Difference between revisions of "1994 AJHSME Problems"

Latest revision as of 12:36, 19 February 2020

1994 AJHSME (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 25-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 ? points for each correct answer, ? points for each problem left unanswered, and ? 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 ? minutes working time to complete the test.
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

Problem 1

Which of the following is the largest?

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{3}{8} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{7}{24}$

Solution

Problem 2

$\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$

$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$

Solution

Problem 3

Each day Maria must work $8$ hours. This does not include the $45$ minutes she takes for lunch. If she begins working at $\text{7:25 A.M.}$ and takes her lunch break at noon, then her working day will end at

$\text{(A)}\ \text{3:40 P.M.} \qquad \text{(B)}\ \text{3:55 P.M.} \qquad \text{(C)}\ \text{4:10 P.M.} \qquad \text{(D)}\ \text{4:25 P.M.} \qquad \text{(E)}\ \text{4:40 P.M.}$

Solution

Problem 4

Which of the following represents the result when the figure shown below is rotated clockwise $120^\circ$ around its center?

$[asy] unitsize(6); draw(circle((0,0),5)); draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle); draw(circle((-2.5,-1.5),1)); draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle); [/asy]$

$[asy] unitsize(6); for (int i = 0; i < 5; ++i) { draw(circle((12*i,0),5)); } draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle); draw(circle((-2.5,-1.5),1)); draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle); draw((14,-2)--(16,-2)--(15,-2+sqrt(3))--cycle); draw(circle((12,3),1)); draw((10.5,-1)--(9,0)--(8,-1.5)--(9.5,-2.5)--cycle); draw((22,-2)--(20,-2)--(21,-2+sqrt(3))--cycle); draw(circle((27,-1),1)); draw((24,1.5)--(22.75,2.75)--(24,4)--(25.25,2.75)--cycle); draw((35,2.5)--(37,2.5)--(36,2.5+sqrt(3))--cycle); draw(circle((39,-1),1)); draw((34.5,-1)--(33,0)--(32,-1.5)--(33.5,-2.5)--cycle); draw((50,-2)--(52,-2)--(51,-2+sqrt(3))--cycle); draw(circle((45.5,-1.5),1)); draw((48,1.5)--(46.75,2.75)--(48,4)--(49.25,2.75)--cycle); label("(A)",(0,5),N); label("(B)",(12,5),N); label("(C)",(24,5),N); label("(D)",(36,5),N); label("(E)",(48,5),N); [/asy]$

Solution

Problem 5

Given that $\text{1 mile} = \text{8 furlongs}$ and $\text{1 furlong} = \text{40 rods}$ , the number of rods in one mile is

$\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280$

Solution

Problem 6

The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

Problem 7

If $\angle A = 60^\circ$ , $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$

$[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]$

$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$

Solution

Problem 8

For how many three-digit whole numbers does the sum of the digits equal $25$ ?

$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution

Problem 9

A shopper buys a $100$ dollar coat on sale for $20\%$ off. An additional $5$ dollars are taken off the sale price by using a discount coupon. A sales tax of $8\%$ is paid on the final selling price. The total amount the shopper pays for the coat is

$\text{(A)}\ \text{81.00 dollars} \qquad \text{(B)}\ \text{81.40 dollars} \qquad \text{(C)}\ \text{82.00 dollars} \qquad \text{(D)}\ \text{82.08 dollars} \qquad \text{(E)}\ \text{82.40 dollars}$

Solution

Problem 10

For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?

$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution

Problem 11

Last summer $100$ students attended basketball camp. Of those attending, $52$ were boys and $48$ were girls. Also, $40$ students were from Jonas Middle School and $60$ were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?

$\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52$

Solution

Problem 12

Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?

$[asy] unitsize(36); fill((0,0)--(1,0)--(1,1)--cycle,gray); fill((1,1)--(1,2)--(2,2)--cycle,gray); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((1,0)--(1,2)); draw((0,0)--(2,2)); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray); draw((3,0)--(5,0)--(5,2)--(3,2)--cycle); draw((4,0)--(4,2)); draw((3,1)--(5,1)); fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray); draw((6,0)--(8,0)--(8,2)--(6,2)--cycle); draw((6,0)--(8,2)); draw((6,2)--(8,0)); draw((7,0)--(6,1)--(7,2)--(8,1)--cycle); label("$I$",(1,2),N); label("$II$",(4,2),N); label("$III$",(7,2),N); [/asy]$

