Difference between revisions of "1954 AHSME Problems"
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− | == Problem 1 == | + | {{AHSC 50 Problems |
+ | |year=1954 | ||
+ | }} | ||
+ | == Problem 1== | ||
− | <math> \ | + | The square of <math>5-\sqrt{y^2-25}</math> is: |
+ | <math>\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 1|Solution]] | [[1954 AHSME Problems/Problem 1|Solution]] | ||
− | == Problem 2 == | + | == Problem 2== |
− | |||
+ | The equation <math>\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0</math> can be transformed by eliminating fractions to the equation <math>x^2-5x+4=0</math>. | ||
+ | The roots of the latter equation are <math>4</math> and <math>1</math>. Then the roots of the first equation are: | ||
+ | |||
+ | <math>\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 2|Solution]] | [[1954 AHSME Problems/Problem 2|Solution]] | ||
− | == Problem 3 == | + | == Problem 3== |
− | |||
+ | If <math>x</math> varies as the cube of <math>y</math>, and <math>y</math> varies as the fifth root of <math>z</math>, then <math>x</math> varies as the nth power of <math>z</math>, where n is: | ||
+ | |||
+ | <math>\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 3|Solution]] | [[1954 AHSME Problems/Problem 3|Solution]] | ||
− | == Problem 4 == | + | == Problem 4== |
− | |||
+ | If the Highest Common Divisor of <math>6432</math> and <math>132</math> is diminished by <math>8</math>, it will equal: | ||
+ | |||
+ | <math>\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 4|Solution]] | [[1954 AHSME Problems/Problem 4|Solution]] | ||
− | == Problem 5 == | + | == Problem 5== |
− | |||
+ | A regular hexagon is inscribed in a circle of radius <math>10</math> inches. Its area is: | ||
+ | |||
+ | <math>\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 5|Solution]] | [[1954 AHSME Problems/Problem 5|Solution]] | ||
− | == Problem 6 == | + | == Problem 6== |
− | <math> \ | + | The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is: |
+ | <math>\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 6|Solution]] | [[1954 AHSME Problems/Problem 6|Solution]] | ||
− | == Problem 7 == | + | == Problem 7== |
− | <math> \ | + | A housewife saved <math>\textdollar{2.50}</math> in buying a dress on sale. If she spent <math>\textdollar{25}</math> for the dress, she saved about: |
+ | <math>\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\% </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 7|Solution]] | [[1954 AHSME Problems/Problem 7|Solution]] | ||
− | == Problem 8 == | + | == Problem 8== |
− | + | The base of a triangle is twice as long as a side of a square and their areas are the same. | |
+ | Then the ratio of the altitude of the triangle to the side of the square is: | ||
+ | <math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 8|Solution]] | [[1954 AHSME Problems/Problem 8|Solution]] | ||
− | == Problem 9 == | + | == Problem 9== |
− | <math> | + | A point <math>P</math> is outside a circle and is <math>13</math> inches from the center. A secant from <math>P</math> cuts the circle at <math>Q</math> and <math>R</math> |
+ | so that the external segment of the secant <math>PQ</math> is <math>9</math> inches and <math>QR</math> is <math>7</math> inches. The radius of the circle is: | ||
+ | <math>\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7" </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 9|Solution]] | [[1954 AHSME Problems/Problem 9|Solution]] | ||
− | == Problem 10 == | + | == Problem 10== |
− | <math> | + | The sum of the numerical coefficients in the expansion of the binomial <math>(a+b)^6</math> is: |
+ | <math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 10|Solution]] | [[1954 AHSME Problems/Problem 10|Solution]] | ||
− | == Problem 11 == | + | == Problem 11== |
− | <math> \ | + | A merchant placed on display some dresses, each with a marked price. He then posted a sign “<math>\frac{1}{3}</math> off on these dresses.” |
+ | The cost of the dresses was <math>\frac{3}{4}</math> of the price at which he actually sold them. Then the ratio of the cost to the marked price was: | ||
