1989 AHSME Problems

Problem 1

$(-1)^{5^{2}}+1^{2^{5}}=$

$\textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57$

Problem 2

$\sqrt{\frac{1}{9}+\frac{1}{16}}=$

$\textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12}$

Solution

Problem 3

A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is

$[asy] draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(3,9), dashed); draw((6,0)--(6,9), dashed);[/asy]$

$\textrm{(A)}\ 24\qquad\textrm{(B)}\ 36\qquad\textrm{(C)}\ 64\qquad\textrm{(D)}\ 81\qquad\textrm{(E)}\ 96$

Solution

Problem 4

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$ , $AB=4$ , and $DC=10$ . The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$ . Then $CF=$

$[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]$

$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad\textrm{(E)}\ 4.25$

Solution

Problem 5

Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is

$[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); for(int i = 1;i<y.length;++i) { pair p = (a,y[i]); pen pen = linewidth(.7); if(y[i]-prev.y!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(xscl*p.x,p.y),pen); prev = p; } }for(int a:y) { pair prev = (x[0],a); for(int i = 1;i<x.length;++i) { pair p = (x[i],a); pen pen = linewidth(.7); if(x[i]-prev.x!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(p.x*xscl,p.y),pen); prev = p; } } path lblx = (0,-.7)--(5*xscl,-.7); draw(lblx); label("$10$",lblx); path lbly = (5*xscl+.7,0)--(5*xscl+.7,5); draw(lbly); label("$20$",lbly);[/asy]$

$\textrm{(A)}\ 30\qquad\textrm{(B)}\ 200\qquad\textrm{(C)}\ 410\qquad\textrm{(D)}\ 420\qquad\textrm{(E)}\ 430$

Solution

Problem 6

If $a,\,b>0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by=6$ has area $6$ , then $ab=$

$\textrm{(A)}\ 3\qquad\textrm{(B)}\ 6\qquad\textrm{(C)}\ 12\qquad\textrm{(D)}\ 108\qquad\textrm{(E)}\ 432$

Solution

Problem 7

In $\triangle ABC$ , $\angle A = 100^\circ$ , $\angle B = 50^\circ$ , $\angle C = 30^\circ$ , $\overline{AH}$ is an altitude, and $\overline{BM}$ is a median. Then $\angle MHC=$

$[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE);[/asy]$