Difference between revisions of "1967 AHSME Problems"
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+ | {{AHSC 40 Problems | ||
+ | |year = 1967 | ||
+ | }} | ||
==Problem 1== | ==Problem 1== | ||
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==Problem 30== | ==Problem 30== | ||
− | A dealer bought <math>n</math> radios for <math>d</math> dollars, <math>d</math> a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of | + | A dealer bought <math>n</math> radios for <math>d</math> dollars, <math>d</math> 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 <math>n</math> for the given information is: |
<math>\textbf{(A)}\ 18\qquad | <math>\textbf{(A)}\ 18\qquad | ||
Line 354: | Line 357: | ||
==Problem 32== | ==Problem 32== | ||
− | In quadrilateral <math>ABCD</math> with diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersecting at <math>O</math>, <math>\overline{BO}=4</math>, <math>\overline{AO}=8</math>, <math>\overline{OC}=3</math>, and <math>\overline{AB}=6</math>. The length of <math>\overline{AD}</math> is: | + | In quadrilateral <math>ABCD</math> with diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersecting at <math>O</math>, <math>\overline{BO}=4</math>, <math>\overline{OD}=6</math>, <math>\overline{AO}=8</math>, <math>\overline{OC}=3</math>, and <math>\overline{AB}=6</math>. The length of <math>\overline{AD}</math> is: |
<math>\textbf{(A)}\ 9\qquad | <math>\textbf{(A)}\ 9\qquad | ||
Line 448: | Line 451: | ||
==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 | ||
Line 471: | Line 474: | ||
== See also == | == See also == | ||
− | + | ||
− | |||
* [[AMC 12 Problems and Solutions]] | * [[AMC 12 Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 40p box|year=1967|before=[[1966 AHSME|1966 AHSC]]|after=[[1968 AHSME|1968 AHSC]]}} | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:14, 20 February 2020
1967 AHSC (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 |
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 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1966 AHSC |
Followed by 1968 AHSC | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.