Difference between revisions of "1995 AHSME Problems"

Latest revision as of 21:44, 26 May 2021

1995 AHSME (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 30-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 5 points for each correct answer, 2 points for each problem left unanswered, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have 90 minutes working time to complete the test.
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Problem 1

Kim earned scores of 87, 83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will

$\mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} }$

@@ Line 1: / Line 1: @@
+{{AHSME Problems
+|year = 1995
+}}
 == Problem 1 ==
-Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will
+Kim earned scores of 87, 83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will
 <math> \mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} }  </math>
-[[1995 AMC 12 Problems/Problem 1|Solution]]
+[[1995 AHSME Problems/Problem 1|Solution]]
 == Problem 2 ==
 If <math>\sqrt {2 + \sqrt {x}} = 3</math>, then <math>x =</math>
 <math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \sqrt{7} } \qquad \mathrm{(C) \ 7 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 121 }  </math>
-[[1995 AMC 12 Problems/Problem 2|Solution]]
+[[1995 AHSME Problems/Problem 2|Solution]]
 == Problem 3 ==
-The total in-store price for an appliance is <math>\</math>99.99. A television commercial advertises the same product for three easy payments of <math>\</math>29.98 and a one-time shipping and handling charge of <math>\</math>9.98. How much is saved by buying the appliance from the television advertiser?
+The total in-store price for an appliance is <math>\textdollar 99.99</math>. A television commercial advertises the same product for three easy payments of <math>\textdollar 29.98</math> and a one-time shipping and handling charge of <math>\textdollar 9.98</math>. How many cents are saved by buying the appliance from the television advertiser?
 <math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  </math>
-[[1995 AMC 12 Problems/Problem 3|Solution]]
+[[1995 AHSME Problems/Problem 3|Solution]]
 == Problem 4 ==
 If <math>M</math> is <math>30 \%</math> of <math>Q</math>, <math>Q</math> is <math>20 \%</math> of <math>P</math>, and <math>N</math> is <math>50 \%</math> of <math>P</math>, then <math>\frac {M}{N} =</math>
 <math> \mathrm{(A) \ \frac {3}{250} } \qquad \mathrm{(B) \ \frac {3}{25} } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} }  </math>
-[[1995 AMC 12 Problems/Problem 4|Solution]]
+[[1995 AHSME Problems/Problem 4|Solution]]
 == Problem 5 ==
 A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is
-A. 500 thousand
+<math> \mathrm{(A) \ \text{500 thousand} } \qquad \mathrm{(B) \ \text{5 million} } \qquad \mathrm{(C) \ \text{50 million} } \qquad \mathrm{(D) \ \text{500 million} } \qquad \mathrm{(E) \ \text{5 billion} }  </math>
-B. 5 million
-C. 50 million
-D. 500 million
-E. 5 billion
-<math> \mathrm{(A) \ 500 \text{thousand} } \qquad \mathrm{(B) \ 5 \text{million} } \qquad \mathrm{(C) \ 50 \text{million} } \qquad \mathrm{(D) \ 500 \text{million} } \qquad \mathrm{(E) \ 5 \text{billion} }  </math>
-[[1995 AMC 12 Problems/Problem 5|Solution]]
+[[1995 AHSME Problems/Problem 5|Solution]]
 == Problem 6 ==
-The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked ?
+The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked <math>x</math>?
-{{image}}
+<asy>
+defaultpen(linewidth(0.7));
+path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3);
+draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin);
+draw(shift(1,0)*p, dashed);
+label("$x$", (0.3,0.5), E);
+label("$A$", (1.3,0.5), E);
+label("$B$", (1.3,1.5), E);
+label("$C$", (2.3,1.5), E);
+label("$D$", (2.3,2.5), E);
+label("$E$", (3.3,2.5), E);</asy>
 <math> \mathrm{(A) \ A } \qquad \mathrm{(B) \ B } \qquad \mathrm{(C) \ C } \qquad \mathrm{(D) \ D } \qquad \mathrm{(E) \ E }  </math>
