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1958 AHSME Problems/Problem 34: Difference between revisions

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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:
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  
Line 6: Line 6:
\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>
\textbf{(E)}\ -2 \le x \le 2</math>
 


== Solution ==
== Solution ==

Latest revision as of 22:22, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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