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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
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 length of the first candle?
$$\mathrm{Two}\:\mathrm{candles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{height}\:\mathrm{are} \\ $$$$\mathrm{lighted}\:\mathrm{together}.\:\mathrm{First}\:\mathrm{one}\:\mathrm{gets}\:\mathrm{burnt}\:\mathrm{up} \\ $$$$\mathrm{completely}\:\mathrm{in}\:\mathrm{3}\:\mathrm{hours}\:\mathrm{while}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{in}\:\mathrm{4}\:\mathrm{hours}.\:\mathrm{At}\:\mathrm{what}\:\mathrm{point}\:\mathrm{of}\:\mathrm{time},\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{second}\:\mathrm{candle}\:\mathrm{will}\:\mathrm{be}\:\mathrm{double} \\ $$$$\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{candle}? \\ $$
Answered by mrW1 last updated on 11/Jul/17
h_1 =((3−t)/3)=1−(t/3) h_2 =((4−t)/4)=1−(t/4) h_2 =h_1 ⇒1−(t/4)=2(1−(t/3)) ((2/3)−(1/4))t=1 ⇒t=((12)/5)=2.4h=2h 24m
$$\mathrm{h}_{\mathrm{1}} =\frac{\mathrm{3}−\mathrm{t}}{\mathrm{3}}=\mathrm{1}−\frac{\mathrm{t}}{\mathrm{3}} \\ $$$$\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{4}−\mathrm{t}}{\mathrm{4}}=\mathrm{1}−\frac{\mathrm{t}}{\mathrm{4}} \\ $$$$\mathrm{h}_{\mathrm{2}} =\mathrm{h}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{t}}{\mathrm{4}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{3}}\right) \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{t}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{t}=\frac{\mathrm{12}}{\mathrm{5}}=\mathrm{2}.\mathrm{4h}=\mathrm{2h}\:\mathrm{24m} \\ $$
Commented by Tinkutara last updated on 11/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 11/Jul/17
Commented by ajfour last updated on 11/Jul/17
height of first candle as function of time is (h_1 /h_0 )+(t/3)=1 of second candle: (h_2 /h_0 )+(t/4)=1 when h_2 =2h_1 , (h_2 /h_0 ) = 2((h_1 /h_0 )) ⇒ 1−(t/4)=2(1−(t/3)) or t((2/3)−(1/4))=1 ⇒ t=((12)/5) h =2.4 h = 2h 24min .
$$\mathrm{height}\:\mathrm{of}\:\mathrm{first}\:\mathrm{candle}\:\mathrm{as}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{time}\:\mathrm{is}\:\:\:\:\:\frac{\mathrm{h}_{\mathrm{1}} }{\mathrm{h}_{\mathrm{0}} }+\frac{\mathrm{t}}{\mathrm{3}}=\mathrm{1} \\ $$$$\mathrm{of}\:\mathrm{second}\:\mathrm{candle}:\:\:\:\frac{\mathrm{h}_{\mathrm{2}} }{\mathrm{h}_{\mathrm{0}} }+\frac{\mathrm{t}}{\mathrm{4}}=\mathrm{1} \\ $$$$\mathrm{when}\:\mathrm{h}_{\mathrm{2}} =\mathrm{2h}_{\mathrm{1}} \:,\:\:\:\frac{\mathrm{h}_{\mathrm{2}} }{\mathrm{h}_{\mathrm{0}} }\:=\:\mathrm{2}\left(\frac{\mathrm{h}_{\mathrm{1}} }{\mathrm{h}_{\mathrm{0}} }\right) \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{1}−\frac{\mathrm{t}}{\mathrm{4}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{3}}\right) \\ $$$$\:\mathrm{or}\:\:\:\:\:\:\:\mathrm{t}\left(\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{1}\:\:\Rightarrow\:\:\boldsymbol{\mathrm{t}}=\frac{\mathrm{12}}{\mathrm{5}}\:\mathrm{h} \\ $$$$\:\:\:\:=\mathrm{2}.\mathrm{4}\:\mathrm{h}\:=\:\mathrm{2h}\:\mathrm{24min}\:. \\ $$
Commented by Tinkutara last updated on 11/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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