1961 AHSME Problems/Problem 20

Problem

The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:

$\textbf{(A)}\ \text{I and II}\qquad \textbf{(B)}\ \text{II and III}\qquad \textbf{(C)}\ \text{I and III}\qquad \textbf{(D)}\ \text{III and IV}\qquad \textbf{(E)}\ \text{I and IV}$

Solution

[asy] fill((-1,5)--(5,5)--(5,-1)--cycle,red); fill((2.5,5)--(-5,5)--(-5,-5)--(-2.5,-5)--cycle,cyan); fill((-1,5)--(4/3,8/3)--(2.5,5)--cycle,magenta); draw((-1,5)--(5,-1),dotted,Arrows); draw((-2.5,-5)--(2.5,5),dotted,Arrows);  import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=5.2,ymin=-5.2,ymax=5.2;  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);  [/asy] Graph both equations. The solution to the system of inequalities is the magenta area because it is in both of the solutions of each individual inequality. From the graph, the magenta area is in quadrants $I$ and $II$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS