Difference between revisions of "1968 AHSME Problems"
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+ | {{AHSC 35 Problems | ||
+ | |year = 1968 | ||
+ | }} | ||
==Problem 1== | ==Problem 1== | ||
Let <math>P</math> units be the increase in circumference of a circle resulting from an increase in <math>\pi</math> units in the diameter. Then <math>P</math> equals: | Let <math>P</math> units be the increase in circumference of a circle resulting from an increase in <math>\pi</math> units in the diameter. Then <math>P</math> equals: | ||
Line 71: | Line 74: | ||
\text{(B) } \frac{9}{2}\quad | \text{(B) } \frac{9}{2}\quad | ||
\text{(C) } 6\quad | \text{(C) } 6\quad | ||
− | \text{(D) } | + | \text{(D) } 9\quad |
\text{(E) } \text{undetermined}</math> | \text{(E) } \text{undetermined}</math> | ||
Line 140: | Line 143: | ||
==Problem 13== | ==Problem 13== | ||
+ | If <math>m</math> and <math>n</math> are the roots of <math>x^2+mx+n=0 ,m \ne 0,n \ne 0</math>, then the sum of the roots is: | ||
+ | <math>\text{(A) } -\frac{1}{2}\quad | ||
+ | \text{(B) } -1\quad | ||
+ | \text{(C) } \frac{1}{2}\quad | ||
+ | \text{(D) } 1\quad | ||
+ | \text{(E) } \text{undetermined}</math> | ||
[[1968 AHSME Problems/Problem 13|Solution]] | [[1968 AHSME Problems/Problem 13|Solution]] | ||
+ | |||
==Problem 14== | ==Problem 14== | ||
+ | If <math>x</math> and <math>y</math> are non-zero numbers such that <math>x=1+\frac{1}{y}</math> and <math>y=1+\frac{1}{x}</math>, then <math>y</math> equals | ||
+ | |||
+ | <math>\text{(A) } x-1\quad | ||
+ | \text{(B) } 1-x\quad | ||
+ | \text{(C) } 1+x\quad | ||
+ | \text{(D) } -x\quad | ||
+ | \text{(E) } x</math> | ||
[[1968 AHSME Problems/Problem 14|Solution]] | [[1968 AHSME Problems/Problem 14|Solution]] | ||
+ | |||
==Problem 15== | ==Problem 15== | ||
+ | Let <math>P</math> be the product of any three consecutive positive odd integers. The largest integer dividing all such <math>P</math> is: | ||
+ | |||
+ | <math>\text{(A) } 15\quad | ||
+ | \text{(B) } 6\quad | ||
+ | \text{(C) } 5\quad | ||
+ | \text{(D) } 3\quad | ||
+ | \text{(E) } 1</math> | ||
[[1968 AHSME Problems/Problem 15|Solution]] | [[1968 AHSME Problems/Problem 15|Solution]] | ||
+ | |||
==Problem 16== | ==Problem 16== | ||
+ | If <math>x</math> is such that <math>\frac{1}{x}<2</math> and <math>\frac{1}{x}>-3</math>, then: | ||
+ | <math>\text{(A) } -\frac{1}{3}<x<\frac{1}{2}\quad | ||
+ | \text{(B) } -\frac{1}{2}<x<3\quad | ||
+ | \text{(C) } x>\frac{1}{2}\quad\\ | ||
+ | \text{(D) } x>\frac{1}{2} \text{ or} -\frac{1}{3}<x<0\quad | ||
+ | \text{(E) } x>\frac{1}{2} \text{ or } x<-\frac{1}{3}</math> | ||
[[1968 AHSME Problems/Problem 16|Solution]] | [[1968 AHSME Problems/Problem 16|Solution]] | ||
+ | |||
==Problem 17== | ==Problem 17== | ||
+ | Let <math>f(n)=\frac{x_1+x_2+\cdots +x_n}{n}</math>, where <math>n</math> is a positive integer. If <math>x_k=(-1)^k, k=1,2,\cdots ,n</math>, the set of possible values of <math>f(n)</math> is: | ||
+ | |||
+ | <math>\text{(A) } \{0\}\quad | ||
+ | \text{(B) } \{\frac{1}{n}\}\quad | ||
+ | \text{(C) } \{0,-\frac{1}{n}\}\quad | ||
+ | \text{(D) } \{0,\frac{1}{n}\}\quad | ||
+ | \text{(E) } \{1,\frac{1}{n}\}</math> | ||
[[1968 AHSME Problems/Problem 17|Solution]] | [[1968 AHSME Problems/Problem 17|Solution]] | ||
+ | |||
==Problem 18== | ==Problem 18== | ||
