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1960 AHSME Problems/Problem 26

Revision as of 12:53, 11 May 2018 by Rockmanex3 (talk | contribs) (→‎Solution 2)

Problem

Find the set of $x$-values satisfying the inequality $|\frac{5-x}{3}|<2$. [The symbol $|a|$ means $+a$ if $a$ is positive, $-a$ if $a$ is negative,$0$ if $a$ is zero. The notation $1<a<2$ means that a can have any value between $1$ and $2$, excluding $1$ and $2$. ]

$\textbf{(A)}\ 1 < x < 11\qquad \textbf{(B)}\ -1 < x < 11\qquad \textbf{(C)}\ x< 11\qquad \textbf{(D)}\ x>11\qquad \textbf{(E)}\ |x| < 6$

Solutions

Solution 1

Break up the absolute value into two cases.

For the first case, let $x < 5$, so $\frac{5-x}{3}$ is positive. That means (for $x<5$) \[\frac{5-x}{3} < 2\] \[5-x<6\] \[-x<1\] \[x>-1\] For the second case, let $x \ge 5$, so $\frac{5-x}{3}$ is negative. That means (for $x \ge 5$) \[\frac{x-5}{3} < 2\] \[x-5<6\] \[x<11\]

Combine both cases to get $-1 < x < 11$, which is answer choice $\boxed{\textbf{(B)}}$.

Solution 2

[asy]import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=14.2,ymin=-1.2,ymax=4.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((5,0)--(-4,3),EndArrow); draw((5,0)--(14,3),EndArrow); draw((-4,2)--(14,2),dotted,Arrows); draw(circle((-1,2),0.2)); draw(circle((11,2),0.2)); [/asy]

Another way to solve this is to graph $y=|\frac{5-x}{3}|$ and $y=2$. The solution is the areas on the graph where the y-values of $y=|\frac{5-x}{3}|$ are lower than $2$. From the graph, $-1<x<11$, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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
All AHSME Problems and Solutions
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