ΒΆ 1974 AHSME Problems and SolutionsIndividual Problems and Solutions
For problems and detailed solutions to each of the 1974 AHSME problems, please refer below:
Problem 1: If x β 0 x ξ = 0 4 4 y β 0 y ξ = 0 6 6 2 x + 3 y = 1 2 x 2 β + y 3 β = 2 1 β
Answer Choices:
A. 4 x + 3 y = x y 4 x + 3 y = x y y = 4 x 6 β y y = 6 β y 4 x β x 2 + y 3 = 2 2 x β + 3 y β = 2 4 y y β 6 = x y β 6 4 y β = x
Solution:
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Problem 2: Let x 1 x 1 β x 2 x 2 β x 1 β x 2 x 1 β ξ = x 2 β 3 x i 2 β h x i = b , i = 1 , 2 3 x i 2 β β h x i β = b , i = 1 , 2 x 1 + x 2 x 1 β + x 2 β
Answer Choices:
A. β h 3 β 3 h β h 3 3 h β b 3 3 b β 2 b 2 b β b 3 β 3 b β
Solution:
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Problem 3: The coefficient of x 7 x 7 ( 1 + 2 x β x 2 ) 4 ( 1 + 2 x β x 2 ) 4
Answer Choices:
A. β 8 β 8 12 1 2 6 6 β 12 β 1 2
Solution:
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Problem 4: What is the remainder when x 51 + 51 x 5 1 + 5 1 x + 1 ? x + 1 ?
Answer Choices:
A. 0 0 1 1 49 4 9 50 5 0 51 5 1
Solution:
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Problem 5: Given a quadrilateral A B C D A B C D A B A B B B E E β B A D = 9 2 β β B A D = 9 2 β β A D C = 6 8 β β A D C = 6 8 β β E B C β E B C
Answer Choices:
A. 6 6 β 6 6 β 6 8 β 6 8 β 7 0 β 7 0 β 8 8 β 8 8 β 9 2 β 9 2 β
Solution:
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Problem 6: For positive real numbers x x y y x β y = x β
y x + y x β y = x + y x β
y β
Answer Choices:
A. β β β β β β β β
Solution:
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Problem 7: A town's population increased by 1 , 200 1 , 2 0 0 11 % 1 1 % 32 3 2 1 , 200 1 , 2 0 0
Answer Choices:
A. 1200 1 2 0 0 11200 1 1 2 0 0 9968 9 9 6 8 10000 1 0 0 0 0
Solution:
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Problem 8: What is the smallest prime number dividing the sum 3 11 + 5 13 ? 3 1 1 + 5 1 3 ?
Answer Choices:
A. 2 2 3 3 5 5 3 11 + 5 13 3 1 1 + 5 1 3
Solution:
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Problem 9: The integers greater than one are arranged in five columns as follows:
2 3 4 5 9 8 7 6 10 11 12 13 17 16 15 14 β
β
β
β
9 1 7 β 2 8 1 0 1 6 β
β 3 7 1 1 1 5 β
β 4 6 1 2 1 4 β
β 5 1 3 β
β
(Four consecutive integers appear in each row; in the first, third and other odd numbered rows, the integers appear in the last four columns and increase from left to right; in the second, fourth and other even numbered rows, the integers appear in the first four columns and increase from right to left.)
In which column will the number 1 , 000 1 , 0 0 0
Answer Choices:
A. first
Solution:
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Problem 10: What is the smallest integral value of k k 2 x ( k x β 4 ) β x 2 + 6 = 0 2 x ( k x β 4 ) β x 2 + 6 = 0
Answer Choices:
A. β 1 β 1 2 2 3 3 4 4 5 5
Solution:
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Problem 11: If ( a , b ) ( a , b ) ( c , d ) ( c , d ) y = m x + k y = m x + k ( a , b ) ( a , b ) ( c , d ) ( c , d ) a , c a , c m m
Answer Choices:
A. β£ a β c β£ 1 + m 2 β£ a β c β£ 1 + m 2 β β£ a + c β£ 1 + m 2 β£ a + c β£ 1 + m 2 β β£ a β c β£ 1 + m 2 1 + m 2 β β£ a β c β£ β β£ a β c β£ ( 1 + m 2 ) β£ a β c β£ ( 1 + m 2 ) β£ a β c β£ β£ m β£ β£ a β c β£ β£ m β£
Solution:
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Problem 12: If g ( x ) = 1 β x 2 g ( x ) = 1 β x 2 f ( g ( x ) ) = 1 β x 2 x 2 f ( g ( x ) ) = x 2 1 β x 2 β x β 0 x ξ = 0 f ( 1 / 2 ) f ( 1 / 2 )
Answer Choices:
A. 3 / 4 3 / 4 1 1 3 3 2 / 2 2 β / 2 2 2 β
Solution:
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Problem 13: Which of the following is equivalent to "If P P Q Q
Answer Choices:
A. P P Q Q Q Q P P P P Q Q Q Q P P Q Q P P
Solution:
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Problem 14: Which statement is correct?