-[[1995 AMC 12 Problems/Problem 6|Solution]]
+[[1995 AHSME Problems/Problem 6|Solution]]
 == Problem 7 ==
-The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:
+The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a negligible height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:
+<math> \mathrm{(A) \ 8 } \qquad \mathrm{(B) \ 25 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 75 } \qquad \mathrm{(E) \ 100 }  </math>
-<math> \mathrm{(A) \ 25 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 75 } \qquad \mathrm{(D) \ 50 } \qquad \mathrm{(E) \ 100 }  </math>
+[[1995 AHSME Problems/Problem 7|Solution]]
-[[1995 AMC 12 Problems/Problem 7|Solution]]
 == Problem 8 ==
 In <math>\triangle ABC</math>, <math>\angle C = 90^\circ, AC = 6</math> and <math>BC = 8</math>. Points <math>D</math> and <math>E</math> are on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, and <math>\angle BED = 90^\circ</math>. If <math>DE = 4</math>, then <math>BD =</math>
+<asy>
-A. 5
+size(100); pathpen = linewidth(0.7); pointpen = black+linewidth(3);
-B. <math>\frac {16}{3}</math>
+pair A = (0,0), C = (6,0), B = (6,8), D = (2*A+B)/3, E = (2*C+B)/3; D(D("A",A,SW)--D("B",B,NW)--D("C",C,SE)--cycle); D(D("D",D,NW)--D("E",E,plain.E)); D(rightanglemark(D,E,B,16)); D(rightanglemark(A,C,B,16));
-C. <math>\frac {20}{3}</math>
+</asy>
-D. <math>\frac {15}{2}</math>
-E. 8
 <math> \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }  </math>
-[[1995 AMC 12 Problems/Problem 8|Solution]]
+[[1995 AHSME Problems/Problem 8|Solution]]
 == Problem 9 ==
 Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is
+<asy>
+size(100); defaultpen(linewidth(0.7)); draw(unitsquare^^(0,0)--(1,1)^^(0,1)--(1,0)^^(.5,0)--(.5,1)^^(0,.5)--(1,.5));
+</asy>
 <math> \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 }  </math>
-[[1995 AMC 12 Problems/Problem 9|Solution]]
+[[1995 AHSME Problems/Problem 9|Solution]]
 == Problem 10 ==
 The area of the triangle bounded by the lines <math>y = x, y = - x</math> and <math>y = 6</math> is
-A. 12
-B. <math>12 \sqrt 2</math>
-C. 24
-D. <math>24 \sqrt 2</math>
-E. 36
 <math> \mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 12\sqrt{2} } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 24\sqrt{2} } \qquad \mathrm{(E) \ 36 }  </math>
-[[1995 AMC 12 Problems/Problem 10|Solution]]
+[[1995 AHSME Problems/Problem 10|Solution]]
 == Problem 11 ==
 How many base 10 four-digit numbers, <math>N = \underline{a} \underline{b} \underline{c} \underline{d}</math>, satisfy all three of the following conditions?
-(i) <math>4,000 \leq N < 6,000;</math> (ii) <math>N</math> is a multiple of 5; (iii) <math>3 \leq b < c \leq 6</math>.
+<math>\text{(i)}</math> <math>4,000 \leq N < 6,000;</math>
+<math>\text{(ii)}</math> <math>N</math> <math>\text{is a multiple of 5}</math>;
+<math>\text{(iii)}</math> <math>3 \leq b < c \leq 6</math>.
-A. 10
+<math> \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \ 48 }  </math>
-B. 18
-C. 24
-D. 36
-E. 48
-[[1995 AMC 12 Problems/Problem 11|Solution]]
+[[1995 AHSME Problems/Problem 11|Solution]]
 == Problem 12 ==
 Let <math>f</math> be a linear function with the properties that <math>f(1) \leq f(2), f(3) \geq f(4),</math> and <math>f(5) = 5</math>. Which of the following is true?
-A. <math>f(0) < 0</math>
+<math> \mathrm{(A) \ f(0) < 0 } \qquad \mathrm{(B) \ f(0) = 0 } \qquad \mathrm{(C) \ f(1) < f(0) < f( - 1) } \qquad \mathrm{(D) \ f(0) = 5 } \qquad \mathrm{(E) \ f(0) > 5 }  </math>
-B. <math>f(0) = 0</math>
-C. <math>f(1) < f(0) < f( - 1)</math>
-D. <math>f(0) = 5</math>
-E. <math>f(0) > 5</math>
-[[1995 AMC 12 Problems/Problem 12|Solution]]
+[[1995 AHSME Problems/Problem 12|Solution]]
 == Problem 13 ==
 The addition below is incorrect. The display can be made correct by changing one digit <math>d</math>, wherever it occurs, to another digit <math>e</math>. Find the sum of <math>d</math> and <math>e</math>.
-{{image}}
+<math>\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}</math>
