Math Problem Statement

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?

Solution

Let's solve this problem step by step.

Step 1: Defining Variables

Let:

  • L1L_1 be the length of the longer candle (in cm),
  • L2L_2 be the length of the shorter candle (in cm),
  • r1r_1 be the burn rate of the longer candle (in cm per hour),
  • r2r_2 be the burn rate of the shorter candle (in cm per hour).

From the problem, we know that:

  • L1=L2+36L_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: L1=7r1L_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: L2=5r2L_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: L16r1=r1(since it burns out after one more hour at 11 p.m.)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: L22r2L_2 - 2r_2 Since the lengths are the same at 10 p.m., we have: r1=L22r2r_1 = L_2 - 2r_2

Step 3: Solving the System of Equations

We now have the following system of equations:

  1. L1=L2+36L_1 = L_2 + 36
  2. L1=7r1L_1 = 7r_1
  3. L2=5r2L_2 = 5r_2
  4. r1=L22r2r_1 = L_2 - 2r_2

Substitute L1=L2+36L_1 = L_2 + 36 and L1=7r1L_1 = 7r_1 into the system. This gives: L2+36=7r1L_2 + 36 = 7r_1 Using r1=L22r2r_1 = L_2 - 2r_2, substitute into the above equation: L2+36=7(L22r2)L_2 + 36 = 7(L_2 - 2r_2) Expanding and simplifying: L2+36=7L214r2L_2 + 36 = 7L_2 - 14r_2 36=6L214r236 = 6L_2 - 14r_2 Now substitute L2=5r2L_2 = 5r_2 into this equation: 36=6(5r2)14r236 = 6(5r_2) - 14r_2 36=30r214r236 = 30r_2 - 14r_2 36=16r236 = 16r_2 Thus: r2=3616=2.25cm/hourr_2 = \frac{36}{16} = 2.25 \, \text{cm/hour}

Step 4: Finding the Lengths of the Candles

Now that we have r2r_2, we can find L2L_2 and L1L_1: L2=5r2=5×2.25=11.25cmL_2 = 5r_2 = 5 \times 2.25 = 11.25 \, \text{cm} L1=L2+36=11.25+36=47.25cmL_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: L1+L2=47.25+11.25=58.5cmL_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?


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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear equations
Rate problems

Formulas

L1 = L2 + 36
L1 = 7r1
L2 = 5r2
r1 = L2 - 2r2

Theorems

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Suitable Grade Level

Grades 9-12