+ | Side <math>AB</math> of triangle <math>ABC</math> has length 8 inches. Line <math>DEF</math> is drawn parallel to <math>AB</math> so that <math>D</math> is on segment <math>AC</math>, and <math>E</math> is on segment <math>BC</math>. Line <math>AE</math> extended bisects angle <math>FEC</math>. If <math>DE</math> has length <math>5</math> inches, then the length of <math>CE</math>, in inches, is: | ||
+ | |||
+ | <math>\text{(A) } \frac{51}{4}\quad | ||
+ | \text{(B) } 13\quad | ||
+ | \text{(C) } \frac{53}{4}\quad | ||
+ | \text{(D) } \frac{40}{3}\quad | ||
+ | \text{(E) } \frac{27}{2}</math> | ||
[[1968 AHSME Problems/Problem 18|Solution]] | [[1968 AHSME Problems/Problem 18|Solution]] | ||
+ | |||
==Problem 19== | ==Problem 19== | ||
+ | Let <math>n</math> be the number of ways <math>10</math> dollars can be changed into dimes and quarters, with at least one of each coin being used. Then <math>n</math> equals: | ||
+ | |||
+ | <math>\text{(A) } 40\quad | ||
+ | \text{(B) } 38\quad | ||
+ | \text{(C) } 21\quad | ||
+ | \text{(D) } 20\quad | ||
+ | \text{(E) } 19</math> | ||
[[1968 AHSME Problems/Problem 19|Solution]] | [[1968 AHSME Problems/Problem 19|Solution]] | ||
+ | |||
==Problem 20== | ==Problem 20== | ||
+ | The measures of the interior angles of a convex polygon of <math>n</math> sides are in arithmetic progression. If the common difference is <math>5^{\circ}</math> and the largest angle is <math>160^{\circ}</math>, then <math>n</math> equals: | ||
+ | |||
+ | <math>\text{(A) } 9\quad | ||
+ | \text{(B) } 10\quad | ||
+ | \text{(C) } 12\quad | ||
+ | \text{(D) } 16\quad | ||
+ | \text{(E) } 32</math> | ||
[[1968 AHSME Problems/Problem 20|Solution]] | [[1968 AHSME Problems/Problem 20|Solution]] | ||
+ | |||
==Problem 21== | ==Problem 21== | ||
+ | If <math>S=1!+2!+3!+\cdots +99!</math>, then the units' digit in the value of S is: | ||
+ | |||
+ | <math>\text{(A) } 9\quad | ||
+ | \text{(B) } 8\quad | ||
+ | \text{(C) } 5\quad | ||
+ | \text{(D) } 3\quad | ||
+ | \text{(E) } 0</math> | ||
[[1968 AHSME Problems/Problem 21|Solution]] | [[1968 AHSME Problems/Problem 21|Solution]] | ||
+ | |||
==Problem 22== | ==Problem 22== | ||
+ | A segment of length <math>1</math> is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is: | ||
+ | |||
+ | <math>\text{(A) equal to } \frac{1}{4}\quad\\ | ||
+ | \text{(B) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ | ||
+ | \text{(C) greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ | ||
+ | \text{(D) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{4}\quad\\ | ||
+ | \text{(E) less than }\frac{1}{2}</math> | ||
[[1968 AHSME Problems/Problem 22|Solution]] | [[1968 AHSME Problems/Problem 22|Solution]] | ||
+ | |||
==Problem 23== | ==Problem 23== | ||
+ | If all the logarithms are real numbers, the equality | ||
+ | <math>log(x+3)+log(x-1)=log(x^2-2x-3)</math> | ||
+ | is satisfied for: | ||
+ | <math>\text{(A) all real values of }x \quad\\ | ||
+ | \text{(B) no real values of } x\quad\\ | ||
+ | \text{(C) all real values of } x \text{ except } x=0\quad\\ | ||
+ | \text{(D) no real values of } x \text{ except } x=0\quad\\ | ||
+ | \text{(E) all real values of } x \text{ except } x=1</math> | ||
[[1968 AHSME Problems/Problem 23|Solution]] | [[1968 AHSME Problems/Problem 23|Solution]] | ||
+ | |||
==Problem 24== | ==Problem 24== | ||
+ | A painting <math>18</math>" X <math>24</math>" is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is: | ||
+ | |||