-Another table... Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=373460#373460
-A. 4
+<math>\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }</math>
-B. 6
-C. 8
-D. 10
-E. More than 10
-[[1995 AMC 12 Problems/Problem 13|Solution]]
+[[1995 AHSME Problems/Problem 13|Solution]]
 == Problem 14 ==
 If <math>f(x) = ax^4 - bx^2 + x + 5</math> and <math>f( - 3) = 2</math>, then <math>f(3) =</math>
-A. -5
+<math> \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }  </math>
-B. -2
-C. 1
-D. 3
-E. 8
-[[1995 AMC 12 Problems/Problem 14|Solution]]
+[[1995 AHSME Problems/Problem 14|Solution]]
 == Problem 15 ==
 Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point
+<asy>
+size(80); defaultpen(linewidth(0.7)+fontsize(10)); draw(unitcircle);
+for(int i = 0; i < 5; ++i) { pair P = dir(90-i*72); dot(P); label("$"+string(i+1)+"$",P,1.4*P); }
+</asy>
+<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }  </math>
-A. 1
+[[1995 AHSME Problems/Problem 15|Solution]]
-B. 2
-C. 3
-D. 4
-E. 5
-[[1995 AMC 12 Problems/Problem 15|Solution]]
 == Problem 16 ==
-Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:
+Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games notes that:
 i. The actual attendance in Atlanta is within <math>10 \%</math> of Anita's estimate.
@@ Line 165: / Line 144: @@
 To the nearest 1,000, the largest possible difference between the numbers attending the two games is
-A. 10,000
+<math> \mathrm{(A) \ 10000 } \qquad \mathrm{(B) \ 11000 } \qquad \mathrm{(C) \ 20000 } \qquad \mathrm{(D) \ 21000 } \qquad \mathrm{(E) \ 22000 }  </math>
-B. 11,000
-C. 20,000
-D. 21,000
-E. 22,000
-[[1995 AMC 12 Problems/Problem 16|Solution]]
+[[1995 AHSME Problems/Problem 16|Solution]]
 == Problem 17 ==
 Given regular pentagon <math>ABCDE</math>, a circle can be drawn that is tangent to <math>\overline{DC}</math> at <math>D</math> and to <math>\overline{AB}</math> at <math>A</math>. The number of degrees in minor arc <math>AD</math> is
-A. 72
+<asy>
-B. 108
+defaultpen(linewidth(0.7));
-C. 120
+draw(rotate(18)*polygon(5));
-D. 135
+real x=0.6180339887;
-E. 144
+draw(Circle((-x,0), 1));
+int i;
+for(i=0; i<5; i=i+1) {
+dot(origin+1*dir(36+72*i));
+}
+label("$B$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0)));
+label("$A$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72)));
+label("$E$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144)));
+label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3)));
+label("$C$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4)));</asy>
-{{image}}
+<math> \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }  </math>
-[[1955 AMC 12 Problems/Problem 17|Solution]]
+[[1995 AHSME Problems/Problem 17|Solution]]
 == Problem 18 ==
-Two rays with common endpoint <math>O</math> forms a <math>30^\circ</math> angle. Point <math>A</math> lies on one ray, point <math>B</math> on the other ray, and <math>AB = 1</math>. The maximum possible length of <math>OB</math> is
+Two rays with common endpoint <math>O</math> form a <math>30^\circ</math> angle. Point <math>A</math> lies on one ray, point <math>B</math> on the other ray, and <math>AB = 1</math>. The maximum possible length of <math>OB</math> is
-A. 1
+<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ \frac{4}{\sqrt{3}} }  </math>
-B. <math>\frac {1 + \sqrt {3}}{\sqrt 2}</math>
-C. <math>\sqrt 3</math>
-D. 2
-E. <math>\frac {4}{\sqrt 3}</math>
-[[1995 AMC 12 Problems/Problem 18|Solution]]
+[[1995 AHSME Problems/Problem 18|Solution]]
 == Problem 19 ==
-Equilateral triangle <math>DEF</math> is inscribed in equilateral triangle <math>ABC</math> such that <math>\overline{DE} \perp \overline{BC}</math>. The reatio of the area of <math>\triangle DEF</math> to the area of <math>\triangle ABC</math> is
+Equilateral triangle <math>DEF</math> is inscribed in equilateral triangle <math>ABC</math> such that <math>\overline{DE} \perp \overline{BC}</math>. The ratio of the area of <math>\triangle DEF</math> to the area of <math>\triangle ABC</math> is
