Chantelle had two candles, one of which was $32 \mathrm{~cm}$ longer than the other.

She lit the longer one at 3 p.m. and lit the shorter one at 7 p.m.

At 9 p.m., they were both the same length.

The longer one was completely burned out at 10 p.m. and the shorter one was completely burned at midnight.

The two candles burned at different, but constant, rates.

What was the sum of the original lengths of the two candles?
(A) $42 \mathrm{~cm}$
(B) $48 \mathrm{~cm}$
(C) $60 \mathrm{~cm}$
(D) $80 \mathrm{~cm}$
(E) $52 \mathrm{~cm}$
Since the length of the candles was equal at 9 p.m., the longer one burned out at 10 p.m., and the shorter one burned out at midnight, then it took 1 hour for the longer candle and 3 hours for the shorter candle to burn this equal length.

Therefore, the longer candle burned 3 times as quickly as the shorter candle.

Suppose that the shorter candle burned $x \mathrm{~cm}$ per hour.

Then the longer candle burned $3 x \mathrm{~cm}$ per hour.

From its lighting at 3 p.m. to 9 p.m., the longer candles burned for 6 hours, so burned $6 \times 3 x$ or $18 \mathrm{x} \mathrm{cm}$.

From its lighting at 7 p.m. to 9 p.m., the shorter candle burned for 2 hours, so burns $2 \times x=2 x$ $\mathrm{cm}$.

But, up to 9 p.m., the longer candle burned $32 \mathrm{~cm}$ more than the shorter candle, since it began $32 \mathrm{~cm}$ longer.

Therefore, $18 x-2 x=32$ or $16 x=32$ or $x=2$.

In summary, the shorter candle burned for 5 hours at $2 \mathrm{~cm}$ per hour, so its initial length was $10 \mathrm{~cm}$.

Also, the longer candle burned for 7 hours at $6 \mathrm{~cm}$ per hour, so its initial length was $42 \mathrm{~cm}$. Thus, the sum of the original lengths is $42+10=52 \mathrm{~cm}$.

ANSWER: (E)