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Difference between revisions of "1959 AHSME Problems/Problem 25"

(Created page with "== Problem 25== The symbol <math>|a|</math> means <math>+a</math> if <math>a</math> is greater than or equal to zero, and <math>-a</math> if a is less than or equal to zero; t...")
 
(→‎Solution)
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<math>\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1</math>   
 
<math>\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1</math>   
 
== Solution ==
 
== Solution ==
−
The equation <math>|3-x| < 4</math> can be solved by splitting it into two inequalities: <math>3-x<4</math> and <math>3-x<-4</math>. The solutions to those inequalities are <math>x<-1</math> and <math>x>7</math>, respectively. The common interval of those two inequalities is <math>\textbf{(D)}\-1<x<7</math>
+
The equation <math>|3-x| < 4</math> can be solved by splitting it into two inequalities: <math>3-x<4</math> and <math>3-x<-4</math>. The solutions to those inequalities are <math>x<-1</math> and <math>x>7</math>, respectively. The common interval of those two inequalities is <math>\textbf{(D)}\ -1<x<7</math>
 +
 
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=24|num-a=26}}
 
{{AHSME 50p box|year=1959|num-b=24|num-a=26}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:29, 21 July 2020

Problem 25

The symbol $|a|$ means $+a$ if $a$ is greater than or equal to zero, and $-a$ if a is less than or equal to zero; the symbol $<$ means "less than"; the symbol $>$ means "greater than." The set of values $x$ satisfying the inequality $|3-x|<4$ consists of all $x$ such that: $\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1$

Solution

The equation $|3-x| < 4$ can be solved by splitting it into two inequalities: $3-x<4$ and $3-x<-4$. The solutions to those inequalities are $x<-1$ and $x>7$, respectively. The common interval of those two inequalities is $\textbf{(D)}\ -1<x<7$

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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All AHSME Problems and Solutions

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