Difference between revisions of "1951 AHSME Problems"
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== Problem 1 == | == Problem 1 == | ||
The percent that <math>M</math> is greater than <math>N</math> is: | The percent that <math>M</math> is greater than <math>N</math> is: | ||
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A rectangular field is half as wide as it is long and is completely enclosed by <math>x</math> yards of fencing. The area in terms of <math>x</math> is: | A rectangular field is half as wide as it is long and is completely enclosed by <math>x</math> yards of fencing. The area in terms of <math>x</math> is: | ||
− | <math>(\ | + | <math>(\textbf{A})\ \frac{x^2}2 \qquad (\textbf{B})\ 2x^2 \qquad (\textbf{C})\ \frac{2x^2}9 \qquad (\textbf{D})\ \frac{x^2}{18} \qquad (\textbf{E})\ \frac{x^2}{72}</math> |
[[1951 AHSME Problems/Problem 2|Solution]] | [[1951 AHSME Problems/Problem 2|Solution]] | ||
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If the length of a diagonal of a square is <math>a + b</math>, then the area of the square is: | If the length of a diagonal of a square is <math>a + b</math>, then the area of the square is: | ||
− | <math> \ | + | <math> \textbf{(A)}\ (a+b)^{2}\qquad\textbf{(B)}\ \frac{1}{2}(a+b)^{2}\qquad\textbf{(C)}\ a^{2}+b^{2} </math> |
+ | <math> \textbf{(D)}\ \frac{1}{2}(a^{2}+b^{2})\qquad\textbf{(E)}\ \text{none of these} </math> | ||
[[1951 AHSME Problems/Problem 3|Solution]] | [[1951 AHSME Problems/Problem 3|Solution]] | ||
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A barn with a flat roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is: | A barn with a flat roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is: | ||
− | <math> \ | + | <math> \textbf{(A) \ } 360 \qquad \textbf{(B) \ } 460 \qquad \textbf{(C) \ } 490 \qquad \textbf{(D) \ } 590 \qquad \textbf{(E) \ } 720 </math> |
[[1951 AHSME Problems/Problem 4|Solution]] | [[1951 AHSME Problems/Problem 4|Solution]] | ||
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Mr. A owns a home worth <math>10,000</math> dollars. He sells it to Mr. B at a <math>10 \%</math> profit based on the worth of the house. Mr. B sells the house back to Mr. A at a <math>10 \%</math> loss. Then: | Mr. A owns a home worth <math>10,000</math> dollars. He sells it to Mr. B at a <math>10 \%</math> profit based on the worth of the house. Mr. B sells the house back to Mr. A at a <math>10 \%</math> loss. Then: | ||
− | <math> \ | + | <math> \textbf{(A)}\ \text{A comes out even} \qquad\textbf{(B)}\ \text{A makes 1100 on the deal} \qquad\textbf{(C)}\ \text{A makes 1000 on the deal}</math> |
− | <math>\ | + | <math>\textbf{(D)}\ \text{A loses 900 on the deal} \qquad\textbf{(E)}\ \text{A loses 1000 on the deal}</math> |
[[1951 AHSME Problems/Problem 5|Solution]] | [[1951 AHSME Problems/Problem 5|Solution]] | ||
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The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to: | The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to: | ||
− | <math> \ | + | <math> \textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}</math> |
− | <math> \ | + | <math> \textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}</math> |
[[1951 AHSME Problems/Problem 6|Solution]] | [[1951 AHSME Problems/Problem 6|Solution]] | ||
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An error of <math>.02"</math> is made in the measurement of a line <math>10"</math> long, while an error of only <math>.2"</math> is made in a measurement of a line <math>100"</math> long. In comparison with the relative error of the first measurement, the relative error of the second measurement is: | An error of <math>.02"</math> is made in the measurement of a line <math>10"</math> long, while an error of only <math>.2"</math> is made in a measurement of a line <math>100"</math> long. In comparison with the relative error of the first measurement, the relative error of the second measurement is: | ||
− | <math> \ | + | <math> \textbf{(A) \ } \text{greater by }.18 \qquad\textbf{(B) \ } \text{the same} \qquad \textbf{(C) \ } \text{less} \qquad\textbf{(D) \ } 10\text{ times as great} \qquad\textbf{(E) \ } \text{correctly described by both} </math> |
[[1951 AHSME Problems/Problem 7|Solution]] | [[1951 AHSME Problems/Problem 7|Solution]] | ||
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The price of an article is cut <math>10 \%.</math> To restore it to its former value, the new price must be increased by: | The price of an article is cut <math>10 \%.</math> To restore it to its former value, the new price must be increased by: | ||
− | <math> \ | + | <math> \textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers} </math> |
[[1951 AHSME Problems/Problem 8|Solution]] | [[1951 AHSME Problems/Problem 8|Solution]] | ||
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An equilateral triangle is drawn with a side length of <math>a.</math> A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is: | An equilateral triangle is drawn with a side length of <math>a.</math> A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is: | ||
