Two-candles-of-the-same-height-are-lighted-together-First-one-gets-burnt-up-completely-in-3-hours-while-the-second-in-4-hours-At-what-point-of-time-the-length-of-second-candle-will-be-double-the-le

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Question Number 17830 by Tinkutara last updated on 11/Jul/17

$$\mathrm{Two}\mathrm{candles}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{height}\mathrm{are}$$$$\mathrm{lighted}\mathrm{together}.\mathrm{First}\mathrm{one}\mathrm{gets}\mathrm{burnt}\mathrm{up}$$$$\mathrm{completely}\mathrm{in}3\mathrm{hours}\mathrm{while}\mathrm{the}\mathrm{second}$$$$\mathrm{in}4\mathrm{hours}.\mathrm{At}\mathrm{what}\mathrm{point}\mathrm{of}\mathrm{time},\mathrm{the}$$$$\mathrm{length}\mathrm{of}\mathrm{second}\mathrm{candle}\mathrm{will}\mathrm{be}\mathrm{double}$$$$\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{candle}?$$

Answered by mrW1 last updated on 11/Jul/17

$${\mathrm{h}}_1=\frac{3-\mathrm{t}}{3}=1-\frac{\mathrm{t}}{3}$$$${\mathrm{h}}_2=\frac{4-\mathrm{t}}{4}=1-\frac{\mathrm{t}}{4}$$$${\mathrm{h}}_2={\mathrm{h}}_1$$$$\Rightarrow 1-\frac{\mathrm{t}}{4}=2(1-\frac{\mathrm{t}}{3})$$$$(\frac{2}{3}-\frac{1}{4})\mathrm{t}=1$$$$\Rightarrow \mathrm{t}=\frac{12}{5}=2.4\mathrm{h}=2\mathrm{h}24\mathrm{m}$$

Commented by Tinkutara last updated on 11/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by ajfour last updated on 11/Jul/17

Commented by ajfour last updated on 11/Jul/17

$$\mathrm{height}\mathrm{of}\mathrm{first}\mathrm{candle}\mathrm{as}\mathrm{function}$$$$\mathrm{of}\mathrm{time}\mathrm{is}\frac{{\mathrm{h}}_1}{{\mathrm{h}}_0}+\frac{\mathrm{t}}{3}=1$$$$\mathrm{of}\mathrm{second}\mathrm{candle}:\frac{{\mathrm{h}}_2}{{\mathrm{h}}_0}+\frac{\mathrm{t}}{4}=1$$$$\mathrm{when}{\mathrm{h}}_2={2\mathrm{h}}_1,\frac{{\mathrm{h}}_2}{{\mathrm{h}}_0}=2\left(\frac{{\mathrm{h}}_1}{{\mathrm{h}}_0}\right)$$$$\Rightarrow 1-\frac{\mathrm{t}}{4}=2(1-\frac{\mathrm{t}}{3})$$$$\mathrm{or}\mathrm{t}(\frac{2}{3}-\frac{1}{4})=1\Rightarrow \mathbf{t}=\frac{12}{5}\mathrm{h}$$$$=2.4\mathrm{h}=2\mathrm{h}24\min .$$

Commented by Tinkutara last updated on 11/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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