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Difference between revisions of "1956 AHSME Problems/Problem 19"

(Created page with "== Problem 19== Two candles of the same height are lighted at the same time. The first is consumed in <math>4</math> hours and the second in <math>3</math> hours. Assuming...")
 
(old solution is incorrect, as the candle doesn't have the same height at x=0)
 
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
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In <math>x</math> hours, the height of the first candle is <math>4-x</math> and the height of the second candle is <math>3-x.</math> Essentially, we are solving the equation
+
In <math>x</math> hours, the height of the first candle is <math>1 - \frac{x}{4}</math> and the height of the second candle is <math>1 - \frac{x}{3}</math> Essentially, we are solving the equation
−
<cmath>4-x = 2(3-x).</cmath>
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<cmath>1 - \frac{x}{4} = 2(1 - \frac{x}{3})</cmath>.
−
This happens after <math>x = \boxed{\textbf{(C)} \quad 2}</math> hours.
 
  
 +
<math>- \frac{x}{4} = 1 - \frac{2x}{3}</math> | <math>-1</math>
  
−
~coolmath34
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<math>- 3x = 12 - 8x</math> | <math>\cdot 12</math>
 +
 
 +
<math>5x = 12</math> | <math>+8x</math>
 +
 
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<math>x = \frac{12}{5}</math> | <math>/5</math>
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 +
 
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<math>x = \boxed{\textbf{(D)} \quad 2 \frac{2}{5}}</math> hours.
  
 
==See Also==
 
==See Also==

Latest revision as of 03:59, 8 August 2023

Problem 19

Two candles of the same height are lighted at the same time. The first is consumed in $4$ hours and the second in $3$ hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second?

$\textbf{(A)}\ \frac{3}{4}\qquad\textbf{(B)}\ 1\frac{1}{2}\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2\frac{2}{5}\qquad\textbf{(E)}\ 2\frac{1}{2}$

Solution

In $x$ hours, the height of the first candle is $1 - \frac{x}{4}$ and the height of the second candle is $1 - \frac{x}{3}$ Essentially, we are solving the equation \[1 - \frac{x}{4} = 2(1 - \frac{x}{3})\].

$- \frac{x}{4} = 1 - \frac{2x}{3}$ | $-1$

$- 3x = 12 - 8x$ | $\cdot 12$

$5x = 12$ | $+8x$

$x = \frac{12}{5}$ | $/5$


$x = \boxed{\textbf{(D)} \quad 2 \frac{2}{5}}$ hours.

See Also

1956 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AHSME Problems and Solutions

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