Answer Choices:
A. If x < 0 x < 0 x 2 > x x 2 > x x 2 > 0 x 2 > 0 x > 0 x > 0 x 2 > x x 2 > x x > 0 x > 0 x 2 > x x 2 > x x < 0 x < 0 x < 1 x < 1 x 2 < x x 2 < x
Solution:
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Problem 15: If x < β 2 x < β 2 β£ 1 β β£ 1 + x β£ β£ β£ 1 β β£ 1 + x β£ β£
Answer Choices:
A. 2 + x 2 + x β 2 β x β 2 β x x x β x β x β 2 β 2
Solution:
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Problem 16: A circle of radius r r R R R r r R β
Answer Choices:
A. 1 + 2 1 + 2 β 2 + 2 2 2 2 + 2 β β 2 β 1 2 2 2 β β 1 β 1 + 2 2 2 1 + 2 β β 2 ( 2 β 2 ) 2 ( 2 β 2 β )
Solution:
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Problem 17: If i 2 = β 1 i 2 = β 1 ( 1 + i ) 20 β ( 1 β i ) 20 ( 1 + i ) 2 0 β ( 1 β i ) 2 0
Answer Choices:
A. β 1024 β 1 0 2 4 β 1024 i β 1 0 2 4 i 0 0 1024 1 0 2 4 1024 i 1 0 2 4 i
Solution:
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Problem 18: If 8 3 = p log 8 β 3 = p 3 5 = q log 3 β 5 = q p p q , 10 5 q , log 1 0 β 5
Answer Choices:
A. p q p q 3 p + q 5 5 3 p + q β 1 + 3 p q p + q p + q 1 + 3 p q β 3 p q 1 + 3 p q 1 + 3 p q 3 p q β p 2 + q 2 p 2 + q 2
Solution:
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Problem 19: In the adjoining figure A B C D A B C D C M N C M N A B C D A B C D C M N C M N
Answer Choices:
A. 2 3 β 3 2 3 β β 3 2 / 3 2 β / 3 1 β 3 / 3 1 β 3 β / 3 4 β 2 3 4 β 2 3 β 3 / 4 3 β / 4
Solution:
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Problem 20: Let T = 1 3 β 8 β 1 8 β 7 + 1 7 β 6 β 1 6 β 5 + 1 5 β 2 T = 3 β 8 β 1 β β 8 β β 7 β 1 β + 7 β β 6 β 1 β β 6 β β 5 β 1 β + 5 β β 2 1 β
Answer Choices:
A. T < 1 T < 1 T = 1 T = 1 1 < T < 2 1 < T < 2 T > 2 T > 2 T = 1 ( 3 β 8 ) ( 8 β 7 ) ( 7 β 6 ) ( 6 β 5 ) ( 5 β 2 ) T = ( 3 β 8 β ) ( 8 β β 7 β ) ( 7 β β 6 β ) ( 6 β β 5 β ) ( 5 β β 2 ) 1 β
Solution:
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Problem 21: In a geometric series of positive terms the difference between the fifth and fourth terms is 576 5 7 6 9 9
Answer Choices:
A. 1061 1 0 6 1 1023 1 0 2 3 1024 1 0 2 4 768 7 6 8
Solution:
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Problem 22: The minimum value of sin β‘ A 2 β 3 cos β‘ A 2 sin 2 A β β 3 β cos 2 A β A A
Answer Choices:
A. β 18 0 β β 1 8 0 β 6 0 β 6 0 β 12 0 β 1 2 0 β 0 β 0 β
Solution:
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Problem 23: In the adjoining figure T P T P T β² Q T β² Q r r T T T β² T β² P T β² β² Q P T β² β² Q T β² β² T β² β² T P = 4 T P = 4 T β² Q = 9 T β² Q = 9 r r
Answer Choices:
A. 25 / 6 2 5 / 6 6 6 25 / 4 2 5 / 4 25 / 6 , 6 , 25 / 4 2 5 / 6 , 6 , 2 5 / 4
Solution:
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Problem 24: A fair die is rolled six times. The probability of rolling at least a five at least five times is
Answer Choices:
A. 13 / 729 1 3 / 7 2 9 12 / 729 1 2 / 7 2 9 2 / 729 2 / 7 2 9 3 / 729 3 / 7 2 9
Solution:
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Problem 25: In parallelogram A B C D A B C D D P D P B C B C N N A B A B P P C C C Q C Q A D A D M M A B A B Q Q D P D P C Q C Q O O A B C D A B C D k k Q P O Q P O
Answer Choices:
A. k k 6 k / 5 6 k / 5 9 k / 8 9 k / 8 5 k / 4 5 k / 4 2 k 2 k
Solution:
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Problem 26: The number of distinct positive integral divisors of ( 30 ) 4 ( 3 0 ) 4 1 1 ( 30 ) 4 ( 3 0 ) 4
Answer Choices:
A. 100 1 0 0 125 1 2 5 123 1 2 3 30 3 0
Solution:
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Problem 27: If f ( x ) = 3 x + 2 f ( x ) = 3 x + 2 x x β£ f ( x ) + 4 β£ < a β£ f ( x ) + 4 β£ < a β£ x + 2 β£ < b β£ x + 2 β£ < b a > 0 a > 0 b > 0 b > 0
Answer Choices:
A. b β©½ a / 3 b β©½ a / 3 b > a / 3 b > a / 3 a β©½ b / 3 a β©½ b / 3 a > b / 3 a > b / 3
Solution:
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Problem 28: Which of the following is satisfied by all numbers x x
x = a 1 3 + a 2 3 2 + β¦ + a 25 3 25 x = 3 a 1 β β + 3 2 a 2 β β + β¦ + 3 2 5 a 2 5 β β
where a 1 a 1 β 0 0 2 , a 2 2 , a 2 β 0 0 2 , β¦ , a 25 2 , β¦ , a 2 5 β 0 0 2 ? 2 ?
Answer Choices:
A. 0 β©½ x < 1 / 3 0 β©½ x < 1 / 3 1 / 3 β©½ x < 2 / 3 1 / 3 β©½ x < 2 / 3 2 / 3 β©½ x < 1 2 / 3 β©½ x < 1 0 β©½ x < 1 / 3 0 β©½ x < 1 / 3 2 / 3 β©½ x < 1 2 / 3 β©½ x < 1 1 / 2 β©½ x β©½ 3 / 4 1 / 2 β©½ x β©½ 3 / 4
Solution:
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Problem 29: For p = 1 , 2 , β¦ , 10 p = 1 , 2 , β¦ , 1 0 S p S p β 40 4 0 p p 2 p β 1 2 p β 1 S 1 + S 2 + β¦ + S 10 S 1 β + S 2 β + β¦ + S 1 0 β
Answer Choices:
A. 80 , 000 8 0 , 0 0 0 80 , 200 8 0 , 2 0 0 80 , 400 8 0 , 4 0 0 80 , 600 8 0 , 6 0 0 80 , 800 8 0 , 8 0 0
Solution:
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Problem 30: A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If R R
R [ R ( R 2 + 1 R ) + 1 R ] + 1 R R [ R ( R 2 + R 1 β ) + R 1 β ] + R 1 β
is
Answer Choices:
A. 2 2
B. 2 R 2 R
C. 1 / R 1 / R
D. 2 + 1 / R 2 + 1 / R
E. 2 + R 2 + R
Solution:
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The problems and solutions on this page are the property of the MAA's American Mathematics Competitions
