ΒΆ 1984 AHSME Problems and SolutionsIndividual Problems and Solutions
For problems and detailed solutions to each of the 1984 AHSME problems, please refer below:
Problem 1: 100 0 2 25 2 2 β 24 8 2 2 5 2 2 β 2 4 8 2 1 0 0 0 2 β
Answer Choices:
A. 62 , 500 6 2 , 5 0 0 1000 1 0 0 0 500 5 0 0 250 2 5 0 1 2 2 1 β
Solution:
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Problem 2: If x , y x , y y β 1 x y β x 1 β 0 0
x β 1 y y β 1 x equals y β x 1 β x β y 1 β β equals
Answer Choices:
A. 1 1 x y y x β y x x y β x y β y x y x β β x y β x y β 1 x y x y β x y 1 β
Solution:
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Problem 3: Let n n 1 1 10 1 0
Answer Choices:
A. 100 < n β€ 110 1 0 0 < n β€ 1 1 0 110 < n β€ 120 1 1 0 < n β€ 1 2 0 120 < n β€ 130 1 2 0 < n β€ 1 3 0 130 < n β€ 140 1 3 0 < n β€ 1 4 0 140 < n β€ 150 1 4 0 < n β€ 1 5 0
Solution:
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Problem 4: A rectangle intersects a circle as shown: A B = 4 , B C = 5 A B = 4 , B C = 5 D E = 3 D E = 3 E F E F
Answer Choices:
A. 6 6 7 7 20 3 3 2 0 β 8 8 9 9
Solution:
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Problem 5: The largest integer n n n 200 < 5 300 n 2 0 0 < 5 3 0 0
Answer Choices:
A. 8 8 9 9 10 1 0 11 1 1 12 1 2
Solution:
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Problem 6: In a certain school, there are three times as many boys as girls and nine times as many girls as teachers. Using the letters b , g , t b , g , t
Answer Choices:
A. 31 b 3 1 b 37 27 b 2 7 3 7 β b 13 g 1 3 g 37 27 g 2 7 3 7 β g 37 27 t 2 7 3 7 β t
Solution:
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Problem 7: When Dave walks to school, he averages 90 9 0 75 7 5 16 1 6 100 1 0 0 60 6 0
Answer Choices:
A. 14 2 9 m i n 1 4 9 2 β m i n 15 1 5 18 1 8 20 2 0 22 2 9 m i n 2 2 9 2 β m i n
Solution:
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Problem 8: Figure A B C D A B C D A B β₯ D C , A B = 5 , B C = 3 2 A B β₯ D C , A B = 5 , B C = 3 2 β β B C D = 4 5 β β B C D = 4 5 β β C D A = 6 0 β β C D A = 6 0 β D C D C
Answer Choices:
A. 7 + 2 3 3 7 + 3 2 β 3 β 8 8 9 1 2 9 2 1 β 8 + 3 8 + 3 β 8 + 3 3 8 + 3 3 β
Solution:
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Problem 9: The number of digits in 4 16 5 25 4 1 6 5 2 5 10 1 0
Answer Choices:
A. 31 3 1 30 3 0 29 2 9 28 2 8 27 2 7
Solution:
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Problem 10: Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are 1 + 2 i , β 2 + i 1 + 2 i , β 2 + i β 1 β 2 i β 1 β 2 i
Answer Choices:
A. 2 + i 2 + i 2 β i 2 β i 1 β 2 i 1 β 2 i β 1 + 2 i β 1 + 2 i β 2 β i β 2 β i
Solution:
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Problem 11: A calculator has a key which replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let y y x β 0 x ξ = 0 n n y y
Answer Choices:
A. x ( ( β 2 ) n ) x ( ( β 2 ) n ) x 2 n x 2 n x β 2 n x β 2 n x β ( 2 n ) x β ( 2 n ) x ( ( β 1 ) n 2 n ) x ( ( β 1 ) n 2 n )
Solution:
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Problem 12: If the sequence { a n } { a n β }
a 1 = 2 a n + 1 = a n + 2 n ( n β₯ 1 ) β a 1 β = 2 a n + 1 β = a n β + 2 n ( n β₯ 1 ) β
then a 100 a 1 0 0 β
Answer Choices:
A. 9900 9 9 0 0 9902 9 9 0 2 9904 9 9 0 4 10100 1 0 1 0 0 10102 1 0 1 0 2
Solution:
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Problem 13: 2 6 2 + 3 + 5 2 β + 3 β + 5 β 2 6 β β
Answer Choices:
A. 2 + 3 β 5 2 β + 3 β β 5 β 4 β 2 β 3 4 β 2 β β 3 β 2 + 3 + 6 β 5 2 β + 3 β + 6 β β 5 1 2 ( 2 + 5 β 3 ) 2 1 β ( 2 β + 5 β β 3 β ) 1 3 ( 3 + 5 β 2 ) 3 1 β ( 3 β + 5 β β 2 β )
Solution:
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Problem 14: The product of all real roots of the equation x 10 x = 10 x l o g 1 0 β x = 1 0
Answer Choices:
A. 1 1 β 1 β 1 10 1 0 1 0 β 1 1 0 β 1
Solution:
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Problem 15: If sin β‘ 2 x sin β‘ 3 x = cos β‘ 2 x cos β‘ 3 x sin 2 x sin 3 x = cos 2 x cos 3 x x x
Answer Choices:
A. 1 8 β 1 8 β 3 0 β 3 0 β 3 6 β 3 6 β 4 5 β 4 5 β 6 0 β 6 0 β
Solution:
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Problem 16: The function f ( x ) f ( x ) f ( 2 + x ) = f ( 2 β x ) f ( 2 + x ) = f ( 2 β x ) x x f ( x ) = 0 f ( x ) = 0
Answer Choices:
A. 0 0 2 2 4 4 6 6 8 8
Solution:
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Problem 17: A right triangle A B C A B C A B A B A C = 15 A C = 1 5 C H C H A B A B A H A H H B H B H B = 16 H B = 1 6 β³ A B C β³ A B C
Answer Choices:
A. 120 1 2 0 144 1 4 4 150 1 5 0 216 2 1 6 144 5 1 4 4 5 β
Solution:
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Problem 18: A point ( x , y ) ( x , y ) x x y y x + y = 2 x + y = 2 x x
Answer Choices:
A. 2 β 1 2 β β 1 1 2 2 1 β 2 β 2 2 β 2 β 1 1
Solution:
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Problem 19: A box contains 11 1 1 1 , 2 , 3 , β¦ , 11 1 , 2 , 3 , β¦ , 1 1 6 6
Answer Choices:
A. 100 231 2 3 1 1 0 0 β 115 231 2 3 1 1 1 5 β 1 2 2 1 β 118 231 2 3 1 1 1 8 β 6 11 1 1 6 β
Solution:
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Problem 20: The number of distinct solutions of the equation
β£ x β β£ 2 x + 1 β£ β£ = 3 is β£ x β β£ 2 x + 1 β£ β£ = 3 is
Answer Choices:
A. 0 0 1 1 2 2 3 3 4 4
Solution:
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Problem 21: The number of triples ( a , b , c ) ( a , b , c )
a b + b c = 44 a c + b c = 23 β a b + b c = 4 4 a c + b c = 2 3 β
is
Answer Choices:
A. 0 0 1 1 2 2 3 3 4 4
Solution:
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Problem 22: Let a a c c t t ( x t , y t ) ( x t β , y t β ) y = a x 2 + t x + c y = a x 2 + t x + c ( x t , y t ) ( x t β , y t β ) t t
Answer Choices:
A. a straight line
Solution:
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Problem 23: sin β‘ 1 0 β + sin β‘ 2 0 β cos β‘ 1 0 β + cos β‘ 2 0 β cos 1 0 β + cos 2 0 β sin 1 0 β + sin 2 0 β β
Answer Choices:
A. tan β‘ 1 0 β + tan β‘ 2 0 β tan 1 0 β + tan 2 0 β tan β‘ 3 0 β tan 3 0 β 1 2 ( tan β‘ 1 0 β + tan β‘ 2 0 β ) 2 1 β ( tan 1 0 β + tan 2 0 β ) tan β‘ 1 5 β tan 1 5 β 1 4 tan β‘ 6 0 β 4 1 β tan 6 0 β
Solution:
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Problem 24: If a a b b x 2 + a x + 2 b = 0 x 2 + a x + 2 b = 0 x 2 + 2 b x + a = 0 x 2 + 2 b x + a = 0 a + b a + b
Answer Choices:
A. 2 2 3 3 4 4 5 5 6 6
Solution:
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Problem 25: The total area of all the faces of a rectangular solid is 22 c m 2 2 2 c m 2 24 2 4
Answer Choices:
A. 11 1 1 β 12 1 2 β 13 1 3 β 14 1 4 β
Solution:
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Problem 26: In the obtuse triangle A B C , A M = M B , M D β₯ B C , E C β₯ B C A B C , A M = M B , M D β₯ B C , E C β₯ B C β³ A B C β³ A B C 24 2 4 β³ B E D β³ B E D
Answer Choices:
A. 9 9 12 1 2 15 1 5 18 1 8
Solution:
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Problem 27: In β³ A B C , D β³ A B C , D A C A C F F B C B C A B β₯ A C , A F β₯ B C A B β₯ A C , A F β₯ B C B D = D C = F C = 1 B D = D C = F C = 1 A C A C
Answer Choices:
A. 2 2 β 3 3 β 2 3 3 2 β 3 3 3 3 β 3 4 4 3 β
Solution:
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Problem 28: The number of distinct pairs of integers ( x , y ) ( x , y ) 0 < x < y 0 < x < y 1984 = x + y 1 9 8 4 β = x β + y β
Answer Choices:
A. 0 0 1 1 3 3 4 4 7 7
Solution:
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Problem 29: Find the largest value of y x x y β ( x , y ) ( x , y ) ( x β 3 ) 2 + ( y β 3 ) 2 = 6 ( x β 3 ) 2 + ( y β 3 ) 2 = 6
Answer Choices:
A. 3 + 2 2 3 + 2 2 β 2 + 3 2 + 3 β 3 3 3 3 β 6 6 6 + 2 3 6 + 2 3 β
Solution:
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Problem 30: For any complex number w = a + b i , β£ w β£ w = a + b i , β£ w β£ a 2 + b 2 a 2 + b 2 β w = cos β‘ 4 0 β + i sin β‘ 4 0 β w = cos 4 0 β + i sin 4 0 β β£ w + 2 w 2 + 3 w 3 + β― + 9 w 9 β£ β 1 β£ β£ β£ β w + 2 w 2 + 3 w 3 + β― + 9 w 9 β£ β£ β£ β β 1
Answer Choices:
A. 1 9 sin β‘ 4 0 β 9 1 β sin 4 0 β 2 9 sin β‘ 2 0 β 9 2 β sin 2 0 β 1 9 cos β‘ 4 0 β 9 1 β cos 4 0 β 1 18 cos β‘ 2 0 β 1 8 1 β cos 2 0 β
Solution:
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The problems on this page are the property of the MAA's American Mathematics Competitions
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