+ | <math>\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 11|Solution]] | [[1954 AHSME Problems/Problem 11|Solution]] | ||
− | == Problem 12 == | + | == Problem 12== |
− | + | The solution of the equations | |
+ | <cmath> | ||
+ | |||
+ | is: | ||
+ | |||
+ | <math>\textbf{(A)}\ x=18, y=12 \qquad | ||
+ | \textbf{(B)}\ x=0, y=0 \qquad | ||
+ | \textbf{(C)}\ \text{There is no solution} \ | ||
+ | \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad | ||
+ | \textbf{(E)}\ x=8, y=5 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 12|Solution]] | [[1954 AHSME Problems/Problem 12|Solution]] | ||
− | == Problem 13 == | + | == Problem 13== |
− | + | A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off | |
+ | by the sides of the quadrilateral, their sum will be: | ||
+ | <math>\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 13|Solution]] | [[1954 AHSME Problems/Problem 13|Solution]] | ||
− | == Problem 14 == | + | == Problem 14== |
− | <math> \ | + | When simplified <math>\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}</math> equals: |
+ | <math>\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 14|Solution]] | [[1954 AHSME Problems/Problem 14|Solution]] | ||
− | == Problem 15 == | + | == Problem 15== |
− | <math> \ | + | <math>\log 125</math> equals: |
+ | <math>\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \ | ||
+ | \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5) </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 15|Solution]] | [[1954 AHSME Problems/Problem 15|Solution]] | ||
− | == Problem 16 == | + | == Problem 16== |
− | <math> | + | If <math>f(x) = 5x^2 - 2x - 1</math>, then <math>f(x + h) - f(x)</math> equals: |
+ | <math>\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 16|Solution]] | [[1954 AHSME Problems/Problem 16|Solution]] | ||
− | == Problem 17 == | + | == Problem 17== |
− | <math> | + | The graph of the function <math>f(x) = 2x^3 - 7</math> goes: |
+ | <math>\textbf{(A)}\ \text{up to the right and down to the left} \ \textbf{(B)}\ \text{down to the right and up to the left}\ \textbf{(C)}\ \text{up to the right and up to the left}\ \textbf{(D)}\ \text{down to the right and down to the left}\ \textbf{(E)}\ \text{none of these ways.} </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 17|Solution]] | [[1954 AHSME Problems/Problem 17|Solution]] | ||
− | == Problem 18 == | + | == Problem 18== |
− | <math> | + | Of the following sets, the one that includes all values of <math>x</math> which will satisfy <math>2x - 3 > 7 - x</math> is: |
+ | <math>\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 18|Solution]] | [[1954 AHSME Problems/Problem 18|Solution]] | ||
− | == Problem 19 == | + | == Problem 19== |
+ | |||
+ | If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other} </math> |
[[1954 AHSME Problems/Problem 19|Solution]] | [[1954 AHSME Problems/Problem 19|Solution]] | ||
− | == Problem 20 == | + | == Problem 20== |
+ | |||
+ | The equation <math>x^3+6x^2+11x+6=0</math> has: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root} </math> |
[[1954 AHSME Problems/Problem 20|Solution]] | [[1954 AHSME Problems/Problem 20|Solution]] | ||
− | == Problem 21 == | + | == Problem 21== |
− | <math> \ | + | The roots of the equation <math>2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5</math> can be found by solving: |
+ | <math> \textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 21|Solution]] | [[1954 AHSME Problems/Problem 21|Solution]] | ||
− | == Problem 22 == | + | == Problem 22== |
+ | |||
+ | The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>, | ||
+ | since division by zero is not allowed. For other values of <math>x</math>: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1. </math> |
[[1954 AHSME Problems/Problem 22|Solution]] | [[1954 AHSME Problems/Problem 22|Solution]] | ||
− | == Problem 23 == | + | == Problem 23== |
+ | |||