$\text{(A)}\ \text{The shaded areas in all three are equal.}$

$\text{(B)}\ \text{Only the shaded areas of }I\text{ and }II\text{ are equal.}$

$\text{(C)}\ \text{Only the shaded areas of }I\text{ and }III\text{ are equal.}$

$\text{(D)}\ \text{Only the shaded areas of }II\text{ and }III\text{ are equal.}$

$\text{(E)}\ \text{The shaded areas of }I, II\text{ and }III\text{ are all different.}$

Solution

Problem 13

The number halfway between $\dfrac{1}{6}$ and $\dfrac{1}{4}$ is

$\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}$

Solution

Problem 14

Two children at a time can play pairball. For $90$ minutes, with only two children playing at a time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36$

Solution

Problem 15

If this path is to continue in the same pattern:

$[asy] unitsize(24); draw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,0)--(5,0)--(5,1)--(6,1)); draw((2/3,1/5)--(1,0)--(2/3,-1/5)); draw((4/5,2/3)--(1,1)--(6/5,2/3)); draw((5/3,6/5)--(2,1)--(5/3,4/5)); draw((9/5,1/3)--(2,0)--(11/5,1/3)); draw((8/3,1/5)--(3,0)--(8/3,-1/5)); draw((14/5,2/3)--(3,1)--(16/5,2/3)); draw((11/3,6/5)--(4,1)--(11/3,4/5)); draw((19/5,1/3)--(4,0)--(21/5,1/3)); draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); draw((17/3,6/5)--(6,1)--(17/3,4/5)); dot((0,0)); dot((1,0)); dot((1,1)); dot((2,1)); dot((2,0)); dot((3,0)); dot((3,1)); dot((4,1)); dot((4,0)); dot((5,0)); dot((5,1)); label("$0$",(0,0),S); label("$1$",(1,0),S); label("$2$",(1,1),N); label("$3$",(2,1),N); label("$4$",(2,0),S); label("$5$",(3,0),S); label("$6$",(3,1),N); label("$7$",(4,1),N); label("$8$",(4,0),S); label("$9$",(5,0),S); label("$10$",(5,1),N); label("$\vdots$",(5.85,0.5),E); label("$\cdots$",(6.5,0.15),S); [/asy]$

then which sequence of arrows goes from point $425$ to point $427$ ?

$[asy] unitsize(24); dot((0,0)); dot((0,1)); dot((1,1)); draw((0,0)--(0,1)--(1,1)); draw((-1/5,2/3)--(0,1)--(1/5,2/3)); draw((2/3,6/5)--(1,1)--(2/3,4/5)); label("(A)",(-1/3,1/3),W); dot((4,0)); dot((5,0)); dot((5,1)); draw((4,0)--(5,0)--(5,1)); draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); label("(B)",(11/3,1/3),W); dot((8,1)); dot((8,0)); dot((9,0)); draw((8,1)--(8,0)--(9,0)); draw((39/5,1/3)--(8,0)--(41/5,1/3)); draw((26/3,1/5)--(9,0)--(26/3,-1/5)); label("(C)",(23/3,1/3),W); dot((12,1)); dot((13,1)); dot((13,0)); draw((12,1)--(13,1)--(13,0)); draw((38/3,6/5)--(13,1)--(38/3,4/5)); draw((64/5,1/3)--(13,0)--(66/5,1/3)); label("(D)",(35/3,1/3),W); dot((17,1)); dot((17,0)); dot((16,0)); draw((17,1)--(17,0)--(16,0)); draw((84/5,1/3)--(17,0)--(86/5,1/3)); draw((49/3,1/5)--(16,0)--(49/3,-1/5)); label("(E)",(47/3,1/3),W); [/asy]$

Solution

Problem 16

The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$

Solution

Problem 17

Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12$

Solution

Problem 18

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip?