− | <math> \ | + | <math> \textbf{(A) \ } \text{Infinite} \qquad\textbf{(B) \ } 5\frac{1}{4}a \qquad \textbf{(C) \ } 2a \qquad\textbf{(D) \ } 6a \qquad\textbf{(E) \ } 4\frac{1}{2}a </math> |
[[1951 AHSME Problems/Problem 9|Solution]] | [[1951 AHSME Problems/Problem 9|Solution]] | ||
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Of the following statements, the one that is incorrect is: | Of the following statements, the one that is incorrect is: | ||
− | <math> \ | + | <math> \textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}\qquad\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}</math> |
− | + | <math> \textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}</math> | |
− | <math> \ | + | <math> \textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}</math> |
− | <math> \ | + | <math> \textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}</math> |
− | <math> \ | ||
[[1951 AHSME Problems/Problem 10|Solution]] | [[1951 AHSME Problems/Problem 10|Solution]] | ||
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== Problem 11 == | == Problem 11 == | ||
− | The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 | + | The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1- r}</math> where <math> a</math> denotes the first term and <math> -1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is: |
− | <math> \ | + | <math> \textbf{(A)}\ \frac {a^2}{(1 -r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1+ r^2} \qquad\textbf{(E)}\ \text{none of these}</math> |
[[1951 AHSME Problems/Problem 11|Solution]] | [[1951 AHSME Problems/Problem 11|Solution]] | ||
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At <math> 2: 15</math> o'clock, the hour and minute hands of a clock form an angle of: | At <math> 2: 15</math> o'clock, the hour and minute hands of a clock form an angle of: | ||
− | <math> \ | + | <math> \textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}</math> |
[[1951 AHSME Problems/Problem 12|Solution]] | [[1951 AHSME Problems/Problem 12|Solution]] | ||
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<math> A</math> can do a piece of work in <math> 9</math> days. <math> B</math> is <math> 50\%</math> more efficient than <math> A</math>. The number of days it takes <math> B</math> to do the same piece of work is: | <math> A</math> can do a piece of work in <math> 9</math> days. <math> B</math> is <math> 50\%</math> more efficient than <math> A</math>. The number of days it takes <math> B</math> to do the same piece of work is: | ||
− | <math> \ | + | <math> \textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}</math> |
[[1951 AHSME Problems/Problem 13|Solution]] | [[1951 AHSME Problems/Problem 13|Solution]] | ||
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In connection with proof in geometry, indicate which one of the following statements is ''incorrect'': | In connection with proof in geometry, indicate which one of the following statements is ''incorrect'': | ||
− | <math> \ | + | <math> \textbf{(A)}\ \text{Some statements are accepted without being proved.} </math> |
− | <math> \ | + | <math> \textbf{(B)}\ \text{In some cases there is more than one correct order in proving certain propositions.} </math> |
− | <math> \ | + | <math> \textbf{(C)}\ \text{Every term used in a proof must have been defined previously.} </math> |
− | <math> \ | + | <math> \textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.} </math> |
− | <math> \ | + | <math> \textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.} </math> |
[[1951 AHSME Problems/Problem 14|Solution]] | [[1951 AHSME Problems/Problem 14|Solution]] | ||
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Indicate in which one of the following equations <math>y</math> is neither directly nor inversely proportional to <math>x</math>: | Indicate in which one of the following equations <math>y</math> is neither directly nor inversely proportional to <math>x</math>: | ||
− | <math> \textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y\qquad\textbf{(D)}\ 3x+y = 10 </math> | + | <math> \textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y\qquad\textbf{(D)}\ 3x+y = 10 \textbf{(E)}\ x/y = sqrt(3)\qquad</math> |
[[1951 AHSME Problems/Problem 17|Solution]] | [[1951 AHSME Problems/Problem 17|Solution]] | ||
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== Problem 18 == | == Problem 18 == | ||
− | The expression <math>21x^2+ax+21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be | + | The expression <math>21x^2+ax+21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be done if <math>a</math> is: |
<math> \textbf{(A)}\ \text{any odd number}\qquad\textbf{(B)}\ \text{some odd number}\qquad\textbf{(C)}\ \text{any even number}\qquad\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero} </math> | <math> \textbf{(A)}\ \text{any odd number}\qquad\textbf{(B)}\ \text{some odd number}\qquad\textbf{(C)}\ \text{any even number}\qquad\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero} </math> | ||