+ | <math>\text{(A) } 1:3\quad | ||
+ | \text{(B) } 1:2\quad | ||
+ | \text{(C) } 2:3\quad | ||
+ | \text{(D) } 3:4\quad | ||
+ | \text{(E) } 1:1</math> | ||
[[1968 AHSME Problems/Problem 24|Solution]] | [[1968 AHSME Problems/Problem 24|Solution]] | ||
+ | |||
==Problem 25== | ==Problem 25== | ||
+ | Ace runs with constant speed and Flash runs <math>x</math> times as fast, <math>x>1</math>. Flash gives Ace a head start of <math>y</math> yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is: | ||
+ | <math>\text{(A) } xy\quad | ||
+ | \text{(B) } \frac{y}{x+y}\quad | ||
+ | \text{(C) } \frac{xy}{x-1}\quad | ||
+ | \text{(D) } \frac{x+y}{x+1}\quad | ||
+ | \text{(E) } \frac{x+y}{x-1}</math> | ||
[[1968 AHSME Problems/Problem 25|Solution]] | [[1968 AHSME Problems/Problem 25|Solution]] | ||
+ | |||
==Problem 26== | ==Problem 26== | ||
+ | Let <math>S=2+4+6+\cdots +2N</math>, where <math>N</math> is the smallest positive integer such that <math>S>1,000,000</math>. Then the sum of the digits of <math>N</math> is: | ||
+ | |||
+ | <math>\text{(A) } 27\quad | ||
+ | \text{(B) } 12\quad | ||
+ | \text{(C) } 6\quad | ||
+ | \text{(D) } 2\quad | ||
+ | \text{(E) } 1</math> | ||
[[1968 AHSME Problems/Problem 26|Solution]] | [[1968 AHSME Problems/Problem 26|Solution]] | ||
+ | |||
==Problem 27== | ==Problem 27== | ||
+ | Let <math>S_n=1-2+3-4+\cdots +(-1)^{n-1}n</math>, where <math>n=1,2,\cdots</math>. Then <math>S_{17}+S_{33}+S_{50}</math> equals: | ||
+ | <math>\text{(A) } 0\quad | ||
+ | \text{(B) } 1\quad | ||
+ | \text{(C) } 2\quad | ||
+ | \text{(D) } -1\quad | ||
+ | \text{(E) } -2</math> | ||
[[1968 AHSME Problems/Problem 27|Solution]] | [[1968 AHSME Problems/Problem 27|Solution]] | ||
+ | |||
==Problem 28== | ==Problem 28== | ||
+ | If the arithmetic mean of <math>a</math> and <math>b</math> is double their geometric mean, with <math>a>b>0</math>, then a possible value for the ratio <math>a/b</math>, to the nearest integer, is: | ||
+ | <math>\text{(A) } 5\quad | ||
+ | \text{(B) } 8\quad | ||
+ | \text{(C) } 11\quad | ||
+ | \text{(D) } 14\quad | ||
+ | \text{(E) none of these}</math> | ||
[[1968 AHSME Problems/Problem 28|Solution]] | [[1968 AHSME Problems/Problem 28|Solution]] | ||
+ | |||
==Problem 29== | ==Problem 29== | ||
+ | Given the three numbers <math>x,y=x^x,z=x^{x^x}</math> with <math>.9<x<1.0</math>. Arranged in order of increasing magnitude, they are: | ||
+ | <math>\text{(A) } x,z,y\quad | ||
+ | \text{(B) } x,y,z\quad | ||
+ | \text{(C) } y,x,z\quad | ||
+ | \text{(D) } y,z,x\quad | ||
+ | \text{(E) } z,x,y</math> | ||
[[1968 AHSME Problems/Problem 29|Solution]] | [[1968 AHSME Problems/Problem 29|Solution]] | ||
+ | |||
==Problem 30== | ==Problem 30== | ||
+ | Convex polygons <math>P_1</math> and <math>P_2</math> are drawn in the same plane with <math>n_1</math> and <math>n_2</math> sides, respectively, <math>n_1\le n_2</math>. If <math>P_1</math> and <math>P_2</math> do not have any line segment in common, then the maximum number of intersections of <math>P_1</math> and <math>P_2</math> is: | ||
+ | <math>\text{(A) } 2n_1\quad | ||
+ | \text{(B) } 2n_2\quad | ||
+ | \text{(C) } n_1n_2\quad | ||
+ | \text{(D) } n_1+n_2\quad | ||
+ | \text{(E) } \text{none of these}</math> | ||
[[1968 AHSME Problems/Problem 30|Solution]] | [[1968 AHSME Problems/Problem 30|Solution]] | ||
+ | |||
==Problem 31== | ==Problem 31== | ||
Line 228: | Line 369: | ||