$[asy] import graph; unitsize(12); real a(real x) {return ((x-15)^2)/2;} real b(real x) {return ((x-25)^2)/2;} real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;} real d(real x) {return ((x-15)^2)*8/25-15;} real e(real x) {return ((x-25)^2)*8/25-15;} draw((0,9)--(0,0)--(11,0)); draw((15,9)--(15,0)--(26,0)); draw((30,9)--(30,0)--(41,0)); draw((0,-6)--(0,-15)--(11,-15)); draw((15,-6)--(15,-15)--(26,-15)); draw((0,0)--(3,8)--(7,8)--(10,0)); draw(graph(a,15,17)); draw((17,2)--(18,8)--(22,8)--(23,2)); draw(graph(b,23,25)); draw(graph(c,30,40)); draw((0,-15)--(5,-7)--(10,-15)); draw(graph(d,15,20)); draw(graph(e,20,25)); for (int k=0; k<3; ++k) { label("d",(15*k-1,8),N); label("i",(15*k-1,7),N); label("s",(15*k-1,6),N); label("t",(15*k-1,5),N); label("a",(15*k-1,4),N); label("n",(15*k-1,3),N); label("c",(15*k-1,2),N); label("e",(15*k-1,1),N); label("time",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label("d",(15*k-1,8-15),N); label("i",(15*k-1,7-15),N); label("s",(15*k-1,6-15),N); label("t",(15*k-1,5-15),N); label("a",(15*k-1,4-15),N); label("n",(15*k-1,3-15),N); label("c",(15*k-1,2-15),N); label("e",(15*k-1,1-15),N); label("time",(15*k+8,0-15),S); } label("(A)",(5,9),N); label("(B)",(20,9),N); label("(C)",(35,9),N); label("(D)",(5,-6),N); label("(E)",(20,-6),N); [/asy]$

Solution

Problem 19

Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$ , has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is

$[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); [/asy]$

$\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64$

Solution

Problem 20

Let $W,X,Y$ and $Z$ be four different digits selected from the set

$\{ 1,2,3,4,5,6,7,8,9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\text{(A)}\ \dfrac{2}{17} \qquad \text{(B)}\ \dfrac{3}{17} \qquad \text{(C)}\ \dfrac{17}{72} \qquad \text{(D)}\ \dfrac{25}{72} \qquad \text{(E)}\ \dfrac{13}{36}$

Solution

Problem 21

A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$

Solution

Problem 22

The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is

$[asy] draw(circle((0,0),3)); draw(circle((7,0),3)); draw((0,0)--(3,0)); draw((0,-3)--(0,3)); draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2)); draw((7,0)--(7-3*sqrt(3)/2,-3/2)); draw((0,5)--(0,3.5)--(-0.5,4)); draw((0,3.5)--(0.5,4)); draw((7,5)--(7,3.5)--(6.5,4)); draw((7,3.5)--(7.5,4)); label("$3$",(-0.75,0),W); label("$1$",(0.75,0.75),NE); label("$2$",(0.75,-0.75),SE); label("$6$",(6,0.5),NNW); label("$5$",(7,-1),S); label("$4$",(8,0.5),NNE); [/asy]$

$\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}$

Solution

Problem 23

If $X$ , $Y$ and $Z$ are different digits, then the largest possible $3-$ digit sum for

$\begin{tabular}{ccc} X & X & X \\ & Y & X \\ + & & X \\ \hline \end{tabular}$

has the form

$\text{(A)}\ XXY \qquad \text{(B)}\ XYZ \qquad \text{(C)}\ YYX \qquad \text{(D)}\ YYZ \qquad \text{(E)}\ ZZY$

Solution

Problem 24

A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.

$\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$

Solution

Problem 25

Find the sum of the digits in the answer to

$\underbrace{9999\ldots 99}_{94\text{ nines}} \times \underbrace{4444\ldots 44}_{94\text{ fours}}$

where a string of $94$ nines is multiplied by a string of $94$ fours.

$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

Solution

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by 1993 AJHSME	Followed by 1995 AJHSME
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All AJHSME/AMC 8 Problems and Solutions

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Difference between revisions of "1994 AJHSME Problems"