+<asy>
-A. <math>\frac {1}{6}</math>
+pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
+pair B = (0,0), C = (1,0), A = dir(60), D = C*2/3, E = (2*A+C)/3, F = (2*B+A)/3;
-B. <math>\frac {1}{4}</math>
+D(D("A",A,N)--D("B",B,SW)--D("C",C,SE)--cycle); D(D("D",D)--D("E",E,NE)--D("F",F,NW)--cycle); D(rightanglemark(C,D,E,1.5));
+</asy>
-C. <math>\frac {1}{3}</math>
+<math> \mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} }  </math>
-D. <math>\frac {2}{5}</math>
-E. <math>\frac {1}{2}</math>
+[[1995 AHSME Problems/Problem 19|Solution]]
-[[1995 AMC 12 Problems/Problem 19|Solution]]
 == Problem 20 ==
 If <math>a,b</math> and <math>c</math> are three (not necessarily different) numbers chosen randomly and with replacement from the set <math>\{1,2,3,4,5 \}</math>, the probability that <math>ab + c</math> is even is
 <math> \mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } \qquad \mathrm{(E) \ \frac {3}{5} }  </math>
-[[1995 AMC 12 Problems/Problem 20|Solution]]
+[[1995 AHSME Problems/Problem 20|Solution]]
 == Problem 21 ==
-Two nonadjacent vertices of a rectangle are (4,3) and (-4,-3), and the coordinates of the other two vertices are integers. The number of such rectangles is
+Two nonadjacent vertices of a rectangle are <math>(4,3)</math> and <math>(-4,-3)</math>, and the coordinates of the other two vertices are integers. The number of such rectangles is
-A. 1
+<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }  </math>
-B. 2
-C. 3
-D. 4
-E. 5
-[[1995 AMC 12 Problems/Problem 21|Solution]]
+[[1995 AHSME Problems/Problem 21|Solution]]
 == Problem 22 ==
-A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths 13,19,20,25 and 31, although this is not necessarily their order around the pentagon. The area of the pentagon is
+A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths 13, 19, 20, 25 and 31, although this is not necessarily their order around the pentagon. The area of the pentagon is
 <math> \mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }  </math>
-[[1995 AMC 12 Problems/Problem 22|Solution]]
+[[1995 AHSME Problems/Problem 22|Solution]]
 == Problem 23 ==
-The sides of a triangle have lengths 11,15, and <math>k</math>, where <math>k</math> is an integer. For how many values of <math>k</math> is the triangle obtuse?
+The sides of a triangle have lengths 11, 15, and <math>k</math>, where <math>k</math> is an integer. For how many values of <math>k</math> is the triangle obtuse?
 <math> \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 }  </math>
-[[1995 AMC 12 Problems/Problem 23|Solution]]
+[[1995 AHSME Problems/Problem 23|Solution]]
 == Problem 24 ==
@@ Line 253: / Line 222: @@
 What is <math>A + B + C</math>?
 <math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }  </math>
-[[1995 AMC 12 Problems/Problem 24|Solution]]
+[[1995 AHSME Problems/Problem 24|Solution]]
 == Problem 25 ==
-A list of five positive integers has mean 12 and range 18. The mode and median are both 8. How many different values are possible for the second largest element of the list?
+A list of five positive integers has mean 12 and range 18. The mode and median are both 8. How many different values are possible for the second- largest element of the list?
 <math> \mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }  </math>
-[[1995 AMC 12 Problems/Problem 25|Solution]]
+[[1995 AHSME Problems/Problem 25|Solution]]
 == Problem 26 ==
 In the figure, <math>\overline{AB}</math> and <math>\overline{CD}</math> are diameters of the circle with center <math>O</math>, <math>\overline{AB} \perp \overline{CD}</math>, and chord <math>\overline{DF}</math> intersects <math>\overline{AB}</math> at <math>E</math>. If <math>DE = 6</math> and <math>EF = 2</math>, then the area of the circle is
+<asy>
-{{image}}
+defaultpen(linewidth(0.7));
+draw(Circle(origin, 5));
-Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62477