==Problem 32== | ==Problem 32== | ||
+ | <math>A</math> and <math>B</math> move uniformly along two straight paths intersecting at right angles in point <math>O</math>. When <math>A</math> is at <math>O</math>, <math>B</math> is <math>500</math> yards short of <math>O</math>. In two minutes they are equidistant from <math>O</math>, and in <math>8</math> minutes more they are again equidistant from <math>O</math>. Then the ratio of <math>A</math>'s speed to <math>B</math>'s speed is: | ||
+ | <math>\text{(A) } 4:5\quad | ||
+ | \text{(B) } 5:6\quad | ||
+ | \text{(C) } 2:3\quad | ||
+ | \text{(D) } 5:8\quad | ||
+ | \text{(E) } 1:2</math> | ||
[[1968 AHSME Problems/Problem 32|Solution]] | [[1968 AHSME Problems/Problem 32|Solution]] | ||
+ | |||
==Problem 33== | ==Problem 33== | ||
+ | A number <math>N</math> has three digits when expressed in base <math>7</math>. When <math>N</math> is expressed in base <math>9</math> the digits are reversed. Then the middle digit is: | ||
+ | <math>\text{(A) } 0\quad | ||
+ | \text{(B) } 1\quad | ||
+ | \text{(C) } 3\quad | ||
+ | \text{(D) } 4\quad | ||
+ | \text{(E) } 5</math> | ||
[[1968 AHSME Problems/Problem 33|Solution]] | [[1968 AHSME Problems/Problem 33|Solution]] | ||
+ | |||
==Problem 34== | ==Problem 34== | ||
+ | With <math>400</math> members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was <math>\frac{12}{11}</math> of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time? | ||
+ | <math>\text{(A) } 75\quad | ||
+ | \text{(B) } 60\quad | ||
+ | \text{(C) } 50\quad | ||
+ | \text{(D) } 45\quad | ||
+ | \text{(E) } 20</math> | ||
[[1968 AHSME Problems/Problem 34|Solution]] | [[1968 AHSME Problems/Problem 34|Solution]] | ||
+ | |||
==Problem 35== | ==Problem 35== | ||
<asy> | <asy> | ||
Line 261: | Line 423: | ||
[[1968 AHSME Problems/Problem 35|Solution]] | [[1968 AHSME Problems/Problem 35|Solution]] | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | * [[AMC 12 Problems and Solutions]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 35p box|year=1968|before=[[1967 AHSME|1967 AHSC]]|after=[[1969 AHSME|1969 AHSC]]}} | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:13, 20 February 2020
1968 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 |
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 See also
Problem 1
Let units be the increase in circumference of a circle resulting from an increase in
units in the diameter. Then
equals:
Problem 2
The real value of such that
divided by
equals
is:
Problem 3
A straight line passing through the point is perpendicular to the line
. Its equation is:
Problem 4
Define an operation for positive real numbers as
. Then
equals:
Problem 5
If , then
equals:
Problem 6
Let side of convex quadrilateral
be extended through
, and let side
be extended through
, to meet in point
Let
be the degree-sum of angles
and
, and let
represent the degree-sum of angles
and
If
, then:
Problem 7
Let be the intersection point of medians
and
of triangle
if
is 3 inches, then
, in inches, is:
Problem 8
A positive number is mistakenly divided by instead of being multiplied by
Based on the correct answer, the error thus committed, to the nearest percent, is :
Problem 9
The sum of the real values of satisfying the equality
is:
Problem 10
Assume that, for a certain school, it is true that
I: Some students are not honest. II: All fraternity members are honest.
