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Difference between revisions of "1951 AHSME Problems"

m (→‎Problem 2)
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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> \mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \frac {1}{2}(a^2+b^2) } \qquad \mathrm{(E) \ \text{none of these} }  </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> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </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> \textrm{(A)}\ \text{A comes out even} \qquad\textrm{(B)}\ \text{A makes 1100 on the deal} \qquad\textrm{(C)}\ \text{A makes 1000 on the deal}</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>\textrm{(D)}\ \text{A loses 900 on the deal} \qquad\textrm{(E)}\ \text{A loses 1000 on the deal}</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> \textrm{(A)}\ \text{the volume of the box} \qquad\textrm{(B)}\ \text{the square root of the volume} \qquad\textrm{(C)}\ \text{twice the volume}</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> \textrm{(D)}\ \text{the square of the volume} \qquad\textrm{(E)}\ \text{the cube of the volume}</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]]
Line 48: Line 49:
 
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> \mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both} </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> \mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers} </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> \mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a </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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The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 \minus{} r}</math> where <math> a</math> denotes the first term and <math> \minus{} 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
 
The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 \minus{} r}</math> where <math> a</math> denotes the first term and <math> \minus{} 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
  
−
<math> \textrm{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textrm{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textrm{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textrm{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textrm{(E)}\ \text{none of these}</math>
+
<math> \textbf{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 \plus{} 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> \textrm{(A)}\ 30^{\circ} \qquad\textrm{(B)}\ 5^{\circ} \qquad\textrm{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textrm{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textrm{(E)}\ 28^{\circ}</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> \textrm{(A)}\ 13\frac {1}{2} \qquad\textrm{(B)}\ 4\frac {1}{2} \qquad\textrm{(C)}\ 6 \qquad\textrm{(D)}\ 3 \qquad\textrm{(E)}\ \text{none of these answers}</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]]

Revision as of 10:00, 28 August 2011

Problem 1

The percent that $M$ is greater than $N$ is:

$\textbf{(A) \ } \frac {100(M - N)}{M} \qquad \textbf{(B) \ } \frac {100(M - N)}{N} \qquad \textbf{(C) \ } \frac {M - N}{N} \qquad \textbf{(D) \ } \frac {M - N}{M} \qquad \textbf{(E) \ } \frac {100(M + N)}{N}$

Solution

Problem 2

A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:

$(\textbf{A})\ \frac{x^2}2 \qquad (\textbf{B})\ 2x^2 \qquad (\textbf{C})\ \frac{2x^2}9 \qquad (\textbf{D})\ \frac{x^2}{18} \qquad (\textbf{E})\ \frac{x^2}{72}$

Solution

Problem 3

If the length of a diagonal of a square is $a + b$, then the area of the square is:

$\textbf{(A)}\ (a+b)^{2}\qquad\textbf{(B)}\ \frac{1}{2}(a+b)^{2}\qquad\textbf{(C)}\ a^{2}+b^{2}$ $\textbf{(D)}\ \frac{1}{2}(a^{2}+b^{2})\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 4

A barn with a flat roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ 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:

$\textbf{(A) \ } 360 \qquad \textbf{(B) \ } 460 \qquad \textbf{(C) \ } 490 \qquad \textbf{(D) \ } 590 \qquad \textbf{(E) \ } 720$

Solution

Problem 5

Mr. A owns a home worth $10,000$ dollars. He sells it to Mr. B at a $10 \%$ profit based on the worth of the house. Mr. B sells the house back to Mr. A at a $10 \%$ loss. Then:

$\textbf{(A)}\ \text{A comes out even} \qquad\textbf{(B)}\ \text{A makes 1100 on the deal} \qquad\textbf{(C)}\ \text{A makes 1000 on the deal}$ $\textbf{(D)}\ \text{A loses 900 on the deal} \qquad\textbf{(E)}\ \text{A loses 1000 on the deal}$

Solution

Problem 6

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

$\textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$ $\textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}$

Solution

Problem 7

An error of $.02"$ is made in the measurement of a line $10"$ long, while an error of only $.2"$ is made in a measurement of a line $100"$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:

$\textbf{(A) \ } \text{greater by }.18 \qquad\textbf{(B) \ } \text{the same} \qquad \textbf{(C) \ } \text{less} \qquad\textbf{(D) \ } 10\text{ times as great} \qquad\textbf{(E) \ } \text{correctly described by both}$