+ | If the margin made on an article costing <math>C</math> dollars and selling for <math>S</math> dollars is <math>M=\frac{1}{n}C</math>, then the margin is given by: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S </math> |
[[1954 AHSME Problems/Problem 23|Solution]] | [[1954 AHSME Problems/Problem 23|Solution]] | ||
− | == Problem 24 == | + | == Problem 24== |
− | <math> \textbf{(A) \ } | + | The values of <math>k</math> for which the equation <math>2x^2-kx+x+8=0</math> will have real and equal roots are: |
+ | |||
+ | <math> \textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9} </math> | ||
[[1954 AHSME Problems/Problem 24|Solution]] | [[1954 AHSME Problems/Problem 24|Solution]] | ||
− | == Problem 25 == | + | == Problem 25== |
− | <math> \textbf{(A) \ } | + | The two roots of the equation <math>a(b-c)x^2+b(c-a)x+c(a-b)=0</math> are <math>1</math> and: |
+ | |||
+ | <math> \textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)} </math> | ||
[[1954 AHSME Problems/Problem 25|Solution]] | [[1954 AHSME Problems/Problem 25|Solution]] | ||
− | == Problem 26 == | + | == Problem 26== |
− | <math> \textbf{(A) \ } | + | The straight line <math>\overline{AB}</math> is divided at <math>C</math> so that <math>AC=3CB</math>. Circles are described on <math>\overline{AC}</math> |
+ | and <math>\overline{CB}</math> as diameters and a common tangent meets <math>AB</math> produced at <math>D</math>. Then <math>BD</math> equals: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii} </math> | ||
[[1954 AHSME Problems/Problem 26|Solution]] | [[1954 AHSME Problems/Problem 26|Solution]] | ||
− | == Problem 27 == | + | == Problem 27== |
− | <math> \textbf{(A) \ } | + | A right circular cone has for its base a circle having the same radius as a given sphere. |
+ | The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{2}{1}\qquad\textbf{(E)}\ \sqrt{\frac{5}{4}} </math> | ||
[[1954 AHSME Problems/Problem 27|Solution]] | [[1954 AHSME Problems/Problem 27|Solution]] | ||
− | == Problem 28 == | + | == Problem 28== |
− | <math> \textbf{(A) \ } | + | If <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, the value of <math>\frac{3mr-nt}{4nt-7mr}</math> is: |
+ | |||
+ | <math> \textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3} </math> | ||
[[1954 AHSME Problems/Problem 28|Solution]] | [[1954 AHSME Problems/Problem 28|Solution]] | ||
− | == Problem 29 == | + | == Problem 29== |
− | <math> \textbf{(A) \ | + | If the ratio of the legs of a right triangle is <math>1: 2</math>, then the ratio of the corresponding segments of |
+ | the hypotenuse made by a perpendicular upon it from the vertex is: | ||
+ | |||
+ | <math> \textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5 </math> | ||
[[1954 AHSME Problems/Problem 29|Solution]] | [[1954 AHSME Problems/Problem 29|Solution]] | ||
− | == Problem 30 == | + | == Problem 30== |
− | <math> | + | <math>A</math> and <math>B</math> together can do a job in <math>2</math> days; <math>B</math> and <math>C</math> can do it in four days; and <math>A</math> and <math>C</math> in <math>2\frac{2}{5}</math> days. |
+ | The number of days required for A to do the job alone is: | ||
+ | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 30|Solution]] | [[1954 AHSME Problems/Problem 30|Solution]] | ||
− | == Problem 31 == | + | == Problem 31== |
− | <math> \ | + | In <math>\triangle ABC</math>, <math>AB=AC</math>, <math>\angle A=40^\circ</math>. Point <math>O</math> is within the triangle with <math>\angle OBC \cong \angle OCA</math>. |
+ | The number of degrees in <math>\angle BOC</math> is: | ||
+ | <math>\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}</math> | ||
+ | |||
[[1954 AHSME Problems/Problem 31|Solution]] | [[1954 AHSME Problems/Problem 31|Solution]] | ||
− | == Problem 32 == | + | == Problem 32== |
+ | |||
+ | The factors of <math>x^4+64</math> are: | ||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8) </math> |
[[1954 AHSME Problems/Problem 32|Solution]] | [[1954 AHSME Problems/Problem 32|Solution]] | ||
− | == Problem 33 == | + | == Problem 33== |
− | <math> \ | + | A bank charges <math>\textdollar{6}</math> for a loan of <math>\textdollar{120}</math>. The borrower receives <math>\textdollar{114}</math> and |
