Difference between revisions of "1992 AHSME Problems"

Revision as of 02:29, 28 September 2014

Problem 1

If $3(4x+\pi)=P$ then $6(8x+10\pi)=$

$\text{(A) } 2P\quad \text{(B) } 3P\quad \text{(C) } 6P\quad \text{(D) } 8P\quad \text{(E) } 18P$

Problem 2

An urn is filled with coins and beads, all of which are either silver or gold. Twenty percent of the objects in the urn are beads. Forty percent of the coins in the urn are silver. What percent of objects in the urn are gold coins?

$\text{(A) } 40\%\quad \text{(B) } 48\%\quad \text{(C) } 52\%\quad \text{(D) } 60\%\quad \text{(E) } 80\%$

Solution

Problem 3

If $m>0$ and the points $(m,3)$ and $(1,m)$ lie on a line with slope $m$ , then $m=$

$\text{(A) } 1\quad \text{(B) } \sqrt{2}\quad \text{(C) } \sqrt{3}\quad \text{(D) } 2\quad \text{(E) } \sqrt{5}$

Solution

Problem 4

If $a,b$ and $c$ are positive integers and $a$ and $b$ are odd, then $3^a+(b-1)^2c$ is

$\text{(A) odd for all choices of c} \quad \text{(B) even for all choices of c} \quad\\ \text{(C) odd if c is even; even if c is odd} \quad\\ \text{(D) odd if c is odd; even if c is even} \quad\\ \text{(E) odd if c is not a multiple of 3;evn if c is a multiple of 3}$ Solution

Problem 5

$6^6+6^6+6^6+6^6+6^6+6^6=$

$\text{(A) } 6^6 \quad \text{(B) } 6^7\quad \text{(C) } 36^6\quad \text{(D) } 6^{36}\quad \text{(E) } 36^{36}$ Solution

Problem 6

If $x>y>0$ , then $\frac{x^y y^x}{y^y x^x}=$

$\text{(A) } (x-y)^{y/x}\quad \text{(B) } \left(\frac{x}{y}\right)^{x-y}\quad \text{(C) } 1\quad \text{(D) } \left(\frac{x}{y}\right)^{y-x}\quad \text{(E) } (x-y)^{x/y}$ Solution

Problem 7

The ratio of $w$ to $x$ is $4:3$ , of $y$ to $z$ is $3:2$ and of $z$ to $x$ is $1:6$ . What is the ratio of $w$ to $y$ ?

$\text{(A) } 1:3\quad \text{(B) } 16:3\quad \text{(C) } 20:3\quad \text{(D) } 27:4\quad \text{(E) } 12:1$

Solution

Problem 8

$[asy] draw((-10,-10)--(-10,10)--(10,10)--(10,-10)--cycle,dashed+linewidth(.75)); draw((-7,-7)--(-7,7)--(7,7)--(7,-7)--cycle,dashed+linewidth(.75)); draw((-10,-10)--(10,10),dashed+linewidth(.75)); draw((-10,10)--(10,-10),dashed+linewidth(.75)); fill((10,10)--(10,9)--(9,9)--(9,10)--cycle,black); fill((9,9)--(9,8)--(8,8)--(8,9)--cycle,black); fill((8,8)--(8,7)--(7,7)--(7,8)--cycle,black); fill((-10,-10)--(-10,-9)--(-9,-9)--(-9,-10)--cycle,black); fill((-9,-9)--(-9,-8)--(-8,-8)--(-8,-9)--cycle,black); fill((-8,-8)--(-8,-7)--(-7,-7)--(-7,-8)--cycle,black); fill((10,-10)--(10,-9)--(9,-9)--(9,-10)--cycle,black); fill((9,-9)--(9,-8)--(8,-8)--(8,-9)--cycle,black); fill((8,-8)--(8,-7)--(7,-7)--(7,-8)--cycle,black); fill((-10,10)--(-10,9)--(-9,9)--(-9,10)--cycle,black); fill((-9,9)--(-9,8)--(-8,8)--(-8,9)--cycle,black); fill((-8,8)--(-8,7)--(-7,7)--(-7,8)--cycle,black); [/asy]$

A square floor is tiled with congruent square tiles. The tiles on the two diagonals of the floor are black. The rest of the tiles are white. If there are 101 black tiles, then the total number of tiles is

