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Difference between revisions of "1985 AJHSME Problems"

(New page: == Problem 1 == Solution == Problem 2 == The percent that <math>M</math> is greater than <math>N</math>, is: <math> \mathrm{(A) \ } \frac {100(M - N)}...)
 
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== Problem 2 ==
 
== Problem 2 ==
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The percent that <math>M</math> is greater than <math>N</math>, is:
 
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<math> \mathrm{(A) \ } \frac {100(M - N)}{M} \qquad \mathrm{(B) \ } \frac {100(M - N)}{N} \qquad \mathrm{(C) \ } \frac {M - N}{N} \qquad \mathrm{(D) \ } \frac {M - N}{M} \qquad \mathrm{(E) \ } \frac {100(M + N)}{N} </math>
 
  
 
[[1985 AJHSME Problems/Problem 2|Solution]]
 
[[1985 AJHSME Problems/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
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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:
 
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<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>
 
  
 
[[1985 AJHSME Problems/Problem 3|Solution]]
 
[[1985 AJHSME Problems/Problem 3|Solution]]
  
 
== Problem 4 ==
 
== Problem 4 ==
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A barn with a 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:
 
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<math> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </math>
 
  
 
[[1985 AJHSME Problems/Problem 4|Solution]]
 
[[1985 AJHSME Problems/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
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Mr. <math>A</math> owns a home worth <math>\</math>10,000.  He sells it to Mr. <math>B</math> at a 10% profit based on the worth of the house. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a 10% loss.  Then:
 
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<math> \mathrm{(A) \ A\ comes\ out\ even  } \qquad</math> <math>\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}</math> <math> \qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad</math> <math>\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }</math> <math>\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }  </math>
 
  
 
[[1985 AJHSME Problems/Problem 5|Solution]]
 
[[1985 AJHSME Problems/Problem 5|Solution]]
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== Problem 16 ==
 
== Problem 16 ==
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If in applying the quadratic formula to a quadratic equation
 
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<cmath>f(x) \equiv ax^2 + bx + c = 0,</cmath>
 
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it happens that <math>c = \frac{b^2}{4a}</math>, then the graph of <math>y = f(x)</math> will certainly:
 
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<math>\mathrm{(A) \ have\ a\ maximum  } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad</math> <math>\mathrm{(C) \ be\ tangent\ to\ the\ xaxis} \qquad</math> <math>\mathrm{(D) \ be\ tangent\ to\ the\ yaxis} \qquad</math> <math>\mathrm{(E) \ lie\ in\ one\ quadrant\ only}</math>
 
  
 
[[1985 AJHSME Problems/Problem 16|Solution]]
 
[[1985 AJHSME Problems/Problem 16|Solution]]

Revision as of 11:14, 2 April 2008

Problem 1

Solution

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

If $a = - 2$, the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is

$A)\quad - 3a \qquad B)\quad 4a \qquad C)\quad \frac {24}{a} \qquad D)\quad a^2 \qquad E)\quad 1$

Solution

Problem 9

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$A)\quad \frac {1}{10} \qquad B)\quad \frac {1}{9} \qquad C)\quad \frac {1}{2} \qquad D)\quad \frac {10}{11} \qquad E)\quad \frac {11}{2}$

Solution

Problem 10

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

Five cards are lying on a table as shown.

\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]

Each card has a letter on one side and a whole number on the other side. Jane said, "If a vowel is on one side of any card, then an even number is on the other side." Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?

$A)\quad 3 \qquad B)\quad 4 \qquad C)\quad 6 \qquad D)\quad P \qquad E)\quad Q$

Solution

See also

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