Latest revision as of 12:36, 19 February 2020

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

See also

@@ Line 1: / Line 1: @@
+{{AJHSME Problems
+|year = 1994
+}}
 ==Problem 1==
+Which of the following is the largest?
+<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{3}{8} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{7}{24}</math>
 [[1994 AJHSME Problems/Problem 1|Solution]]
 == Problem 2 ==
+<math>\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=</math>
+<math>\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math>
 [[1994 AJHSME Problems/Problem 2|Solution]]
 == Problem 3 ==
+Each day Maria must work <math>8</math> hours.  This does not include the <math>45</math> minutes she takes for lunch.  If she begins working at <math>\text{7:25 A.M.}</math> and takes her lunch break at noon, then her working day will end at
+<math>\text{(A)}\ \text{3:40 P.M.} \qquad \text{(B)}\ \text{3:55 P.M.} \qquad \text{(C)}\ \text{4:10 P.M.} \qquad \text{(D)}\ \text{4:25 P.M.} \qquad \text{(E)}\ \text{4:40 P.M.}</math>
 [[1994 AJHSME Problems/Problem 3|Solution]]
 == Problem 4 ==
+Which of the following represents the result when the figure shown below is rotated clockwise <math>120^\circ</math> around its center?
+<center>
+<asy>
+unitsize(6);
+draw(circle((0,0),5));
+draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle);
+draw(circle((-2.5,-1.5),1));
+draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle);
+</asy>
+</center>
+<asy>
+unitsize(6);
+for (int i = 0; i < 5; ++i)
+{
+draw(circle((12*i,0),5));
+}
+draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle);
+draw(circle((-2.5,-1.5),1));
+draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle);
+draw((14,-2)--(16,-2)--(15,-2+sqrt(3))--cycle);
+draw(circle((12,3),1));
+draw((10.5,-1)--(9,0)--(8,-1.5)--(9.5,-2.5)--cycle);
+draw((22,-2)--(20,-2)--(21,-2+sqrt(3))--cycle);
+draw(circle((27,-1),1));
+draw((24,1.5)--(22.75,2.75)--(24,4)--(25.25,2.75)--cycle);
+draw((35,2.5)--(37,2.5)--(36,2.5+sqrt(3))--cycle);
+draw(circle((39,-1),1));
+draw((34.5,-1)--(33,0)--(32,-1.5)--(33.5,-2.5)--cycle);
+draw((50,-2)--(52,-2)--(51,-2+sqrt(3))--cycle);
+draw(circle((45.5,-1.5),1));
+draw((48,1.5)--(46.75,2.75)--(48,4)--(49.25,2.75)--cycle);
+label("(A)",(0,5),N);
+label("(B)",(12,5),N);
+label("(C)",(24,5),N);
+label("(D)",(36,5),N);
+label("(E)",(48,5),N);
+</asy>
 [[1994 AJHSME Problems/Problem 4|Solution]]
 == Problem 5 ==
+Given that <math>\text{1 mile} = \text{8 furlongs}</math> and <math>\text{1 furlong} = \text{40 rods}</math>, the number of rods in one mile is
+<math>\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280</math>
 [[1994 AJHSME Problems/Problem 5|Solution]]
 == Problem 6 ==
+The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
+<math>\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
 [[1994 AJHSME Problems/Problem 6|Solution]]
 == Problem 7 ==
+If <math>\angle A = 60^\circ </math>, <math>\angle E = 40^\circ </math> and <math>\angle C = 30^\circ </math>, then <math>\angle BDC = </math>
+<center>
+<asy>
+pair A,B,C,D,EE;
+A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5);
+draw(A--C--D); draw(B--EE--A);
+dot(A); dot(B); dot(C); dot(D); dot(EE);
+label("$A$",A,SW);
+label("$B$",B,S);
+label("$C$",C,SE);
+label("$D$",D,NE);
+label("$E$",EE,N);
+</asy>
+</center>
+<math>\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ</math>
 [[1994 AJHSME Problems/Problem 7|Solution]]
 == Problem 8 ==
+For how many three-digit whole numbers does the sum of the digits equal <math>25</math>?
+<math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math>
 [[1994 AJHSME Problems/Problem 8|Solution]]
 == Problem 9 ==
+A shopper buys a <math>100</math> dollar coat on sale for <math>20\% </math> off.  An additional <math>5</math> dollars are taken off the sale price by using a discount coupon.  A sales tax of <math>8\% </math> is paid on the final selling price.  The total amount the shopper pays for the coat is
