Question

A candle burns in 6 hrs. Another candle of the same height and width burns in 8 hrs. After how many
hours the height of the first candle will be half of the second Candle?

Answer 1

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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) 

❤hope it helps you ❤

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