Assuming the candle continues to burn at this rate, how many hours after it is lit will the candle's height be 6.0 centimeters?

Question

A candle burns at a constant rate. Its height is 18.0 centimeters 3 hours after it is lit and 12.0 centimeters 7 hours after it is lit.

Accepted Answer

B. 11. The burn rate is (18.0 − 12.0)/(7 − 3) = 6.0/4 = 1.5 centimeters per hour, so the height t hours after lighting is h(t) = 18.0 − 1.5(t − 3) = 22.5 − 1.5t. Setting 22.5 − 1.5t = 6.0 gives 1.5t = 16.5, so t = 11 hours. Check: 22.5 − 1.5(11) = 6.0. Choice A comes from computing (18.0 − 6.0)/1.5 = 8 and forgetting that this counts hours after the 3-hour mark, not after lighting (the offset must be added: 8 + 3 = 11). Choice C is 22.5/1.5, the time at which the candle would burn all the way down to height 0 — the wrong target height. Choice D comes from inverting the rate as 4/6 ≈ 0.67 cm per hour, giving an initial height of 20 and t = (20 − 6)/0.67 ≈ 21.

Answer

A. 8 — You found (18.0 − 6.0)/1.5 = 8, but that's 8 hours after the 3-hour mark — add the offset to get 11 hours after lighting.

Answer

C. 15 — You computed 22.5/1.5, the time the candle reaches height 0; you want height 6.0, so solve 22.5 − 1.5t = 6.0.

Answer

D. 21 — You inverted the rate as 4/6 ≈ 0.67 cm per hour instead of 6.0/4 = 1.5; the rate is centimeters per hour, so height change goes on top.

Algebra practice question

Answer: B. 11

Why the other answers are wrong

More Algebra questions