$\text{(A) } 121\quad \text{(B) } 625\quad \text{(C) } 676\quad \text{(D) } 2500\quad \text{(E) } 2601$

Solution

Problem 9

$[asy] draw((-7,0)--(7,0),black+linewidth(.75)); draw((-3*sqrt(3),0)--(-2*sqrt(3),3)--(-sqrt(3),0)--(0,3)--(sqrt(3),0)--(2*sqrt(3),3)--(3*sqrt(3),0),black+linewidth(.75)); draw((-2*sqrt(3),0)--(-1*sqrt(3),3)--(0,0)--(sqrt(3),3)--(2*sqrt(3),0),black+linewidth(.75)); [/asy]$

Five equilateral triangles, each with side $2\sqrt{3}$ , are arranged so they are all on the same side of a line containing one side of each vertex. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is

$\text{(A) 10} \quad \text{(B) } 12\quad \text{(C) } 15\quad \text{(D) } 10\sqrt{3}\quad \text{(E) } 12\sqrt{3}$

Solution

Problem 10

The number of positive integers $k$ for which the equation $kx-12=3k$ has an integer solution for $x$ is

$\text{(A) } 3\quad \text{(B) } 4\quad \text{(C) } 5\quad \text{(D) } 6\quad \text{(E) } 7$

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cubeoctahedron. The ratio of the volume of the cubeoctahedron to the volume of the original cube is closest to which of these?

$\text{(A) } 75\%\quad \text{(B) } 78\%\quad \text{(C) } 81\%\quad \text{(D) } 84\%\quad \text{(E) } 87\%$ Solution

Problem 20

$[asy] draw((1,0)--(2*cos(pi/8),2*sin(pi/8))--(cos(pi/4),sin(pi/4))--(2*cos(3*pi/8),2*sin(3*pi/8))--(cos(pi/2),sin(pi/2))--(2*cos(5*pi/8),2*sin(5*pi/8))--(cos(3*pi/4),sin(3*pi/4))--(2*cos(7*pi/8),2*sin(7*pi/8))--(-1,0),black+linewidth(.75)); MP("A_1",(2*cos(5*pi/8),2*sin(5*pi/8)),N);MP("A_2",(2*cos(3*pi/8),2*sin(3*pi/8)),N);MP("A_3",(2*cos(1*pi/8),2*sin(1*pi/8)),N); MP("A_n",(2*cos(7*pi/8),2*sin(7*pi/8)),N); MP("B_1",(cos(4*pi/8),sin(4*pi/8)),S);MP("B_2",(cos(2*pi/8),sin(2*pi/8)),S);MP("B_n",(cos(6*pi/8),sin(6*pi/8)),S); [/asy]$ Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,A_2,\cdots,A_n$ are congruent, and angles $B_1,B_2,\cdots,B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1$ , then $n=$

$\text{(A) } 12\quad \text{(B) } 18\quad \text{(C) } 24\quad \text{(D) } 36\quad \text{(E) } 60$ Solution

Problem 21

For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$ ?

$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$ Solution

Problem 22

Ten points are selected on the positive $x$ -axis, $X^+$ , and five points are selected on the positive $y$ -axis, $Y^+$ . The fifty segments connecting the ten points on $X^+$ to the five points on $Y^+$ are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant?

$\text{(A) } 250\quad \text{(B) } 450\quad \text{(C) } 500\quad \text{(D) } 1250\quad \text{(E) } 2500$ Solution

Problem 23

Let $S$ be a subset of $\{1,2,3,...,50\}$ such that no pair of distinct elements in $S$ has a sum divisible by $7$ . What is the maximum number of elements in $S$ ?

$\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23$ Solution

Problem 24

Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$ . Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$ , respectively, with $AE=BF=AG=2$ . Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$ . The area of quadrilateral $EFGH$ is

$\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$ Solution

Problem 25

In $\triangle{ABC}$ , $\angle{ABC=120^\circ,AB=3$ (Error compiling LaTeX. Unknown error_msg) and $BC=4$ . If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$ , then $CD=$

