$2 \cdot 6$ Two candles of equal height are lit at the same time, the first one burns out in 4 hours, the second one in 3 hours, assuming both candles burn at a constant rate, when the height of the first candle is twice that of the second, the time since they were lit is
(A) $\frac{3}{4}$ hours.
(B) $1 \frac{1}{2}$ hours.
(C) 2 hours.
(D) $2 \frac{2}{5}$ hours.
(E) $2 \frac{1}{2}$ hours.
(7th American High School Mathematics Examination, 1956)
［Solution］Let the candles burn for $x$ hours, after which the height ratio of the two candles is $2: 1$, then
$$
1-\frac{x}{4}=2\left(1-\frac{x}{3}\right) \text {, }
$$

Solving for $x$ gives $x=2 \frac{2}{5}$ (hours).
Therefore, the answer is $(D)$.