Question

2 candles, A and B, of the same height were lighted at the same time. Candle A will be fully consumed in 7 hours, while B in 5 hours. In how many hours will Candle A be thrice the height of Candle B?

Answer 1

Can someone explain how to get the answer? Thanks!

Answer 2

candle A loses 1/7 of its height every hour, and candle B loses 1/5th of its height every hour.
in x hours, candle A will lose x*1/7 of its height.
in x hours, candle B will lose x*1/5 of its height

Answer 3

if candles A and B are of height H, then their heights after x hours are H-x/7 and H-x/5 respectively.

Answer 4

it is given that H-x/7=3(H-x/5)

Answer 5

I seem to have hit a brick wall at this point.

Answer 6

if you assume both the candles are of unit length, then 1-x/7 = 3-3x/5
3x/5-x/7=2
16x =70
x = 70/16 = 4.375 hours

Answer 7

this looks good to me.
i assumed that they were both 35 inches tall so that one had rate of 5 inches per hour and the other 7 inches per hour.

then one height is given by 35 - 5t and the other by 35 - 7t and we need
\[3(35-7t)=35-5t\] giving
\[t=\frac{35}{8}=4.375\]

the only reason i am posting is to make the point that since the height of the candles would not alter the outcome of the problems, you can pick any number to work with, and therefore are free to pick something nice. of course 1 is nice too

Answer 8

thanks, guys!

Answer 9

Another way is...
\[A=1-\frac{t}{5} and B=1-\frac{t}{7}\]\[A=\frac{1}{3}B\]\[1-\frac{t}{5}=\frac{1}{3}(1-\frac{t}{7})\] now solve for t
\[\frac{1}{3}=1-\frac{16t}{105} or \frac{105}{3}=105-16t\] this gives\[(-70=-16t) = 4.375 hours\] I aonly add this as I did it while I was out and so I may as well add it in :D