$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$

Solution

Problem 26

$[asy] fill((1,0)--arc((1,0),2,180,225)--cycle,grey); fill((-1,0)--arc((-1,0),2,315,360)--cycle,grey); fill((0,-1)--arc((0,-1),2-sqrt(2),225,315)--cycle,grey); fill((0,0)--arc((0,0),1,180,360)--cycle,white); draw((1,0)--arc((1,0),2,180,225)--(1,0),black+linewidth(1)); draw((-1,0)--arc((-1,0),2,315,360)--(-1,0),black+linewidth(1)); draw((0,0)--arc((0,0),1,180,360)--(0,0),black+linewidth(1)); draw(arc((0,-1),2-sqrt(2),225,315),black+linewidth(1)); draw((0,0)--(0,-1),black+linewidth(1)); MP("C",(0,0),N);MP("A",(-1,0),N);MP("B",(1,0),N); MP("D",(0,-.8),NW);MP("E",(1-sqrt(2),-sqrt(2)),SW);MP("F",(-1+sqrt(2),-sqrt(2)),SE); [/asy]$

Semicircle $\widehat{AB}$ has center $C$ and radius $1$ . Point $D$ is on $\widehat{AB}$ and $\overline{CD}\perp\overline{AB}$ . Extend $\overline{BD}$ and $\overline{AD}$ to $E$ and $F$ , respectively, so that circular arcs $\widehat{AE}$ and $\widehat{BF}$ have $B$ and $A$ as their respective centers. Circular arc $\widehat{EF}$ has center $D$ . The area of the shaded "smile" $AEFBDA$ , is

$\text{(A) } (2-\sqrt{2})\pi\quad \text{(B) } 2\pi-\pi \sqrt{2}-1\quad \text{(C) } (1-\frac{\sqrt{2}}{2})\pi\quad\\ \text{(D) } \frac{5\pi}{2}-\pi\sqrt{2}-1\quad \text{(E) } (3-2\sqrt{2})\pi$

Solution

Problem 27

A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length 7. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$ , respectively, they intersect at $P$ , which is outside of the circle. If $\angle{APD}=60^\circ$ and $BP=8$ , then $r^2=$

$\text{(A) } 70\quad \text{(B) } 71\quad \text{(C) } 72\quad \text{(D) } 73\quad \text{(E) } 74$ Solution

Problem 28

Let $i=\sqrt{-1}$ . The product of the real parts of the roots of $z^2-z=5-5i$ is

$\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$

Solution

Problem 29

An "unfair" coin has a $2/3$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?

$\text{(A) } 25(\frac{2}{3})^{50}\quad \text{(B) } \frac{1}{2}(1-\frac{1}{3^{50}})\quad \text{(C) } \frac{1}{2}\quad \text{(D) } \frac{1}{2}(1+\frac{1}{3^{50}})\quad \text{(E) } \frac{2}{3}$

Solution

Problem 30

Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19$ . Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$ . If $m$ is the smallest possible value of $x$ , then $m^2$ =

$\text{(A) } 1369\quad \text{(B) } 1679\quad \text{(C) } 1748\quad \text{(D) } 2109\quad \text{(E) } 8825$ Solution

