An Algebraic Challenge -- Part 5

Fortunately, I do know how to use $\LaTeX$ .

Let's set some variables first, $L$ for the larger candle and $S$ for the smaller candle.

We can immediately set our first equation for the first sentence: $L=S+3$ .

We know that $L$ burned out in $6$ hours and $S$ burned out in $4$ . After $4$ hours of burning for $L$ and $\frac{5}{2}$ hours of burning for $S$ , the candles' lengths are equal. This gives us our second equation:

$\frac{4-\frac{5}{2}}{4}S = \frac{6-4}{6}L \Rightarrow \frac{\frac{3}{2}}{4}S = \frac{2}{6}L \Rightarrow \frac{3}{8}S = \frac{1}{3}L$

Substituting the first equation into the second we can find $S$ :

$\frac{3}{8}S = \frac{1}{3}(S+3) \Rightarrow \frac{3}{8}S = \frac{1}{3}S+1 \Rightarrow \frac{1}{24}S = 1 \Rightarrow S = 24$

Using our first equation we can find that $L = 27$ .

So the final answer is $L+S = 27+24 = \boxed{51}$ .

Let A be the shorter of length x. It's rate of burning is x(11 - 7) = x/4. Let B be the larger of length x + 3. It,s rate of burning is (x + 3)/6. In 4 hours, Length of B is (x + 3) -4(x + 3)/6. In 5/2 hours, length of A is x - 5x/8. They then have the same length, so (x + 3) - (2/3)(x + 3) = x - 5x/8, or 9x = 8x + 24. So x = 24, x + 3 = 27, and A + B = 51.

(x+3)- $\frac{(x+3)t_1}{6}$ = x - $\frac{x(t_2)}{4}$

since t1 = 4 and t2 = 2.5. we can then substitute the values in and solve for x

let A=x+3 and B = x

A = 27 and B = 24

therefore, A+B = 51

We define four variables in this problem. Let A be the length of the longer candle. Let B be the length of the shorter candle. Let x be the rate at which the longer candle burns out, in centimeters per hour, And let y be the rate at which the shorter candle burns out, in centimeters per hour.

NOTE: It is implicitly assumed in the problem that both candles burn at uniform rates.

Candle A is longer than candle B by 3 cm. This can be expressed as A = B + 3 (Eqn. 1).

Let's take a look at the situation at 9:30 pm: It has been four hours since the longer candle has been lit, so the remaining length of the longer candle at this time can be expressed as (A-4x). For the shorter candle, 2 hours and 30 minutes have elapsed since it was lighted, so the remaining length of the shorter candle is: B - 2.5y. The remaining lengths are equal during this time, so A - 4x = B - 2.5y (Eqn. 2).

At 11:00 pm, it took the shorter candle 4 hours to completely burn out. This can be expressed as: B = 4y (Eqn. 3).

At 11:30 pm, it took the longer candle 6 hours to completely burn out. This can be expressed as: A = 6x (Eqn. 4).

We have four equations in four unknowns (A,B,x,y). However, we only need to know the values of A & B. In eqn. 3, y = B/4. Likewise, in equation 4, x = A/6.

By substituting the manipulated equations 3 & 4 and simplifying, we get a modified form of equation 2 as A = 9B/8 (Eqn. 5).

Substitute eqn. 5 into eqn. 1 and simplify. We obtain A = 27 cm. and B = 24 cm. Finally, A+B = 27 cm + 24 cm = 51 cm (Final Answer).

P.S. Sorry, I don't know how to use LaTeX. :((