+ | repays the loan in <math>12</math> installments of <math>\textdollar{10}</math> a month. The interest rate is approximately: | ||
+ | <math>\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \% </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 33|Solution]] | [[1954 AHSME Problems/Problem 33|Solution]] | ||
− | == Problem 34 == | + | == Problem 34== |
+ | |||
+ | The fraction <math>\frac{1}{3}</math>: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9} </math> |
[[1954 AHSME Problems/Problem 34|Solution]] | [[1954 AHSME Problems/Problem 34|Solution]] | ||
− | == Problem 35 == | + | == Problem 35== |
− | <math> \textbf{(A) \ } | + | In the right triangle shown the sum of the distances <math>BM</math> and <math>MA</math> is equal to the sum of the distances <math>BC</math> and <math>CA</math>. |
+ | If <math>MB = x, CB = h</math>, and <math>CA = d</math>, then <math>x</math> equals: | ||
+ | |||
+ | <asy> | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | draw((0,0)--(8,0)--(0,5)--cycle); | ||
+ | label("C",(0,0),SW); | ||
+ | label("A",(8,0),SE); | ||
+ | label("M",(0,5),N); | ||
+ | dot((0,3.5)); | ||
+ | label("B",(0,3.5),W); | ||
+ | label("$x$",(0,4.25),W); | ||
+ | label("$h$",(0,1),W); | ||
+ | label("$d$",(4,0),S);</asy> | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h </math> | ||
[[1954 AHSME Problems/Problem 35|Solution]] | [[1954 AHSME Problems/Problem 35|Solution]] | ||
− | == Problem 36 == | + | == Problem 36== |
− | <math> \textbf{(A) \ } | + | A boat has a speed of <math>15</math> mph in still water. In a stream that has a current of <math>5</math> mph it travels a certain |
+ | distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is: | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8} </math> | ||
[[1954 AHSME Problems/Problem 36|Solution]] | [[1954 AHSME Problems/Problem 36|Solution]] | ||
− | == Problem 37 == | + | == Problem 37== |
− | <math> \textbf{(A) \ } | + | Given <math>\triangle PQR</math> with <math>\overline{RS}</math> bisecting <math>\angle R</math>, <math>PQ</math> extended to <math>D</math> and <math>\angle n</math> a right angle, then: |
+ | |||
+ | <asy> | ||
+ | path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) | ||
+ | { | ||
+ | pair M,N; | ||
+ | path mark; | ||
+ | M=t*0.03*unit(A-B)+B; | ||
+ | N=t*0.03*unit(C-B)+B; | ||
+ | if(flip) | ||
+ | mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B)); | ||
+ | else | ||
+ | mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B)); | ||
+ | return mark; | ||
+ | } | ||
+ | unitsize(1.5cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | pair P=(0,0), R=(3,2), Q=(4,0); | ||
+ | pair S0=bisectorpoint(P,R,Q); | ||
+ | pair Sp=extension(P,Q,S0,R); | ||
+ | pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp); | ||
+ | pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R); | ||
+ | draw(P--R--Q); | ||
+ | draw(R--Sp); | ||
+ | draw(P--D--M); | ||
+ | draw(anglemark2(Sp,P,R,17)); | ||
+ | label("$p$",P+(0.35,0.1)); | ||
+ | draw(anglemark2(R,Q,P,11)); | ||
+ | label("$q$",Q+(-0.17,0.1)); | ||
+ | draw(anglemark2(R,Np,D,8,true)); | ||
+ | label("$n$",Np+(+0.12,0.07)); | ||
+ | draw(anglemark2(R,M,D,13,true)); | ||
+ | label("$m$",M+(+0.25,0.03)); | ||
+ | draw(anglemark2(M,D,P,29)); | ||
+ | label("$d$",D+(-0.75,0.095)); | ||
+ | pen f=fontsize(10pt); | ||
+ | label("$R$",R,N,f); | ||
+ | label("$P$",P,S,f); | ||
+ | label("$S$",Sp,S,f); | ||
+ | label("$Q$",Q,S,f); | ||
+ | label("$D$",D,S,f);</asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad | ||
+ | \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) </math> | ||
+ | <math> \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad | ||
+ | \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad | ||
+ | \textbf{(E)}\ \text{none of these is correct} </math> | ||
[[1954 AHSME Problems/Problem 37|Solution]] | [[1954 AHSME Problems/Problem 37|Solution]] | ||
− | == Problem 38 == | + | == Problem 38== |
− | If <math>\log 2 = .3010</math> and <math>\log 3 = .4771</math>, the value of <math>x</math> when <math>3^{x+3} = 135</math> is approximately | + | If <math>\log 2=.3010</math> and <math>\log 3=.4771</math>, the value of <math>x</math> when <math>3^{x+3}=135</math> is approximately: |
− | <math> \textbf{(A) \ | + | <math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63 </math> |
+ | |||
+ | [[1954 AHSME Problems/Problem 38|Solution]] | ||
− | + | == Problem 39== | |
− | + | The locus of the midpoint of a line segment that is drawn from a given external point <math>P</math> to a given circle with center <math>O</math> and radius <math>r</math>, is: | |
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r </math> |
[[1954 AHSME Problems/Problem 39|Solution]] | [[1954 AHSME Problems/Problem 39|Solution]] | ||
− | == Problem 40 == | + | == Problem 40== |
+ | |||
+ | If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3} </math> |
[[1954 AHSME Problems/Problem 40|Solution]] | [[1954 AHSME Problems/Problem 40|Solution]] | ||
− | == Problem 41 == | + | == Problem 41== |
− | <math> | + | The sum of all the roots of <math>4x^3-8x^2-63x-9=0</math> is: |
+ | <math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 41|Solution]] | [[1954 AHSME Problems/Problem 41|Solution]] | ||
− | == Problem 42 == | + | == Problem 42== |
+ | |||
+ | Consider the graphs of | ||
+ | <cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath> | ||
+ | and | ||
+ | <cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath> | ||
+ | on the same set of axis. | ||
+ | These parabolas are exactly the same shape. Then: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math> |
[[1954 AHSME Problems/Problem 42|Solution]] | [[1954 AHSME Problems/Problem 42|Solution]] | ||
− | == Problem 43 == | + | == Problem 43== |
− | <math> | + | The hypotenuse of a right triangle is <math>10</math> inches and the radius of the inscribed circle is <math>1</math> inch. The perimeter of the triangle in inches is: |
+ | <math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 43|Solution]] | [[1954 AHSME Problems/Problem 43|Solution]] | ||
− | == Problem 44 == | + | == Problem 44== |
− | <math> | + | A man born in the first half of the nineteenth century was <math>x</math> years old in the year <math>x^2</math>. He was born in: |
+ | <math>\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 44|Solution]] | [[1954 AHSME Problems/Problem 44|Solution]] | ||
− | == Problem 45 == | + | == Problem 45== |
− | <math> \textbf{(A) \ } | + | In a rhombus, <math>ABCD</math>, line segments are drawn within the rhombus, parallel to diagonal <math>BD</math>, |
+ | and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment | ||
+ | as a function of its distance from vertex <math>A</math>. The graph is: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.} </math> | ||
[[1954 AHSME Problems/Problem 45|Solution]] | [[1954 AHSME Problems/Problem 45|Solution]] | ||
− | == Problem 46 == | + | == Problem 46== |
− | <math> \textbf{(A) \ } | + | In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> equals: |
+ | |||
+ | <asy> | ||
+ | unitsize(5cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | dotfactor=3; | ||
+ | pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); | ||
+ | pair O=(0,3/8); | ||
+ | draw((-2/3,9/16)--(2/3,9/16)); | ||
+ | draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); | ||
+ | draw(Circle(O,3/16)); | ||
+ | draw((-2/3,0)--(2/3,0)); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,N); | ||
+ | label("$\frac{3}{8}$",O); | ||
+ | draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3)); | ||
+ | draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3)); | ||
+ | label("$\frac{1}{2}$",(.5,.25)); | ||
+ | draw((.5,.33)--(.5,.5),EndArrow(3)); | ||
+ | draw((.5,.17)--(.5,0),EndArrow(3)); | ||
+ | label("$x$",midpoint((.5,.5)--(.5,9/16))); | ||
+ | draw((.5,5/8)--(.5,9/16),EndArrow(3)); | ||
+ | label("$60^{\circ}$",(0.01,0.12)); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C);</asy> | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}" </math> | ||