+<math>\text{(A)}\ \text{81.00 dollars} \qquad \text{(B)}\ \text{81.40 dollars} \qquad \text{(C)}\ \text{82.00 dollars} \qquad \text{(D)}\ \text{82.08 dollars} \qquad \text{(E)}\ \text{82.40 dollars}</math>
 [[1994 AJHSME Problems/Problem 9|Solution]]
 == Problem 10 ==
+For how many positive integer values of <math>N</math> is the expression <math>\dfrac{36}{N+2}</math> an integer?
+<math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12</math>
 [[1994 AJHSME Problems/Problem 10|Solution]]
 == Problem 11 ==
+Last summer <math>100</math> students attended basketball camp.  Of those attending, <math>52</math> were boys and <math>48</math> were girls.  Also, <math>40</math> students were from Jonas Middle School and <math>60</math> were from Clay Middle School.  Twenty of the girls were from Jonas Middle School.  How many of the boys were from Clay Middle School?
+<math>\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52</math>
 [[1994 AJHSME Problems/Problem 11|Solution]]
 == Problem 12 ==
+Each of the three large squares shown below is the same size.  Segments that intersect the sides of the squares intersect at the midpoints of the sides.  How do the shaded areas of these squares compare?
+<center>
+<asy>
+unitsize(36);
+fill((0,0)--(1,0)--(1,1)--cycle,gray);
+fill((1,1)--(1,2)--(2,2)--cycle,gray);
+draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
+draw((1,0)--(1,2));
+draw((0,0)--(2,2));
+fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray);
+draw((3,0)--(5,0)--(5,2)--(3,2)--cycle);
+draw((4,0)--(4,2));
+draw((3,1)--(5,1));
+fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray);
+draw((6,0)--(8,0)--(8,2)--(6,2)--cycle);
+draw((6,0)--(8,2));
+draw((6,2)--(8,0));
+draw((7,0)--(6,1)--(7,2)--(8,1)--cycle);
+label("$I$",(1,2),N);
+label("$II$",(4,2),N);
+label("$III$",(7,2),N);
+</asy>
+</center>
+<math>\text{(A)}\ \text{The shaded areas in all three are equal.}</math>
+<math>\text{(B)}\ \text{Only the shaded areas of }I\text{ and }II\text{ are equal.}</math>
+<math>\text{(C)}\ \text{Only the shaded areas of }I\text{ and }III\text{ are equal.}</math>
+<math>\text{(D)}\ \text{Only the shaded areas of }II\text{ and }III\text{ are equal.}</math>
+<math>\text{(E)}\ \text{The shaded areas of }I, II\text{ and }III\text{ are all different.}</math>
 [[1994 AJHSME Problems/Problem 12|Solution]]
 == Problem 13 ==
+The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is
+<math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math>
 [[1994 AJHSME Problems/Problem 13|Solution]]
 == Problem 14 ==
+Two children at a time can play pairball.  For <math>90</math> minutes, with only two children playing at a time, five children take turns so that each one plays the same amount of time.  The number of minutes each child plays is
+<math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36</math>
 [[1994 AJHSME Problems/Problem 14|Solution]]
 == Problem 15 ==
+If this path is to continue in the same pattern:
+<center>
+<asy>
+unitsize(24);
+draw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,0)--(5,0)--(5,1)--(6,1));
+draw((2/3,1/5)--(1,0)--(2/3,-1/5)); draw((4/5,2/3)--(1,1)--(6/5,2/3)); draw((5/3,6/5)--(2,1)--(5/3,4/5)); draw((9/5,1/3)--(2,0)--(11/5,1/3));
+draw((8/3,1/5)--(3,0)--(8/3,-1/5)); draw((14/5,2/3)--(3,1)--(16/5,2/3)); draw((11/3,6/5)--(4,1)--(11/3,4/5)); draw((19/5,1/3)--(4,0)--(21/5,1/3));
+draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); draw((17/3,6/5)--(6,1)--(17/3,4/5));
+dot((0,0)); dot((1,0)); dot((1,1)); dot((2,1)); dot((2,0)); dot((3,0)); dot((3,1)); dot((4,1)); dot((4,0)); dot((5,0)); dot((5,1));
+label("$0$",(0,0),S);
+label("$1$",(1,0),S);
+label("$2$",(1,1),N);
+label("$3$",(2,1),N);
