Rates and ratios problem

Let the length of the shorter candle be $x$ cm, then the length of the longer candle is $x+32$ cm. Let the rate of burning of the longer and shorter candles be $a$ and $b$ cm/h respectively. Then at 9 PM:

$\begin{aligned} x+32 - (\text{9:00}-\text{3:00})a & = x - (\text{9:00}-\text{7:00})b \\ x+32-6a & = x-2b \\ \implies b & = 3a-16 & ...(1) \end{aligned}$

And at 10 PM and 12 PM:

$\begin{cases} x + 32 = (\text{10:00}-\text{3:00})a & \implies x = 7a - 32 &...(2) \\ x = (\text{12:00}-\text{7:00})b & \implies x = 5b &...(3) \end{cases}$

Substituting $(1): \ b = 3a-16$ in $(2): \ x = 5(3a-16) = 15a - 80\quad ...(4)$

$\begin{aligned} (4)-(1): \quad 8a - 48 & = 0 \\ \implies a & = 6 \end{aligned}$

$\begin{aligned} (2): \quad x = 7(6)-32 = 10 \end{aligned}$

Therefore, the sum of original lengths of the two candle: $x+32+x = 2x+32 = \boxed{52}$

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 $x$ cm per hour. Then the longer candle burned $3x$ cm per hour. From its lighting at 3 p.m. to 9 p.m., the longer candles burned for 6 hours, so burned $6 × 3x$ or $18x$ cm.

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

But, up to 9 p.m., the longer candle burned 32 cm more than the shorter candle, since it began 32 cm longer. Therefore, $18x - 2x = 32$ or $16x = 32$ or $x = 2$ . I n summary, the shorter candle burned for 5 hours at 2 cm per hour, so its initial length was 10 cm.

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

Source of question: http://www.cemc.uwaterloo.ca/contests/past_contests.html