Difference between revisions of "1967 AHSME Problems/Problem 27"

Latest revision as of 00:40, 16 August 2023

Problem

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?

$\textbf{(A) 1:24}\qquad \textbf{(B) 1:28}\qquad \textbf{(C) 1:36}\qquad \textbf{(D) 1:40}\qquad \textbf{(E) 1:48}$

Solution

If the candles both have length $\ell$ , then the candle that burns in $3$ hours has a stub of $\ell$ at $0$ minutes, and a stub of $0$ at $180$ minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle $t$ minutes after being lit is $f(t) = \frac{\ell}{180}(180 - t)$ , since $f(0) = \ell$ and $f(180) = 0$ .

Similarly, for the second candle that burns out in $240$ minutes, $g(t) = \frac{\ell}{240}(240 - t)$

Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub. The problem asks for when $g(t) = 2f(t)$ . Solving this equation gives:

$\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)$

$240 - t = \frac{480}{180}(180 - t)$

$240 - t = 480 - \frac{480}{180}t$

$\frac{8}{3} t - t = 480 - 240$

$t = \frac{3}{5} \cdot 240$

$t = 144$

So, the second candle will have a stub twice as big as the first candle $144$ minutes after they are both lit. If we want this to happen at $4$ PM, the candles have to be lit $144$ minutes earlier, or $2$ hours and $24$ minutes earlier. This is at $\text{1:36 PM}$ , which is option $\fbox{C}$

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 == Solution ==
-<math>\fbox{C}</math>
+If the candles both have length <math>\ell</math>, then the candle that burns in <math>3</math> hours has a stub of <math>\ell</math> at <math>0</math> minutes, and a stub of <math>0</math> at <math>180</math> minutes.  Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle <math>t</math> minutes after being lit is <math>f(t) = \frac{\ell}{180}(180 - t)</math>, since <math>f(0) = \ell</math> and <math>f(180) = 0</math>.
+Similarly, for the second candle that burns out in <math>240</math> minutes, <math>g(t) = \frac{\ell}{240}(240 - t)</math>
+Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub.  The problem asks for when <math>g(t) = 2f(t)</math>.  Solving this equation gives:
+<math>\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)</math>
+<math>240 - t = \frac{480}{180}(180 - t)</math>
+<math>240 - t = 480 - \frac{480}{180}t</math>
+<math>\frac{8}{3} t - t = 480 - 240</math>
+<math>t = \frac{3}{5} \cdot 240</math>
+<math>t = 144</math>
+So, the second candle will have a stub twice as big as the first candle <math>144</math> minutes after they are both lit.  If we want this to happen at <math>4</math> PM, the candles have to be lit <math>144</math> minutes earlier, or <math>2</math> hours and <math>24</math> minutes earlier.  This is at <math>\text{1:36 PM}</math>, which is option <math>\fbox{C}</math>
 == See also ==
-{{AHSME box|year=1967|num-b=26|num-a=28}}
+{{AHSME 40p box|year=1967|num-b=26|num-a=28}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "1967 AHSME Problems/Problem 27"

Latest revision as of 00:40, 16 August 2023

Problem

Solution

See also