Solution

Problem 8

The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:

$\textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers}$

Solution

Problem 9

An equilateral triangle is drawn with a side length of $a.$ 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:

$\textbf{(A) \ } \text{Infinite} \qquad\textbf{(B) \ } 5\frac{1}{4}a \qquad \textbf{(C) \ } 2a \qquad\textbf{(D) \ } 6a \qquad\textbf{(E) \ } 4\frac{1}{2}a$

Solution

Problem 10

Of the following statements, the one that is incorrect is:

$\textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}\qquad\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$ $\textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$ $\textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$ $\textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

Solution

Problem 11

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 \minus{} r}$ (Error compiling LaTeX. Unknown error_msg) where $a$ denotes the first term and $\minus{} 1 < r < 1$ (Error compiling LaTeX. Unknown error_msg) denotes the common ratio. The limit of the sum of their squares is:

$\textbf{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textbf{(E)}\ \text{none of these}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 12

At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:

$\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$

Solution

Problem 13

$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:

$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$

Solution

Problem 14

In connection with proof in geometry, indicate which one of the following statements is incorrect:

$\textbf{(A)}\ \text{Some statements are accepted without being proved.}$ $\textbf{(B)}\ \text{In some cases there is more than one correct order in proving certain propositions.}$ $\textbf{(C)}\ \text{Every term used in a proof must have been defined previously.}$ $\textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.}$ $\textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.}$

Solution

Problem 15

The largest number by which the expression $n^3-n$ is divisible for all possible integral values of $n$, is:

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution

Problem 16

If in applying the quadratic formula to a quadratic equation

\[f(x) \equiv ax^2 + bx + c = 0,\]

it happens that $c = \frac{b^2}{4a}$, then the graph of $y = f(x)$ will certainly:

$\textbf{(A)}\ \text{have a maximum}\qquad\textbf{(B)}\ \text{have a minimum}\qquad\textbf{(C)}\ \text{be tangent to the x-axis}\\ \qquad\textbf{(D)}\ \text{be tangent to the y-axis}\qquad\textbf{(E)}\ \text{lie in one quadrant only}$

Solution

Problem 17

Indicate in which one of the following equations $y$ is neither directly nor inversely proportional to $x$:

$\textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y\qquad\textbf{(D)}\ 3x+y = 10$

Solution

Problem 18

The expression $21x^2+ax+21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be one if $a$ is:

$\textbf{(A)}\ \text{any odd number}\qquad\textbf{(B)}\ \text{some odd number}\qquad\textbf{(C)}\ \text{any even number}\qquad\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero}$

Solution

Problem 19

A six place number is formed by repeating a three place number; for example, $256256$ or $678678$, etc. Any number of this form is always exactly divisible by:

$\textbf{(A)}\ 7\text{ only}\qquad\textbf{(B)}\ 11\text{ only}\qquad\textbf{(C)}\ 13\text{ only}\qquad\textbf{(D)}\ 101\qquad\textbf{(E)}\ 1001$

Solution

Problem 20

When simplified and expressed with negative exponents, the expression $(x+y)^{-1}(x^{-1}+y^{-1})$ is equal to:

$\textbf{(A)}\ x^{-2}+2x^{-1}y^{-1}+y^{-2}\qquad\textbf{(B)}\ x^{-2}+2^{-1}x^{-1}y^{-1}+y^{-2}\qquad\textbf{(C)}\ x^{-1}y^{-1}$ $\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero}$

Solution

Problem 21

Given: $x > 0, y > 0, x > y$ and $z\not = 0$. The inequality which is not always correct is:

$\textbf{(A)}\ x+z > y+z\qquad\textbf{(B)}\ x-z > y-z\qquad\textbf{(C)}\ xz > yz$ $\textbf{(D)}\ \frac{x}{z^{2}}>\frac{y}{z^{2}}\qquad\textbf{(E)}\ xz^{2}> yz^{2}$

Solution

Problem 22

The values of $a$ in the equation: $\log_{10}(a^{2}-15a) = 2$ are:

$\textbf{(A)}\ \frac{15\pm\sqrt{233}}{2}\qquad\textbf{(B)}\ 20,-5\qquad\textbf{(C)}\ \frac{15\pm\sqrt{305}}{2}\qquad\textbf{(D)}\ \pm20$ $\textbf{(E)}\ \text{none of these}$

