Difference between revisions of "1967 AHSME Problems"
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==Problem 39== | ==Problem 39== | ||
− | Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math> | + | Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let <math>S_n</math> be the sum of the elements in the nth set. Then <math>S_{21}</math> equals: |
<math>\textbf{(A)}\ 1113\qquad | <math>\textbf{(A)}\ 1113\qquad |
Revision as of 23:30, 6 July 2018
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 See also
Problem 1
The three-digit number is added to the number to give the three-digit number . If is divisible by 9, then equals
Problem 2
An equivalent of the expression
, ,
is:
Problem 3
The side of an equilateral triangle is . A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:
Problem 4
Given , all logarithms to the same base and . If , then is:
Problem 5
A triangle is circumscribed about a circle of radius inches. If the perimeter of the triangle is inches and the area is square inches, then is:
Problem 6
If then equals:
Problem 7
If where , , , and are real numbers and , then:
Problem 8
To ounces of a solution of acid, ounces of water are added to yield a solution. If , then is
Problem 9
Let , in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:
Problem 10
If is an identity for positive rational values of , then the value of is:
Problem 11
If the perimeter of rectangle is inches, the least value of diagonal , in inches, is:
Problem 12
If the (convex) area bounded by the x-axis and the lines , , and is , then equals:
Problem 13
A triangle is to be constructed given a side (oppisite angle ). angle , and , the altitude from . If is the number of noncongruent solutions, then
Problem 14
Let , . If , then can be expressed as
Problem 15
The difference in the areas of two similar triangles is square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is feet. The corresponding side of the larger triangle, in feet, is:
Problem 16
Let the product , each factor written in base , equal in base . Let , each term expressed in base . Then , in base , is
Problem 17
If and are the distinct real roots of , then it must follow that:
Problem 18
If and then
Problem 19
The area of a rectangle remains unchanged when it is made inches longer and inch narrower, or when it is made inches shorter and inch wider. Its area, in square inches, is:
Problem 20
A circle is inscribed in a square of side , then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If is the sum of the areas of the first circles so inscribed, then, as grows beyond all bounds, approaches:
Problem 21
In right triangle the hypotenuse and leg . The bisector of angle meets the opposite side in . A second right triangle is then constructed with hypotenuse and leg . If the bisector of angle meets the opposite side in , the length of is:
Problem 22
For natural numbers, when is divided by , the quotient is and the remainder is . When is divided by , the quotient is and the remainder is . Then, when is divided by , the remainder is:
Problem 23
If is real and positive and grows beyond all bounds, then approaches:
Problem 24
The number of solution-pairs in the positive integers of the equation is:
Problem 25
For every odd number we have:
Problem 26
If one uses only the tabular information , , , , , , then the strongest statement one can make for is that it lies between:
Problem 27
Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in hours and the other in hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?
Problem 28
Given the two hypotheses: Some Mems are not Ens and No Ens are Veens. If "some" means "at least one," we can conclude that:
Problem 29
is a diameter of a circle. Tangents and are drawn so that and intersect in a point on the circle. If and , , the diameter of the circle is:
Problem 30
A dealer bought radios for dollars, a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8 on each radio sold. If the overall profit was $72, then the least possible value of for the given information is:
Problem 31
Let , where , , are consecutive integers and . Then is:
Problem 32
In quadrilateral with diagonals and intersecting at , , , , and . The length of is:
Problem 33
In this diagram semi-circles are constructed on diameters , , and , so that they are mutually tangent. If , then the ratio of the shaded area to the area of a circle with as radius is:
Problem 34
Points , , are taken respectively on sides , , and of triangle so that . The ratio of the area of triangle to that of triangle is:
Problem 35
The roots of are in arithmetic progression. The difference between the largest and smallest roots is:
Problem 36
Given a geometric progression of five terms, each a positive integer less than . The sum of the five terms is . If is the sum of those terms in the progression which are squares of integers, then is:
Problem 37
Segments , , are drawn from the vertices of triangle , each perpendicular to a straight line , not intersecting the triangle. Points , , are the intersection points of with the perpendiculars. If is the length of the perpendicular segment drawn to from the intersection point of the medians of the triangle, then is:
Problem 38
Given a set consisting of two undefined elements "pib" and "maa", and the four postulates: : Every pib is a collection of maas, : Any two distinct pibs have one and only one maa in common, : Every maa belongs to two and only two pibs, : There are exactly four pibs. Consider the three theorems: : There are exactly six maas, : There are exactly three maas in each pib, : For each maa there is exactly one other maa not in the same pid with it. The theorems which are deducible from the postulates are:
Problem 39
Given the sets of consecutive integers ,,,,, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let be the sum of the elements in the nth set. Then equals:
Problem 40
Located inside equilateral triangle is a point such that , , and . To the nearest integer the area of triangle is:
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by 1966 AHSME |
Followed by 1968 AHSME | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.