Word problems are used to see whether you understand the algebraic concepts. Anyone can solve an equation. But if you encounter a problem and can then set up the equations to solve, that shows a deeper understanding.
So, guess what? In real life, no one hands you an equation; instead, you get a problem to solve.
So, here goes:
Let the candles start at height h. The units are not given, so I'll assume inches just to have a reference.
If candle x takes 6 hours to burn down, it burns at h/6 in/hr. So, at time t minutes, its height is
x = h - h/360 * t
The other candle, burning down at h/9 in/hr,
y = h - h/540 * t
When is the slow burner twice as tall as the fast burner?
h - h/9*t = 2(h - h/6*t)
you can divide all the h's out, giving
1 - t/9 = 2(1 - t/6)
1 - t/9 = 2 - t/3
2t/9 = 1
t = 9/2
So, after 4.5 hours,
x = h/2 (burned halfway)
y = t/4 (half as tall as x)
6 hour have to change to minute. i hate words problem.
so here the problem:
Two candles are the same length. One candle takes 6 hours to burn all the way down while the other candle takes 9 hours. If the two candles are lit at the same time, how long will it take for the two candles to burn so that one candle is twice the length of the other?
2 answers
Oops cut/paste error. Left some units in minutes, but the final equations are the solution.