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Difference between revisions of "1970 AHSME Problems/Problem 10"

(Created page with "== Problem == Let <math>F=.48181\cdots</math> be an infinite repeating decimal with the digits <math>8</math> and *1* repeating. When <math>F</math> is written as a fraction in ...")
 
 
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== Problem ==
 
== Problem ==
  
−
Let <math>F=.48181\cdots</math> be an infinite repeating decimal with the digits <math>8</math> and *1* repeating. When <math>F</math> is written as a fraction in lowest terms, the denominator exceeds the numerator by
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Let <math>F=.48181\cdots</math> be an infinite repeating decimal with the digits <math>8</math> and <math>1</math> repeating. When <math>F</math> is written as a fraction in lowest terms, the denominator exceeds the numerator by
  
 
<math>\text{(A) } 13\quad
 
<math>\text{(A) } 13\quad
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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
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Multiplying by <math>100</math> gives <math>100F = 48.181818...</math>.  Subtracting the first equation from the second gives <math>99F = 47.7</math>, and all the other repeating parts cancel out.  This gives <math>F = \frac{47.7}{99} = \frac{477}{990} = \frac{159}{330} = \frac{53}{110}</math>.  Subtracting the numerator from the denominator gives <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1970|num-b=9|num-a=11}}   
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{{AHSME 35p box|year=1970|num-b=9|num-a=11}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:21, 13 July 2019

Problem

Let $F=.48181\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by

$\text{(A) } 13\quad \text{(B) } 14\quad \text{(C) } 29\quad \text{(D) } 57\quad \text{(E) } 126$

Solution

Multiplying by $100$ gives $100F = 48.181818...$. Subtracting the first equation from the second gives $99F = 47.7$, and all the other repeating parts cancel out. This gives $F = \frac{47.7}{99} = \frac{477}{990} = \frac{159}{330} = \frac{53}{110}$. Subtracting the numerator from the denominator gives $\fbox{D}$.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AHSME Problems and Solutions

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