A necessary conclusion is:
Problem 11
If an arc of on circle
has the same length as an arc of
on circle
, the ratio of the area of circle
to that of circle
is:
Problem 12
A circle passes through the vertices of a triangle with side-lengths The radius of the circle is:
Problem 13
If and
are the roots of
, then the sum of the roots is:
Problem 14
If and
are non-zero numbers such that
and
, then
equals
Problem 15
Let be the product of any three consecutive positive odd integers. The largest integer dividing all such
is:
Problem 16
If is such that
and
, then:
Problem 17
Let , where
is a positive integer. If
, the set of possible values of
is:
Problem 18
Side of triangle
has length 8 inches. Line
is drawn parallel to
so that
is on segment
, and
is on segment
. Line
extended bisects angle
. If
has length
inches, then the length of
, in inches, is:
Problem 19
Let be the number of ways
dollars can be changed into dimes and quarters, with at least one of each coin being used. Then
equals:
Problem 20
The measures of the interior angles of a convex polygon of sides are in arithmetic progression. If the common difference is
and the largest angle is
, then
equals:
Problem 21
If , then the units' digit in the value of S is:
Problem 22
A segment of length is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:
Problem 23
If all the logarithms are real numbers, the equality
is satisfied for:
Problem 24
A painting " X
" is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:
Problem 25
Ace runs with constant speed and Flash runs times as fast,
. Flash gives Ace a head start of
yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:
Problem 26
Let , where
is the smallest positive integer such that
. Then the sum of the digits of
is:
Problem 27
Let , where
. Then
equals:
Problem 28
If the arithmetic mean of and
is double their geometric mean, with
, then a possible value for the ratio
, to the nearest integer, is:
Problem 29
Given the three numbers with
. Arranged in order of increasing magnitude, they are:
Problem 30
Convex polygons and
are drawn in the same plane with
and
sides, respectively,
. If
and
do not have any line segment in common, then the maximum number of intersections of
and
is:
Problem 31
In this diagram, not drawn to scale, Figures and
are equilateral triangular regions with respective areas of
and
square inches. Figure
is a square region with area
square inches. Let the length of segment
be decreased by
% of itself, while the lengths of
and
remain unchanged. The percent decrease in the area of the square is:
Problem 32
and
move uniformly along two straight paths intersecting at right angles in point
. When
is at
,
is
yards short of
. In two minutes they are equidistant from
, and in
minutes more they are again equidistant from
. Then the ratio of
's speed to
's speed is:
Problem 33
A number has three digits when expressed in base
. When
is expressed in base
the digits are reversed. Then the middle digit is:
Problem 34
With members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was
of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?
Problem 35
In this diagram the center of the circle is , the radius is
inches, chord
is parallel to chord
.
,
,
,
are collinear, and
is the midpoint of
. Let
(sq. in.) represent the area of trapezoid
and let
(sq. in.) represent the area of rectangle
Then, as
and
are translated upward so that
increases toward the value
, while
always equals
, the ratio
becomes arbitrarily close to:
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1967 AHSC |
Followed by 1969 AHSC | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.