+label("$4$",(2,0),S);
+label("$5$",(3,0),S);
+label("$6$",(3,1),N);
+label("$7$",(4,1),N);
+label("$8$",(4,0),S);
+label("$9$",(5,0),S);
+label("$10$",(5,1),N);
+label("$\vdots$",(5.85,0.5),E);
+label("$\cdots$",(6.5,0.15),S);
+</asy>
+</center>
+then which sequence of arrows goes from point <math>425</math> to point <math>427</math>?
+<asy>
+unitsize(24);
+dot((0,0)); dot((0,1)); dot((1,1));
+draw((0,0)--(0,1)--(1,1));
+draw((-1/5,2/3)--(0,1)--(1/5,2/3));
+draw((2/3,6/5)--(1,1)--(2/3,4/5));
+label("(A)",(-1/3,1/3),W);
+dot((4,0)); dot((5,0)); dot((5,1));
+draw((4,0)--(5,0)--(5,1));
+draw((14/3,1/5)--(5,0)--(14/3,-1/5));
+draw((24/5,2/3)--(5,1)--(26/5,2/3));
+label("(B)",(11/3,1/3),W);
+dot((8,1)); dot((8,0)); dot((9,0));
+draw((8,1)--(8,0)--(9,0));
+draw((39/5,1/3)--(8,0)--(41/5,1/3));
+draw((26/3,1/5)--(9,0)--(26/3,-1/5));
+label("(C)",(23/3,1/3),W);
+dot((12,1)); dot((13,1)); dot((13,0));
+draw((12,1)--(13,1)--(13,0));
+draw((38/3,6/5)--(13,1)--(38/3,4/5));
+draw((64/5,1/3)--(13,0)--(66/5,1/3));
+label("(D)",(35/3,1/3),W);
+dot((17,1)); dot((17,0)); dot((16,0));
+draw((17,1)--(17,0)--(16,0));
+draw((84/5,1/3)--(17,0)--(86/5,1/3));
+draw((49/3,1/5)--(16,0)--(49/3,-1/5));
+label("(E)",(47/3,1/3),W);
+</asy>
 [[1994 AJHSME Problems/Problem 15|Solution]]
 == Problem 16 ==
+The perimeter of one square is <math>3</math> times the perimeter of another square.  The area of the larger square is how many times the area of the smaller square?
+<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9</math>
 [[1994 AJHSME Problems/Problem 16|Solution]]
 == Problem 17 ==
+Pauline Bunyan can shovel snow at the rate of <math>20</math> cubic yards for the first hour, <math>19</math> cubic yards for the second, <math>18</math> for the third, etc., always shoveling one cubic yard less per hour than the previous hour.  If her driveway is <math>4</math> yards wide, <math>10</math> yards long, and covered with snow <math>3</math> yards deep, then the number of hours it will take her to shovel it clean is closest to
+<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12</math>
 [[1994 AJHSME Problems/Problem 17|Solution]]
 == Problem 18 ==
+Mike leaves home and drives slowly east through city traffic.  When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops.  He shops at the mall for an hour.  Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic.  Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis.  Which graph is the best representation of Mike's trip?
+<center>
+<asy>
+import graph;
+unitsize(12);
+real a(real x) {return ((x-15)^2)/2;}
+real b(real x) {return ((x-25)^2)/2;}
+real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;}
+real d(real x) {return ((x-15)^2)*8/25-15;}
+real e(real x) {return ((x-25)^2)*8/25-15;}
+draw((0,9)--(0,0)--(11,0));
+draw((15,9)--(15,0)--(26,0));
+draw((30,9)--(30,0)--(41,0));
+draw((0,-6)--(0,-15)--(11,-15));
+draw((15,-6)--(15,-15)--(26,-15));
+draw((0,0)--(3,8)--(7,8)--(10,0));
+draw(graph(a,15,17));
+draw((17,2)--(18,8)--(22,8)--(23,2));
+draw(graph(b,23,25));
+draw(graph(c,30,40));
+draw((0,-15)--(5,-7)--(10,-15));
+draw(graph(d,15,20));
+draw(graph(e,20,25));
+for (int k=0; k<3; ++k)
+{
+label("d",(15*k-1,8),N); label("i",(15*k-1,7),N); label("s",(15*k-1,6),N); label("t",(15*k-1,5),N); label("a",(15*k-1,4),N); label("n",(15*k-1,3),N); label("c",(15*k-1,2),N); label("e",(15*k-1,1),N);
+label("time",(15*k+8,0),S);
+}
+for (int k=0; k<2; ++k)
+{
+label("d",(15*k-1,8-15),N); label("i",(15*k-1,7-15),N); label("s",(15*k-1,6-15),N); label("t",(15*k-1,5-15),N); label("a",(15*k-1,4-15),N); label("n",(15*k-1,3-15),N); label("c",(15*k-1,2-15),N); label("e",(15*k-1,1-15),N);
+label("time",(15*k+8,0-15),S);
+}
+label("(A)",(5,9),N); label("(B)",(20,9),N); label("(C)",(35,9),N); label("(D)",(5,-6),N); label("(E)",(20,-6),N);
+</asy>
+</center>
