Melting Candles

Algebra Level 2

Consider two candles which have the same height. Each candle burns at a consistent rate (but the two candles have different burn rate). The first one melts completely in 5 hours, and the second one melts completely in 3 hours.

If they are lit at the same time, after how many minutes will the first one be thrice as long as the second one?

150 90 60 30

2 solutions

Benny Joseph
Nov 6, 2016

Let us assume that the length of the candles is 1.
The first candle melts $\frac{1}{300}$ in one minute.
The second candle melts $\frac{1}{180}$ in one minute.

If after $t$ minutes, the length of the first candle is thrice the length of the second candle, then we have

$3 ( 1 - \frac{t}{180}) = (1 - \frac{t}{300} ).$

Solving this equation, we get that $\frac{4t}{300} = 2$ , or that $t = 150$ .

Nice solution. Fixing the length reduces the need for another variable.

FYI There is no need to place everything into the Latex brackets. You can use it just for the equations, and keep the text separated out. I've edited your solution for reference.

Calvin Lin Staff - 4 years, 7 months ago

Anubhav Tyagi
Nov 14, 2016

Let the length of candles be 'L' units.

Candle 1 : It takes 3 hrs to completely melt down i.e. "L" length of candle melts down in 3 hrs
Thus in time "t" hrs length of candle melted: $\frac{L}{3} \times t$
Length of candle left: $L-\frac{L}{3} \times t$

Candle 2 : It takes 5 hrs to completely melt down i.e. "L" length of candle melts down in 5 hrs
Thus in time "t" hrs length of candle melted: $\frac{L}{5} \times t$
Length of candle left: $L-\frac{L}{5} \times t$

It is given that at time "t" length of candle 2 is 3 times the length of candle 1. Hence we write,

$L-\frac{L}{5} \times t= 3(L-\frac{L}{3} \times t)$

On solving, we get $t=\frac{10}{4} \text{ hrs }= \boxed{\SI{150}{\ mins}}$

Very clear solution! +1

Prakhar Bindal - 4 years, 7 months ago

Melting Candles

2 solutions

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