Solution

Problem 23

The radius of a cylindrical box is $8$ inches and the height is $3$ 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:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 5\frac{1}{3}\qquad\textbf{(C)}\ \text{any number}\qquad\textbf{(D)}\ \text{non-existent}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 24

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:

$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$

Solution

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:

$\textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf{(C)}\ \frac{2}{\sqrt{3}}\text{ times the second}\\ \textbf{(D)}\ \frac{\sqrt{2}}{\sqrt{3}}\text{ times the second}\qquad\textbf{(E)}\ \text{indeterminately related to the second}$

Solution

Problem 26

In the equation $\frac{x(x-1)-(m+1)}{(x-1)(m-1)}=\frac{x}{m}$ the roots are equal when:

$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$

Solution

Problem 27

Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:

$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$ $\textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed}$ $\textbf{(E)}\ \text{none of the above relations are true}$

Solution

Problem 28

The pressure $(P)$ of wind on a sail varies jointly as the area $(A)$ of the sail and the square of the velocity $(V)$ of the wind. The pressure on a square foot is $1$ pound when the velocity is $16$ miles per hour. The velocity of the wind when the pressure on a square yard is $36$ pounds is:

$\textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph}$

Solution

Problem 29

Of the following sets of data the only one that does not determine the shape of a triangle is:

$\textbf{(A)}\ \text{the ratio of two sides and the inc{}luded angle}\\ \qquad\textbf{(B)}\ \text{the ratios of the three altitudes}\\ \qquad\textbf{(C)}\ \text{the ratios of the three medians}\\ \qquad\textbf{(D)}\ \text{the ratio of the altitude to the corresponding base}\\ \qquad\textbf{(E)}\ \text{two angles}$

Solution

Problem 30

If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 31

A total of $28$ 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:

$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$

Solution

Problem 32

If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$, then $AC+BC$ must be

$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$

Solution

Problem 33

The roots of the equation $x^{2}-2x = 0$ can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:

$\textbf{(A)}\ y = x^{2}, y = 2x\qquad\textbf{(B)}\ y = x^{2}-2x, y = 0\qquad\textbf{(C)}\ y = x, y = x-2\qquad\textbf{(D)}\ y = x^{2}-2x+1, y = 1$ $\textbf{(E)}\ y = x^{2}-1, y = 2x-1$

[Note: Abscissas means x-coordinate.]

Solution

Problem 34

The value of $10^{\log_{10}7}$ is:

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$

Solution

Problem 35

If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$, then

$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$ $\textbf{(E)}\ x^{y}= q^{z}$

Solution

Problem 36

Which of the following methods of proving a geometric figure a locus is not correct?

$\textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.}$ $\textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.}$ $\textbf{(C)}\ \text{Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.}$ $\textbf{(D)}\ \text{Every point not on the locus does not satisfy the conditions and every point not satisfying}\\ \text{the conditions is not on the locus.}$ $\textbf{(E)}\ \text{Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.}$

Solution

Problem 37

A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves a remainder of $8$, by $8$ leaves a remainder of $7$, etc., down to where, when divided by $2$, it leaves a remainder of $1$, is:

$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$

Solution

Problem 38

A rise of $600$ 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 $3\%$ to $2\%$ is approximately:

$\textbf{(A)}\ 10000\text{ ft.}\qquad\textbf{(B)}\ 20000\text{ ft.}\qquad\textbf{(C)}\ 30000\text{ ft.}\qquad\textbf{(D)}\ 12000\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 39

A stone is dropped into a well and the report of the stone striking the bottom is heard $7.7$ seconds after it is dropped. Assume that the stone falls $16t^2$ feet in t seconds and that the velocity of sound is $1120$ feet per second. The depth of the well is:

$\textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ft.}\qquad\textbf{(C)}\ 1568\text{ ft.}\qquad\textbf{(D)}\ 156.8\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 40

$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:

$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$

Solution

Problem 41

The formula expressing the relationship between $x$ and $y$ in the table is: \[\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}\]

$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$ $\textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4$

Solution

Problem 42

Solution

Problem 43

Solution

Problem 44

Solution

Problem 45

Solution

Problem 46

Solution

Problem 47

Solution

Problem 48

Solution

Problem 49

The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 50

Tom, Dick and Harry started out on a $100$-mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ 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:

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ \text{none of these answers}$

Solution

See also

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