[[1954 AHSME Problems/Problem 46|Solution]] | [[1954 AHSME Problems/Problem 46|Solution]] | ||
− | == Problem 47 == | + | == Problem 47== |
+ | |||
+ | At the midpoint of line segment <math>AB</math> which is <math>p</math> units long, a perpendicular <math>MR</math> is erected with length <math>q</math> units. | ||
+ | An arc is described from <math>R</math> with a radius equal to <math>\frac{1}{2}AB</math>, meeting <math>AB</math> at <math>T</math>. Then <math>AT</math> and <math>TB</math> are the roots of: | ||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ x^2+px+q^2=0\ \textbf{(B)}\ x^2-px+q^2=0\ \textbf{(C)}\ x^2+px-q^2=0\ \textbf{(D)}\ x^2-px-q^2=0\ \textbf{(E)}\ x^2-px+q=0 </math> |
[[1954 AHSME Problems/Problem 47|Solution]] | [[1954 AHSME Problems/Problem 47|Solution]] | ||
− | == Problem 48 == | + | == Problem 48== |
− | <math> \ | + | A train, an hour after starting, meets with an accident which detains it a half hour, after which it |
+ | proceeds at <math>\frac{3}{4}</math> of its former rate and arrives <math>3\tfrac{1}{2}</math> hours late. | ||
+ | Had the accident happened <math>90</math> miles farther along the line, it would have arrived only <math>3</math> hours late. The length of the trip in miles was: | ||
+ | <math>\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550 </math> | ||
+ | |||
[[1954 AHSME Problems/Problem 48|Solution]] | [[1954 AHSME Problems/Problem 48|Solution]] | ||
− | == Problem 49 == | + | == Problem 49== |
+ | |||
+ | The difference of the squares of two odd numbers is always divisible by <math>8</math>. If <math>a>b</math>, and <math>2a+1</math> and <math>2b+1</math> are the odd numbers, | ||
+ | to prove the given statement we put the difference of the squares in the form: | ||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ (2a+1)^2-(2b+1)^2\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\ \textbf{(D)}\ 4(a-b)(a+b+1)\ \textbf{(E)}\ 4(a^2+a-b^2-b) </math> |
[[1954 AHSME Problems/Problem 49|Solution]] | [[1954 AHSME Problems/Problem 49|Solution]] | ||
− | == Problem 50 == | + | == Problem 50== |
+ | |||
+ | The times between <math>7</math> and <math>8</math> o'clock, correct to the nearest minute, when the hands of a clock will form an angle of <math>84^{\circ}</math> are: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{7: 23 and 7: 53}\qquad |
+ | \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad | ||
+ | \textbf{(C)}\ \text{7: 22 and 7: 53}\\ | ||
+ | \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad | ||
+ | \textbf{(E)}\ \text{7: 21 and 7: 49} </math> | ||
[[1954 AHSME Problems/Problem 50|Solution]] | [[1954 AHSME Problems/Problem 50|Solution]] | ||
== See also == | == See also == | ||
− | + | ||
− | * [[ | + | * [[AMC 12 Problems and Solutions]] |
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 50p box|year=1954|before=[[1953 AHSME|1953 AHSC]]|after=[[1955 AHSME|1955 AHSC]]}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 18:16, 8 November 2024
1954 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
[hide]- 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 square of is:
Problem 2
The equation can be transformed by eliminating fractions to the equation . The roots of the latter equation are and . Then the roots of the first equation are:
Problem 3
If varies as the cube of , and varies as the fifth root of , then varies as the nth power of , where n is:
Problem 4
If the Highest Common Divisor of and is diminished by , it will equal:
Problem 5
A regular hexagon is inscribed in a circle of radius inches. Its area is:
Problem 6
The value of is:
Problem 7
A housewife saved in buying a dress on sale. If she spent for the dress, she saved about:
Problem 8
The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:
Problem 9
A point is outside a circle and is inches from the center. A secant from cuts the circle at and so that the external segment of the secant is inches and is inches. The radius of the circle is:
Problem 10
The sum of the numerical coefficients in the expansion of the binomial is:
Problem 11