@@ Line 185: / Line 185: @@
+For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cubeoctahedron. The ratio of the volume of the cubeoctahedron to the volume of the original cube is closest to which of these?
+<math>\text{(A) } 75\%\quad
+\text{(B) } 78\%\quad
+\text{(C) } 81\%\quad
+\text{(D) } 84\%\quad
+\text{(E) } 87\%</math>
 [[1992 AHSME Problems/Problem 19|Solution]]
 == Problem 20 ==
+<asy>
+draw((1,0)--(2*cos(pi/8),2*sin(pi/8))--(cos(pi/4),sin(pi/4))--(2*cos(3*pi/8),2*sin(3*pi/8))--(cos(pi/2),sin(pi/2))--(2*cos(5*pi/8),2*sin(5*pi/8))--(cos(3*pi/4),sin(3*pi/4))--(2*cos(7*pi/8),2*sin(7*pi/8))--(-1,0),black+linewidth(.75));
+MP("A_1",(2*cos(5*pi/8),2*sin(5*pi/8)),N);MP("A_2",(2*cos(3*pi/8),2*sin(3*pi/8)),N);MP("A_3",(2*cos(1*pi/8),2*sin(1*pi/8)),N);
+MP("A_n",(2*cos(7*pi/8),2*sin(7*pi/8)),N);
+MP("B_1",(cos(4*pi/8),sin(4*pi/8)),S);MP("B_2",(cos(2*pi/8),sin(2*pi/8)),S);MP("B_n",(cos(6*pi/8),sin(6*pi/8)),S);
+</asy>
+Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all <math>2n</math> edges are congruent, angles <math>A_1,A_2,\cdots,A_n</math> are congruent, and angles <math>B_1,B_2,\cdots,B_n</math> are congruent. If the acute angle at <math>A_1</math> is <math>10^\circ</math> less than the acute angle at <math>B_1</math>, then <math>n=</math>
+<math>\text{(A) } 12\quad
+\text{(B) } 18\quad
+\text{(C) } 24\quad
+\text{(D) } 36\quad
+\text{(E) } 60</math>
 [[1992 AHSME Problems/Problem 20|Solution]]
 == Problem 21 ==
+For a finite sequence <math>A=(a_1,a_2,...,a_n)</math> of numbers, the ''Cesáro sum'' of A is defined to be
+<math>\frac{S_1+\cdots+S_n}{n}</math> , where <math>S_k=a_1+\cdots+a_k</math> and <math>1\leq k\leq n</math>. If the Cesáro sum of
+the 99-term sequence <math>(a_1,...,a_{99})</math> is 1000,  what is the Cesáro sum of the 100-term sequence
+<math>(1,a_1,...,a_{99})</math>?
+<math>\text{(A) } 991\quad
+\text{(B) } 999\quad
+\text{(C) } 1000\quad
+\text{(D) } 1001\quad
+\text{(E) } 1009</math>
 [[1992 AHSME Problems/Problem 21|Solution]]
 == Problem 22 ==
+Ten points are selected on the positive <math>x</math>-axis,<math>X^+</math>, and five points are selected on the positive <math>y</math>-axis,<math>Y^+</math>. The fifty segments connecting the ten points on <math>X^+</math> to the five points on <math>Y^+</math> are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant?
+<math>\text{(A) } 250\quad
+\text{(B) } 450\quad
+\text{(C) } 500\quad
+\text{(D) } 1250\quad
+\text{(E) } 2500</math>
 [[1992 AHSME Problems/Problem 22|Solution]]
@@ Line 205: / Line 239: @@
 == Problem 23 ==
+Let <math>S</math> be a subset of <math>\{1,2,3,...,50\}</math> such that no pair of distinct elements in <math>S</math> has a sum divisible by <math>7</math>. What is the maximum number of elements in <math>S</math>?
+<math>\text{(A) } 6\quad
+\text{(B) } 7\quad
+\text{(C) } 14\quad
+\text{(D) } 22\quad
+\text{(E) } 23</math>
 [[1992 AHSME Problems/Problem 23|Solution]]
 == Problem 24 ==
+Let <math>ABCD</math> be a parallelogram of area <math>10</math> with <math>AB=3</math> and <math>BC=5</math>. Locate <math>E,F</math> and <math>G</math> on segments <math>\overline{AB},\overline{BC}</math> and <math>\overline{AD}</math>, respectively, with <math>AE=BF=AG=2</math>. Let the line through <math>G</math> parallel to <math>\overline{EF}</math> intersect <math>\overline{CD}</math> at <math>H</math>. The area of quadrilateral <math>EFGH</math> is
+<math>\text{(A) } 4\quad
+\text{(B) } 4.5\quad
+\text{(C) } 5\quad
+\text{(D) } 5.5\quad
+\text{(E) } 6</math>
 [[1992 AHSME Problems/Problem 24|Solution]]
 == Problem 25 ==
+In <math>\triangle{ABC}</math>, <math>\angle{ABC=120^\circ,AB=3</math> and <math>BC=4</math>. If perpendiculars constructed to <math>\overline{AB}</math> at <math>A</math> and to <math>\overline{BC}</math> at <math>C</math> meet at <math>D</math>, then <math>CD=</math>
