Question

Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at $4$ P.M., one stub is twice the length of the other?

$1{:}24$

$1{:}28$

$1{:}36$

$1{:}40$

$1{:}48$

Answer: C

Concepts:rate linear equation date and time

Difficulty rating: 1500

Small Hint:

Let $t$ be the number of hours the candles burn before $4$ P.M.

Big Hint:

Their remaining fractions are $1-\frac{t}{3}$ and $1-\frac{t}{4}$

Solution:

After $t$ hours, the faster and slower candles have fractions $1-\frac{t}{3}$ and $1-\frac{t}{4}$ remaining. The slower stub must be twice the faster: $1-\frac t4=2\left(1-\frac t3\right).$ Thus $t=\frac{12}{5}=2$ hours $24$ minutes. Counting back from $4$ P.M. gives $1{:}36$ P.M.

Therefore, the correct answer is C.

1967 AMC 12 Problem 27

Answer: C

Difficulty rating: 1500

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years