+pair O=origin, A=(-5,0), B=(5,0), C=(0,5), D=(0,-5), F=5*dir(40), E=intersectionpoint(A--B, F--D);
+draw(A--B^^C--D--F);
+dot(O^^A^^B^^C^^D^^E^^F);
+markscalefactor=0.05;
+draw(rightanglemark(B, O, D));
+label("$A$", A, dir(O--A));
+label("$B$", B, dir(O--B));
+label("$C$", C, dir(O--C));
+label("$D$", D, dir(O--D));
+label("$F$", F, dir(O--F));
+label("$O$", O, NW);
+label("$E$", E, SE);</asy>
 <math> \mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi }  </math>
+[[1995 AHSME Problems/Problem 26|Solution]]
-[[1995 AMC 12 Problems/Problem 26|Solution]]
 == Problem 27 ==
 Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.
-{{image}}
+<cmath>\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\
+& & & & 1 & & 1 & & & & \\
-A table really needs to go there... Link: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1526011879&t=62471
+& & & 2 & & 2 & & 2 & & & \\
+& & 3 & & 4 & & 4 & & 3 & & \\
+& 4 & & 7 & & 8 & & 7 & & 4 & \\
+& & 11 & & 15 & & 15 & & 11 & & 5  \end{tabular}</cmath>
 Let <math>f(n)</math> denote the sum of the numbers in row <math>n</math>. What is the remainder when <math>f(100)</math> is divided by 100?
 <math> \mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 }  </math>
-[[1995 AMC 12 Problems/Problem 27|Solution]]
+[[1995 AHSME Problems/Problem 27|Solution]]
 == Problem 28 ==
 Two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6. The chord parallel to these chords and midway between them is of length <math>\sqrt {a}</math> where <math>a</math> is
+<asy>
+// note: diagram deliberately not to scale -- azjps
+void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0));  }
+size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3);
+real min = -0.6, step = 0.5;
+pair[] A, B; D(unitcircle);
+for(int i = 0; i < 3; ++i) {
+ A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[0]); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[1]);
+ D(D(A[i])--D(B[i]));
+}
+MP("10",(A[0]+B[0])/2,N);
+MP("\sqrt{a}",(A[1]+B[1])/2,N);
+MP("14",(A[2]+B[2])/2,N);
+htick((B[1].x+0.1,B[0].y),(B[1].x+0.1,B[2].y),0.06); MP("6",(B[1].x+0.1,B[0].y/2+B[2].y/2),E);
+</asy>
 <math> \mathrm{(A) \ 144 } \qquad \mathrm{(B) \ 156 } \qquad \mathrm{(C) \ 168 } \qquad \mathrm{(D) \ 176 } \qquad \mathrm{(E) \ 184 }  </math>
-[[1995 AMC 12 Problems/Problem 28|Solution]]
+[[1995 AHSME Problems/Problem 28|Solution]]
 == Problem 29 ==
-For how many three-element sets of positive integers <math>\{a,b,c\}</math> is it true that <math>a \times b \times c = 2310</math>?
+For how many three-element sets of distinct positive integers <math>\{a,b,c\}</math> is it true that <math>a \times b \times c = 2310</math>?
 <math> \mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }  </math>
-[[1995 AMC 12 Problems/Problem 29|Solution]]
+[[1995 AHSME Problems/Problem 29|Solution]]
 == Problem 30 ==
 A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
+<asy>
+size(150); defaultpen(linewidth(0.7)); pair slant = (2,1);
+for(int i = 0; i < 4; ++i)
+ draw((0,i)--(3,i)^^(i,0)--(i,3)^^(3,i)--(3,i)+slant^^(i,3)--(i,3)+slant);
+for(int i = 1; i < 4; ++i)
+ draw((0,3)+slant*i/3--(3,3)+slant*i/3^^(3,0)+slant*i/3--(3,3)+slant*i/3);
+</asy>
 <math> \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }  </math>
-[[1995 AMC 12 Problems/Problem 30|Solution]]
+[[1995 AHSME Problems/Problem 30|Solution]]
 == See also ==
-* [[AMC 12]]
 * [[AMC 12 Problems and Solutions]]
-* [[1995 AMC 12]]
 * [[Mathematics competition resources]]
+{{AHSME box|year=1995|before=[[1994 AHSME]]|after=[[1996 AHSME]]}}
+{{MAA Notice}}

1995 AHSME (Problems • Answer Key • Resources)
Preceded by 1994 AHSME	Followed by 1996 AHSME
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All AHSME Problems and Solutions

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Difference between revisions of "1995 AHSME Problems"

Latest revision as of 21:44, 26 May 2021

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