A merchant placed on display some dresses, each with a marked price. He then posted a sign “ off on these dresses.” The cost of the dresses was of the price at which he actually sold them. Then the ratio of the cost to the marked price was:
Problem 12
The solution of the equations
is:
Problem 13
A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, their sum will be:
Problem 14
When simplified equals:
Problem 15
equals:
Problem 16
If , then equals:
Problem 17
The graph of the function goes:
Problem 18
Of the following sets, the one that includes all values of which will satisfy is:
Problem 19
If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:
Problem 20
The equation has:
Problem 21
The roots of the equation can be found by solving:
Problem 22
The expression cannot be evaluated for or , since division by zero is not allowed. For other values of :
Problem 23
If the margin made on an article costing dollars and selling for dollars is , then the margin is given by:
Problem 24
The values of for which the equation will have real and equal roots are:
Problem 25
The two roots of the equation are and:
Problem 26
The straight line is divided at so that . Circles are described on and as diameters and a common tangent meets produced at . Then equals:
Problem 27
A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
Problem 28
If and , the value of is:
Problem 29
If the ratio of the legs of a right triangle is , then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:
Problem 30
and together can do a job in days; and can do it in four days; and and in days. The number of days required for A to do the job alone is:
Problem 31
In , , . Point is within the triangle with . The number of degrees in is:
Problem 32
The factors of are:
Problem 33
A bank charges for a loan of . The borrower receives and repays the loan in installments of a month. The interest rate is approximately:
Problem 34
The fraction :
Problem 35
In the right triangle shown the sum of the distances and is equal to the sum of the distances and . If , and , then equals:
Problem 36
A boat has a speed of mph in still water. In a stream that has a current of mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:
Problem 37
Given with bisecting , extended to and a right angle, then:
Problem 38
If and , the value of when is approximately:
Problem 39
The locus of the midpoint of a line segment that is drawn from a given external point to a given circle with center and radius , is:
Problem 40
If , then equals:
Problem 41
The sum of all the roots of is:
Problem 42
Consider the graphs of and on the same set of axis. These parabolas are exactly the same shape. Then:
Problem 43
The hypotenuse of a right triangle is inches and the radius of the inscribed circle is inch. The perimeter of the triangle in inches is:
Problem 44
A man born in the first half of the nineteenth century was years old in the year . He was born in:
Problem 45
In a rhombus, , line segments are drawn within the rhombus, parallel to diagonal , and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex . The graph is:
Problem 46
In the diagram, if points and are points of tangency, then equals:
Problem 47
At the midpoint of line segment which is units long, a perpendicular is erected with length units. An arc is described from with a radius equal to , meeting at . Then and are the roots of:
Problem 48
A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at of its former rate and arrives hours late. Had the accident happened miles farther along the line, it would have arrived only hours late. The length of the trip in miles was:
Problem 49
The difference of the squares of two odd numbers is always divisible by . If , and and are the odd numbers, to prove the given statement we put the difference of the squares in the form:
Problem 50
The times between and o'clock, correct to the nearest minute, when the hands of a clock will form an angle of are:
See also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1953 AHSC |
Followed by 1955 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.