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Difference between revisions of "1966 AHSME Problems"

m (1966 AHSME moved to 1966 AHSME Problems: problme listing)
m (→‎Problem 3)
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== Problem 3 ==
 
== Problem 3 ==
−
If the arithmetic mean of two numbers is <math>6</math> and thier geometric mean is <math>10</math>, then an equation with the given two numbers as roots is:
+
If the arithmetic mean of two numbers is <math>6</math> and their geometric mean is <math>10</math>, then an equation with the given two numbers as roots is:
  
 
<math>\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0</math>
 
<math>\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0</math>

Revision as of 09:30, 2 April 2008

Problem 1

Given that the ratio of $3x - 4$ to $y + 15$ is constant, and $y = 3$ when $x = 2$, then, when $y = 12$, $x$ equals:

$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$

Solution

Problem 2

Solution

Problem 3

If the arithmetic mean of two numbers is $6$ and their geometric mean is $10$, then an equation with the given two numbers as roots is:

$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$ $\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$

Solution

Problem 4

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle $I$ to that of circle $II$, then $r$ equals:

$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt 3 \qquad \text{(D)} \ 2\sqrt 2 \qquad \text{(E)} \ 2\sqrt 3$

Solution

Problem 5

The number of values of $x$ satisfying the equation \[ \frac {2x^2 - 10x}{x^2 - 5x} = x - 3 \] is:

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$

Solution

Problem 6

$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

Solution

Problem 7

Let $\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}$ be an identity in $x$. The numerical value of $N_1N_2$ is:

$\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$

Solution

Problem 8

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of teh circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

Problem 9

If $x = (\log_82)^{(\log_28)}$, then $\log_3x$ equals:

$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$

Solution

Problem 10

If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:

$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution

See also

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