February 26, 2017

Homework Help: math (rate)

Posted by Kasie on Wednesday, October 25, 2006 at 4:36pm.

I need help! Ok here's the problem.
Melissa lights 2 candles that are the exact same height at the exact same time. The blue candle takes 6 hours to burn completely. The purple candle takes 9 hours to burn completely. After how many hours will the slower burning candle be twice the height of the faster candle? How tall were the candles originally?
Please help. I'm in 7th grade and normally a very good math student but these are really hard. PLEASE RESPOND QUICKLY! I need this by tomarrow!!!

Purple rate=rp=1candle/9hrs.
Now, the length remaining is the original Length minus the burned part.
Length= Lorig - rburning*Time

Lengthblue= Lorig-Lorig/6*T
When the blue is one half the purple,
length blue= 1/2 Lp
Lorig (1- 1/6 T)= 1/2 Lorig(1-1/9 T)
1-T/6= 1/2 - T/18
solve for T.

Maria lights two candles of equal length at the same time. One candle takes 6
>hours to burn out and the other takes 9. How much time will pass until the
>slower-burning candle is exactly twice as long as the faster-burning one?
>Explain how you got the answer.

1--The faster burning candle, Fb, burns at the rate of L/6 inches per hour.
2--The slower burning candle, Sb, burns at the rate of L/9 inches per hour.
3--In t hours, Fb burns a distance of tL/6 inches.
4--In the same t hours, Fs burns a distance of tL/9 inches.
5--The unburned candle lengths at t hours are then (L - tL/6) or (6L - tL)/6 for Fb and (L - tL/9) or (6L - tL)/9 for Sb.
6--Since we are looking for the t where Sb's length is twice Fb's length, we can write (9L - tL)/9 = 2(6L - tL)/6.
7--Expanding, 54L - 6tL = 108L - 18tL.
8--Eliminating the L's throughout, we have 12t = 54 making t = 4.5 hours.

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