2014/15 AHSMC Part 1 Problem 13

Count the number of minutes in which each candle is burning and add them together. The ten minutes of simultaneous burning contributes $30$ minutes to this, and there are $20$ minutes of individual burning. The total must be $30+40+50=120$ minutes, so the time when exactly two candles are burning must contribute the remaining $120-30-20=70$ minutes. Thus the answer must be $70/2=35$.

Alternative Solution: Let the first two candles burn simultaneously for $a$ minutes, the first and the third for $b$ minutes and the last two for $c$ minutes. Then the first candle burns alone for $20 - a - b$ minutes, the second for $30 - a - c$ minutes and the third for $40 - b - c$ minutes. Hence $90 - 2(a+b+c) = 20$ so that $a + b + c = 35$. This may be realized if $a = 0$, $b = 20$ and $c = 15$. The three candles are lit at minutes $30$, $0$ and $15$ respectively. The answer is (A).