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Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

 

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

 

 

 

 

cool cool cool

CPhill  Dec 10, 2015

Best Answer 

 #3
avatar+18714 
+15

Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

 

H = height to begin.

h = height during the burning

 

Candle 1:

\(\begin{array}{rcl} h_1 = -\frac {H} {4\ \mathrm{hours}} \cdot t + H \end{array}\) 

 

Candle 2:

\(\begin{array}{rcl} h_2 = -\frac {H} {7\ \mathrm{hours}} \cdot t + H \end{array}\)

 

condition:

\(\begin{array}{rcll} 2\cdot h_1 &=& h_2 \\ 2\cdot \left( -\frac {H} {4\ \mathrm{hours}} \cdot t + H \right) &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \\\\ -\frac {2\cdot H} {4\ \mathrm{hours}} \cdot t + 2\cdot H &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \qquad &| \qquad :H \\\\ -\frac {2} {4\ \mathrm{hours}} \cdot t + 2 &=& -\frac {1} {7\ \mathrm{hours}} \cdot t + 1 \\\\ -\frac {t} {2} + 2 &=& -\frac {t} {7} + 1 \\\\ -\frac {t} {2} + 1 &=& -\frac {t} {7} \\\\ \frac {t} {2}-\frac {t} {7} &=& 1 \\\\ t\left ( \frac {1} {2}-\frac {1} {7} \right) &=& 1 \\\\ t &=& \frac{1}{ \frac {1} {2}-\frac {1} {7} } \\\\ t &=& \frac{14}{ 7-2 } \\\\ t &=& \frac{14}{ 5 } \\\\ t &=& 2.8\ \mathrm{hours} \end{array}\)

 

\(\begin{array}{rcll} h_1 &=& -\frac {H} {4\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H \\ h_1 &=& -0.7 H + H \\ h_1 &=& 0.3H \\\\ h_2 &=& -\frac {H} {7\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H\\ h_2 &=& -0.4 H + H \\ h_2 &=& 0.6H \\\\ \dfrac{h_2}{h_1} &=& \dfrac{0.6H}{0.3H}= 2 \end{array}\)

 

laugh

heureka  Dec 10, 2015
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5+0 Answers

 #1
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0

Are they a different width? Forgive me if my answer makes absolutely no sense it is 3:00am over here and I just woke up.

Guest Dec 10, 2015
 #2
avatar+78643 
0

They are of different widths, but.....that doesn't really factor in to the answer.....

 

 

cool cool cool

CPhill  Dec 10, 2015
 #3
avatar+18714 
+15
Best Answer

Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

 

H = height to begin.

h = height during the burning

 

Candle 1:

\(\begin{array}{rcl} h_1 = -\frac {H} {4\ \mathrm{hours}} \cdot t + H \end{array}\) 

 

Candle 2:

\(\begin{array}{rcl} h_2 = -\frac {H} {7\ \mathrm{hours}} \cdot t + H \end{array}\)

 

condition:

\(\begin{array}{rcll} 2\cdot h_1 &=& h_2 \\ 2\cdot \left( -\frac {H} {4\ \mathrm{hours}} \cdot t + H \right) &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \\\\ -\frac {2\cdot H} {4\ \mathrm{hours}} \cdot t + 2\cdot H &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \qquad &| \qquad :H \\\\ -\frac {2} {4\ \mathrm{hours}} \cdot t + 2 &=& -\frac {1} {7\ \mathrm{hours}} \cdot t + 1 \\\\ -\frac {t} {2} + 2 &=& -\frac {t} {7} + 1 \\\\ -\frac {t} {2} + 1 &=& -\frac {t} {7} \\\\ \frac {t} {2}-\frac {t} {7} &=& 1 \\\\ t\left ( \frac {1} {2}-\frac {1} {7} \right) &=& 1 \\\\ t &=& \frac{1}{ \frac {1} {2}-\frac {1} {7} } \\\\ t &=& \frac{14}{ 7-2 } \\\\ t &=& \frac{14}{ 5 } \\\\ t &=& 2.8\ \mathrm{hours} \end{array}\)

 

\(\begin{array}{rcll} h_1 &=& -\frac {H} {4\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H \\ h_1 &=& -0.7 H + H \\ h_1 &=& 0.3H \\\\ h_2 &=& -\frac {H} {7\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H\\ h_2 &=& -0.4 H + H \\ h_2 &=& 0.6H \\\\ \dfrac{h_2}{h_1} &=& \dfrac{0.6H}{0.3H}= 2 \end{array}\)

 

laugh

heureka  Dec 10, 2015
 #4
avatar+91038 
0

Thanks Chris and Heureka,

This looks like a really interesting question :)

Melody  Dec 10, 2015
 #5
avatar+78643 
+10

Good job, heureka.....here is how I solved it.....

 

Each hour, the candle that burns down more quickly loses  7/28  of its height.....and the other candle loses 4/28 of its height.....so......we want to know when.....

 

2 ( h  - [7/28]h* t )  =  (h - [4/28]h* t)       where h is the original height and t is the time we are looking for, in hours........

 

2h - [14/28]h*t =  h - [4/28]h*t

 

1h =  [10/28]h * t       divide through by h

 

1 = [10/28] * t

 

28/10 = t   = 2.8 hours

 

 

 

cool cool cool

CPhill  Dec 10, 2015
edited by CPhill  Dec 10, 2015

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