 [[1994 AJHSME Problems/Problem 18|Solution]]
 == Problem 19 ==
+Around the outside of a <math>4</math> by <math>4</math> square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters.  Another square, <math>ABCD</math>, has its sides parallel to the corresponding sides of the original square, and each side of <math>ABCD</math> is tangent to one of the semicircles.  The area of the square <math>ABCD</math> is
+<center>
+<asy>
+pair A,B,C,D;
+A = origin; B = (4,0); C = (4,4); D = (0,4);
+draw(A--B--C--D--cycle);
+draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW));
+draw((1,1)--(3,1)--(3,3)--(1,3)--cycle);
+dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3));
+label("$A$",A,SW);
+label("$B$",B,SE);
+label("$C$",C,NE);
+label("$D$",D,NW);
+</asy>
+</center>
+<math>\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64</math>
 [[1994 AJHSME Problems/Problem 19|Solution]]
 == Problem 20 ==
+Let <math>W,X,Y</math> and <math>Z</math> be four different digits selected from the set
+<center><math>\{ 1,2,3,4,5,6,7,8,9\}.</math></center>
+If the sum <math>\dfrac{W}{X} + \dfrac{Y}{Z}</math> is to be as small as possible, then <math>\dfrac{W}{X} + \dfrac{Y}{Z}</math> must equal
+<math>\text{(A)}\ \dfrac{2}{17} \qquad \text{(B)}\ \dfrac{3}{17} \qquad \text{(C)}\ \dfrac{17}{72} \qquad \text{(D)}\ \dfrac{25}{72} \qquad \text{(E)}\ \dfrac{13}{36}</math>
 [[1994 AJHSME Problems/Problem 20|Solution]]
 == Problem 21 ==
+A gumball machine contains <math>9</math> red, <math>7</math> white, and <math>8</math> blue gumballs.  The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
+<math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math>
 [[1994 AJHSME Problems/Problem 21|Solution]]
 == Problem 22 ==
+The two wheels shown below are spun and the two resulting numbers are added.  The probability that the sum is even is
+<center>
+<asy>
+draw(circle((0,0),3));
+draw(circle((7,0),3));
+draw((0,0)--(3,0));
+draw((0,-3)--(0,3));
+draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2));
+draw((7,0)--(7-3*sqrt(3)/2,-3/2));
+draw((0,5)--(0,3.5)--(-0.5,4));
+draw((0,3.5)--(0.5,4));
+draw((7,5)--(7,3.5)--(6.5,4));
+draw((7,3.5)--(7.5,4));
+label("$3$",(-0.75,0),W);
+label("$1$",(0.75,0.75),NE);
+label("$2$",(0.75,-0.75),SE);
+label("$6$",(6,0.5),NNW);
+label("$5$",(7,-1),S);
+label("$4$",(8,0.5),NNE);
+</asy>
+</center>
+<math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math>
 [[1994 AJHSME Problems/Problem 22|Solution]]
 == Problem 23 ==
+If <math>X</math>, <math>Y</math> and <math>Z</math> are different digits, then the largest possible <math>3-</math>digit sum for
+<center>
+<math>\begin{tabular}{ccc}
+X & X & X \\
+ & Y & X \\
++ & & X \\ \hline
+\end{tabular}</math>
+</center>
+has the form
+<math>\text{(A)}\ XXY \qquad \text{(B)}\ XYZ \qquad \text{(C)}\ YYX \qquad \text{(D)}\ YYZ \qquad \text{(E)}\ ZZY</math>
 [[1994 AJHSME Problems/Problem 23|Solution]]
 == Problem 24 ==
+A <math>2</math> by <math>2</math> square is divided into four <math>1</math> by <math>1</math> squares.  Each of the small squares is to be painted either green or red.  In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square?  There may be as few as zero or as many as four small green squares.
+<math>\text{(A)}\  4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16</math>
 [[1994 AJHSME Problems/Problem 24|Solution]]
 == Problem 25 ==
+Find the sum of the digits in the answer to
+<center>
+<math>\underbrace{9999\ldots 99}_{94\text{ nines}} \times \underbrace{4444\ldots 44}_{94\text{ fours}}</math>
+</center>
+where a string of <math>94</math> nines is multiplied by a string of <math>94</math> fours.
+<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math>
 [[1994 AJHSME Problems/Problem 25|Solution]]
@@ Line 104: / Line 435: @@
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}