+<math>\text{(A) } 3\quad
+\text{(B) } \frac{8}{\sqrt{3}}\quad
+\text{(C) } 5\quad
+\text{(D) } \frac{11}{2}\quad
+\text{(E) } \frac{10}{\sqrt{3}}</math>
 [[1992 AHSME Problems/Problem 25|Solution]]
 == Problem 26 ==
+<asy>
+fill((1,0)--arc((1,0),2,180,225)--cycle,grey);
+fill((-1,0)--arc((-1,0),2,315,360)--cycle,grey);
+fill((0,-1)--arc((0,-1),2-sqrt(2),225,315)--cycle,grey);
+fill((0,0)--arc((0,0),1,180,360)--cycle,white);
+draw((1,0)--arc((1,0),2,180,225)--(1,0),black+linewidth(1));
+draw((-1,0)--arc((-1,0),2,315,360)--(-1,0),black+linewidth(1));
+draw((0,0)--arc((0,0),1,180,360)--(0,0),black+linewidth(1));
+draw(arc((0,-1),2-sqrt(2),225,315),black+linewidth(1));
+draw((0,0)--(0,-1),black+linewidth(1));
+MP("C",(0,0),N);MP("A",(-1,0),N);MP("B",(1,0),N);
+MP("D",(0,-.8),NW);MP("E",(1-sqrt(2),-sqrt(2)),SW);MP("F",(-1+sqrt(2),-sqrt(2)),SE);
+</asy>
+Semicircle <math>\widehat{AB}</math> has center <math>C</math> and radius <math>1</math>. Point <math>D</math> is on <math>\widehat{AB}</math> and <math>\overline{CD}\perp\overline{AB}</math>. Extend <math>\overline{BD}</math> and <math>\overline{AD}</math> to <math>E</math> and <math>F</math>, respectively, so that circular arcs <math>\widehat{AE}</math> and <math>\widehat{BF}</math> have <math>B</math> and <math>A</math> as their respective centers. Circular arc <math>\widehat{EF}</math> has center <math>D</math>. The area of the shaded "smile" <math>AEFBDA</math>, is
+<math>\text{(A) } (2-\sqrt{2})\pi\quad
+\text{(B) } 2\pi-\pi \sqrt{2}-1\quad
+\text{(C) } (1-\frac{\sqrt{2}}{2})\pi\quad\\
+\text{(D) } \frac{5\pi}{2}-\pi\sqrt{2}-1\quad
+\text{(E) } (3-2\sqrt{2})\pi</math>
@@ Line 226: / Line 300: @@
+A circle of radius <math>r</math> has chords <math>\overline{AB}</math> of length <math>10</math> and <math>\overline{CD}</math> of length 7. When <math>\overline{AB}</math> and <math>\overline{CD}</math> are extended through <math>B</math> and <math>C</math>, respectively, they intersect at <math>P</math>, which is outside of the circle. If <math>\angle{APD}=60^\circ</math> and <math>BP=8</math>, then <math>r^2=</math>
+<math>\text{(A) } 70\quad
+\text{(B) } 71\quad
+\text{(C) } 72\quad
+\text{(D) } 73\quad
+\text{(E) } 74</math>
 [[1992 AHSME Problems/Problem 27|Solution]]
 == Problem 28 ==
+Let <math>i=\sqrt{-1}</math>. The product of the real parts of the roots of <math>z^2-z=5-5i</math> is
+<math>\text{(A) } -25\quad
+\text{(B) } -6\quad
+\text{(C) } -5\quad
+\text{(D) } \frac{1}{4}\quad
+\text{(E) } 25</math>
 [[1992 AHSME Problems/Problem 28|Solution]]
 == Problem 29 ==
+An "unfair" coin has a <math>2/3</math> probability of turning up heads. If this coin is tossed <math>50</math> times, what is the probability that the total number of heads is even?
+<math>\text{(A) } 25(\frac{2}{3})^{50}\quad
+\text{(B) } \frac{1}{2}(1-\frac{1}{3^{50}})\quad
+\text{(C) } \frac{1}{2}\quad
+\text{(D) } \frac{1}{2}(1+\frac{1}{3^{50}})\quad
+\text{(E) } \frac{2}{3}</math>
 [[1992 AHSME Problems/Problem 29|Solution]]
@@ Line 240: / Line 334: @@
 == Problem 30 ==
+Let <math>ABCD</math> be an isosceles trapezoid with bases <math>AB=92</math> and <math>CD=19</math>. Suppose <math>AD=BC=x</math> and a circle with center on <math>\overline{AB}</math> is tangent to segments <math>\overline{AD}</math> and <math>\overline{BC}</math>. If <math>m</math> is the smallest possible value of <math>x</math>, then <math>m^2</math>=
+<math>\text{(A) } 1369\quad
+\text{(B) } 1679\quad
+\text{(C) } 1748\quad
+\text{(D) } 2109\quad
+\text{(E) } 8825</math>
 [[1992 AHSME Problems/Problem 30|Solution]]

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Difference between revisions of "1992 AHSME Problems"

Revision as of 02:29, 28 September 2014

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

See also