S. had two candles, one of which was 36cm longer than the other. He lit the longer one at 4 p.m. and lit the shorter one at 8 p.m.  At 10 p.m., they were both the same length. The longer one was completely burned out at 11 p.m. and the shorter one completely burned at 1 a.m. The two candles burned at different, but constant, rates. What was the sum of the original lengths of the two candles?

Let's solve this problem step by step.

### Step 1: Defining Variables
Let:
- \( L_1 \) be the length of the longer candle (in cm),
- \( L_2 \) be the length of the shorter candle (in cm),
- \( r_1 \) be the burn rate of the longer candle (in cm per hour),
- \( r_2 \) be the burn rate of the shorter candle (in cm per hour).

From the problem, we know that:
- \( L_1 = L_2 + 36 \) (the longer candle is 36 cm longer than the shorter candle).
- The longer candle was lit at 4 p.m. and completely burned out at 11 p.m., so it burned for 7 hours. Thus:
  \[
  L_1 = 7r_1
  \]
- The shorter candle was lit at 8 p.m. and completely burned out at 1 a.m., so it burned for 5 hours. Thus:
  \[
  L_2 = 5r_2
  \]

### Step 2: Setting up the Lengths at 10 p.m.
At 10 p.m., both candles had the same remaining length. Let’s calculate the lengths of the candles at that time:
- The longer candle had been burning for 6 hours (from 4 p.m. to 10 p.m.), so its remaining length at 10 p.m. was:
  \[
  L_1 - 6r_1 = r_1 \quad \text{(since it burns out after one more hour at 11 p.m.)}
  \]
- The shorter candle had been burning for 2 hours (from 8 p.m. to 10 p.m.), so its remaining length at 10 p.m. was:
  \[
  L_2 - 2r_2
  \]
Since the lengths are the same at 10 p.m., we have:
\[
r_1 = L_2 - 2r_2
\]

### Step 3: Solving the System of Equations
We now have the following system of equations:
1. \( L_1 = L_2 + 36 \)
2. \( L_1 = 7r_1 \)
3. \( L_2 = 5r_2 \)
4. \( r_1 = L_2 - 2r_2 \)

Substitute \( L_1 = L_2 + 36 \) and \( L_1 = 7r_1 \) into the system. This gives:
\[
L_2 + 36 = 7r_1
\]
Using \( r_1 = L_2 - 2r_2 \), substitute into the above equation:
\[
L_2 + 36 = 7(L_2 - 2r_2)
\]
Expanding and simplifying:
\[
L_2 + 36 = 7L_2 - 14r_2
\]
\[
36 = 6L_2 - 14r_2
\]
Now substitute \( L_2 = 5r_2 \) into this equation:
\[
36 = 6(5r_2) - 14r_2
\]
\[
36 = 30r_2 - 14r_2
\]
\[
36 = 16r_2
\]
Thus:
\[
r_2 = \frac{36}{16} = 2.25 \, \text{cm/hour}
\]

### Step 4: Finding the Lengths of the Candles
Now that we have \( r_2 \), we can find \( L_2 \) and \( L_1 \):
\[
L_2 = 5r_2 = 5 \times 2.25 = 11.25 \, \text{cm}
\]
\[
L_1 = L_2 + 36 = 11.25 + 36 = 47.25 \, \text{cm}
\]

### Step 5: Finding the Sum of the Lengths
The sum of the original lengths of the two candles is:
\[
L_1 + L_2 = 47.25 + 11.25 = 58.5 \, \text{cm}
\]

Thus, the sum of the original lengths of the two candles is **58.5 cm**.

Would you like a more detailed explanation, or do you have any questions?

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Here are 5 related questions you could explore:
1. What if the longer candle burned out at 12 a.m.? How would that change the solution?
2. How does the burn rate impact the total burn time if both candles had the same initial length?
3. What if both candles were lit at the same time? How long would they take to burn out?
4. How would the problem change if the candles burned at variable rates over time?
5. What would happen if a third candle, burning slower than the others, was introduced?

**Tip**: Always define your variables clearly when solving a word problem involving multiple unknowns.

S. had two candles, one of which was 36cm longer than the other. He lit the longer one at 4 p.m. and lit the shorter one at 8 p.m. At 10 p.m., they were both the same length. The longer one was completely burned out at 11 p.m. and the shorter one completely burned at 1 a.m. The two candles burned at different, but constant, rates. What was the sum of the original lengths of the two candles?

Learn how to solve a candle length problem where two candles burn at constant but different rates. The longer candle is 36 cm longer and the problem involves setting up algebraic equations to find the sum of the original lengths of the candles. Suitable for high school students.

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