Candles burning at different rates

[Natasha1]

Homework Statement

Candles burning at different rates

Relevant Equations

See my answer sheet attached

Please see photos attached for problem and my answers.

Can someone just check my work and tell me if the candle heights I get are correct?

 

[HallsofIvy]

Let "x" be the length of the shorter candle in inches. The longer candle is x+ 4 inches long. Let y be the distance the shorter candle was reduced by in the two hours from 7:30 to 9:30 (at a rate of y/2 inches per hour). The longer candle was reduced by y+4 inches in the three and a half hours from 6:00 to 9:30 (at a rate of (y+4)/3.5 inches per hour).

The longer candle burned out at 11:00, 5 hours after being lit. Since it was shortening at the rate of (y+4)/3.5 inches per hour, we have x+ 4= ((y+4)/3.5)(5)= 10(y+4)/7.

The shorter candle burned out at 11:30, 4 hours after being lit. Since it was shortening at the rate of y/2 inches per hour, we have x= (y/2)(4)= 2y.

Solve the equations x+ 4= 10(y+4)/7 and x= 2y for x and y. The value of x is the answer to this question.

 

[SammyS]

Problem Statement: Candles burning at different rates
Relevant Equations: See my answer sheet attached

Please see photos attached for problem and my answers.

Can someone just check my work and tell me if the candle heights I get are correct?

Problem statement:

Last portion of your solution:

Your solution is correct.

 

[neilparker62]

Alternate method. Let length of longer candle be A and shorter A-4. The longer candle burns at a rate of $r_1$ percent per hour and the shorter at $r_2$ percent per hour. Longer candle burns out in 5 hours and the shorter one in 4 hours.

$A(1-5\times \frac{r_1}{100})=0$ and $(A-4)(1-4\times \frac{r_2}{100})=0$.

From which $r_1$ and $r_2$ can easily be determined. Then solve for A:

$A(1-3,5\times \frac{r_1}{100})=(A-4)(1-2\times \frac{r_2}{100})$