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== Problem 42 == | == Problem 42 == | ||
+ | |||
+ | If <math> x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} </math>, then: | ||
+ | |||
+ | <math> \textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\textbf{(D)}\ x\text{ is infinite} </math> | ||
+ | <math> \textbf{(E)}\ x > 2\text{ but finite} </math> | ||
[[1951 AHSME Problems/Problem 42|Solution]] | [[1951 AHSME Problems/Problem 42|Solution]] | ||
== Problem 43 == | == Problem 43 == | ||
+ | |||
+ | Of the following statements, the only one that is incorrect is: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{An inequality will remain true after each side is increased,} </math> <math>\text{ decreased, multiplied or divided zero excluded by the same positive quantity.} </math> | ||
+ | |||
+ | <math> \textbf{(B)}\ \text{The arithmetic mean of two unequal positive quantities is greater than their geometric mean.} </math> | ||
+ | |||
+ | <math> \textbf{(C)}\ \text{If the sum of two positive quantities is given, ther product is largest when they are equal.} </math> | ||
+ | |||
+ | <math> \textbf{(D)}\ \text{If }a\text{ and }b\text{ are positive and unequal, }\frac{1}{2}(a^{2}+b^{2})\text{ is greater than }[\frac{1}{2}(a+b)]^{2}. </math> | ||
+ | |||
+ | <math> \textbf{(E)}\ \text{If the product of two positive quantities is given, their sum is greatest when they are equal.} </math> | ||
[[1951 AHSME Problems/Problem 43|Solution]] | [[1951 AHSME Problems/Problem 43|Solution]] | ||
== Problem 44 == | == Problem 44 == | ||
+ | |||
+ | If <math> \frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c </math>, where <math> a, b, c </math> are other than zero, then <math>x</math> equals: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc} </math> | ||
+ | <math> \textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab} </math> | ||
[[1951 AHSME Problems/Problem 44|Solution]] | [[1951 AHSME Problems/Problem 44|Solution]] | ||
== Problem 45 == | == Problem 45 == | ||
+ | |||
+ | If you are given <math> \log 8\approx .9031 </math> and <math> \log 9\approx .9542 </math>, then the only logarithm that cannot be found without the use of tables is: | ||
+ | |||
+ | <math> \textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4 </math> | ||
[[1951 AHSME Problems/Problem 45|Solution]] | [[1951 AHSME Problems/Problem 45|Solution]] | ||
== Problem 46 == | == Problem 46 == | ||
+ | |||
+ | <math>AB</math> is a fixed diameter of a circle whose center is <math>O</math>. From <math>C</math>, any point on the circle, a chord <math>CD</math> is drawn perpendicular to <math>AB</math>. Then, as <math>C</math> moves over a semicircle, the bisector of angle <math>OCD</math> cuts the circle in a point that always: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies} </math> | ||
+ | <math> \textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C </math> | ||
[[1951 AHSME Problems/Problem 46|Solution]] | [[1951 AHSME Problems/Problem 46|Solution]] | ||
== Problem 47 == | == Problem 47 == | ||
+ | |||
+ | If <math>r</math> and <math>s</math> are the roots of the equation <math>ax^2+bx+c=0</math>, the value of <math> \frac{1}{r^{2}}+\frac{1}{s^{2}} </math> is: | ||
+ | |||
+ | <math> \textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}} </math> | ||
+ | <math> \textbf{(E)}\ \text{none of these} </math> | ||
[[1951 AHSME Problems/Problem 47|Solution]] | [[1951 AHSME Problems/Problem 47|Solution]] | ||
== Problem 48 == | == Problem 48 == | ||
+ | |||
+ | The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as: | ||
+ | |||
+ | <math> \textbf{(A)}\ 1: 2\qquad\textbf{(B)}\ 2: 3\qquad\textbf{(C)}\ 2: 5\qquad\textbf{(D)}\ 3: 4\qquad\textbf{(E)}\ 3: 5 </math> | ||
[[1951 AHSME Problems/Problem 48|Solution]] | [[1951 AHSME Problems/Problem 48|Solution]] | ||
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== See also == | == See also == | ||
− | + | ||
− | * [[ | + | * [[AMC 12 Problems and Solutions]] |
− | |||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 50p box|year=1951|before=[[1950 AHSME|1950 AHSC]]|after=[[1952 AHSME|1952 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 13:47, 23 July 2021
1951 AHSC (Answer Key) Printable version: | 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 |
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 Problem 41
- 42 Problem 42
- 43 Problem 43
- 44 Problem 44
- 45 Problem 45
- 46 Problem 46
- 47 Problem 47
- 48 Problem 48
- 49 Problem 49
- 50 Problem 50
- 51 See also
Problem 1
The percent that is greater than is:
Problem 2
A rectangular field is half as wide as it is long and is completely enclosed by yards of fencing. The area in terms of is:
Problem 3
If the length of a diagonal of a square is , then the area of the square is:
Problem 4
A barn with a flat roof is rectangular in shape, yd. wide, yd. long and yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
Problem 5
Mr. A owns a home worth dollars. He sells it to Mr. B at a profit based on the worth of the house. Mr. B sells the house back to Mr. A at a loss. Then:
Problem 6
The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
Problem 7
An error of is made in the measurement of a line long, while an error of only is made in a measurement of a line long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
Problem 8
The price of an article is cut To restore it to its former value, the new price must be increased by:
Problem 9
An equilateral triangle is drawn with a side length of A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:
Problem 10
Of the following statements, the one that is incorrect is:
Problem 11
The limit of the sum of an infinite number of terms in a geometric progression is where denotes the first term and denotes the common ratio. The limit of the sum of their squares is:
Problem 12
At o'clock, the hour and minute hands of a clock form an angle of:
Problem 13
can do a piece of work in days. is more efficient than . The number of days it takes to do the same piece of work is:
Problem 14
In connection with proof in geometry, indicate which one of the following statements is incorrect:
Problem 15
The largest number by which the expression is divisible for all possible integral values of , is:
Problem 16
If in applying the quadratic formula to a quadratic equation
it happens that , then the graph of will certainly:
Problem 17
Indicate in which one of the following equations is neither directly nor inversely proportional to :
Problem 18
The expression is to be factored into two linear prime binomial factors with integer coefficients. This can be done if is:
Problem 19
A six place number is formed by repeating a three place number; for example, or , etc. Any number of this form is always exactly divisible by:
Problem 20
When simplified and expressed with negative exponents, the expression is equal to:
Problem 21
Given: and . The inequality which is not always correct is:
Problem 22
The values of in the equation: are:
Problem 23
The radius of a cylindrical box is inches and the height is inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:
Problem 24
when simplified is:
Problem 25
The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:
Problem 26
In the equation the roots are equal when:
Problem 27
Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:
Problem 28
The pressure of wind on a sail varies jointly as the area of the sail and the square of the velocity of the wind. The pressure on a square foot is pound when the velocity is miles per hour. The velocity of the wind when the pressure on a square yard is pounds is:
Problem 29
Of the following sets of data the only one that does not determine the shape of a triangle is:
Problem 30
If two poles and high are apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
Problem 31
A total of handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:
Problem 32
If is inscribed in a semicircle whose diameter is , then must be
Problem 33
The roots of the equation can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:
[Note: Abscissas means x-coordinate.]
Problem 34
The value of is:
Problem 35
If and , then
Problem 36
Which of the following methods of proving a geometric figure a locus is not correct?
Problem 37
A number which when divided by leaves a remainder of , when divided by leaves a remainder of , by leaves a remainder of , etc., down to where, when divided by , it leaves a remainder of , is:
Problem 38
A rise of feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from to is approximately:
Problem 39
A stone is dropped into a well and the report of the stone striking the bottom is heard seconds after it is dropped. Assume that the stone falls feet in t seconds and that the velocity of sound is feet per second. The depth of the well is:
Problem 40
equals:
Problem 41
The formula expressing the relationship between and in the table is:
Problem 42
If , then:
Problem 43
Of the following statements, the only one that is incorrect is:
Problem 44
If , where are other than zero, then equals:
Problem 45
If you are given and , then the only logarithm that cannot be found without the use of tables is:
Problem 46
is a fixed diameter of a circle whose center is . From , any point on the circle, a chord is drawn perpendicular to . Then, as moves over a semicircle, the bisector of angle cuts the circle in a point that always:
Problem 47
If and are the roots of the equation , the value of is:
Problem 48
The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:
Problem 49
The medians of a right triangle which are drawn from the vertices of the acute angles are and . The value of the hypotenuse is:
Problem 50
Tom, Dick and Harry started out on a -mile journey. Tom and Harry went by automobile at the rate of mph, while Dick walked at the rate of mph. After a certain distance, Harry got off and walked on at mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:
See also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1950 AHSC |
Followed by 1952 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.