The length of candle B is 3 times the length of candle A. The speed of burning of Candle B is 4 times the speed of burning of Candle A. A party begins with the lighting of these two candles. The party ends when the heights of these candles become equal. If the numerical value of length (in metre) of the burnt away portion of Candle A and the speed (in metre per hour) of its burning are same, how long, in hours, did the party continue?
- A1/2
- B1
- C1¼
- D1½
Explanation
Let candle A's length be L and burning speed v; then B has length 3L and speed 4v. The party ends when heights are equal: L - vt = 3L - 4vt, giving L = 1.5vt. The burnt portion of A (vt metres) numerically equals its speed (v m/hr), so vt = v and t = 1 hour. The heights then both equal 0.5v, confirming the solution.