)L
L = t(5/24)L
t = L * 24/(5L)
t = 24/5
t = 4.8 hours
(In the question we have assumed that both the candles burn uniformly)