Difference between revisions of "1958 AHSME Problems/Problem 34"

(Created page with "== Problem == The numerator of a fraction is <math> 6x \plus{} 1</math>, then denominator is <math> 7 \minus{} 4x</math>, and <math> x</math> can have any value between <math> \m...")
 
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== Problem ==
 
== Problem ==
The numerator of a fraction is <math> 6x \plus{} 1</math>, then denominator is <math> 7 \minus{} 4x</math>, and <math> x</math> can have any value between <math> \minus{}2</math> and <math> 2</math>, both included. The values of <math> x</math> for which the numerator is greater than the denominator are:
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The numerator of a fraction is <math> 6x + 1</math>, then denominator is <math> 7 - 4x</math>, and <math> x</math> can have any value between <math> -2</math> and <math> 2</math>, both included. The values of <math> x</math> for which the numerator is greater than the denominator are:
  
 
<math> \textbf{(A)}\ \frac{3}{5} < x \le 2\qquad  
 
<math> \textbf{(A)}\ \frac{3}{5} < x \le 2\qquad  
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\textbf{(C)}\ 0 < x \le 2\qquad \\
 
\textbf{(C)}\ 0 < x \le 2\qquad \\
 
\textbf{(D)}\ 0 \le x \le 2\qquad  
 
\textbf{(D)}\ 0 \le x \le 2\qquad  
\textbf{(E)}\ \minus{}2 \le x \le 2</math>
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\textbf{(E)}\ -2 \le x \le 2</math>
 
 
  
 
== Solution ==
 
== Solution ==

Latest revision as of 23:22, 13 March 2015

Problem

The numerator of a fraction is $6x + 1$, then denominator is $7 - 4x$, and $x$ can have any value between $-2$ and $2$, both included. The values of $x$ for which the numerator is greater than the denominator are:

$\textbf{(A)}\ \frac{3}{5} < x \le 2\qquad  \textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad  \textbf{(C)}\ 0 < x \le 2\qquad \\ \textbf{(D)}\ 0 \le x \le 2\qquad  \textbf{(E)}\ -2 \le x \le 2$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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All AHSME Problems and Solutions

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