Difference between revisions of "1974 AHSME Problems/Problem 15"

Latest revision as of 12:43, 5 July 2013

If $x<-2$ , then $|1-|1+x||$ equals

$\mathrm{(A)\ } 2+x \qquad \mathrm{(B) \ }-2-x \qquad \mathrm{(C) \ } x \qquad \mathrm{(D) \ } -x \qquad \mathrm{(E) \ }-2$

Notice that, for $a<0$ , $|a|=-a$ , and for $a>0$ , $|a|=a$ .

Since $x<-2$ , $1+x<1-2<0$ , so $|1+x|=-x-1$ . Therefore, $1-|1+x|=1+x+1=x+2<-2+2=0$ , and so $|1-|1+x||=|x+2|=-2-x, \boxed{\text{B}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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
All AHSME Problems and Solutions

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 {{AHSME box|year=1974|num-b=14|num-a=16}}
 [[Category:Introductory Algebra Problems]]
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