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1985 AJHSME Problems

Revision as of 15:42, 7 January 2009 by 5849206328x (talk | contribs) (→‎Problem 2)

Problem 1

$\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=$

$\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50$

Solution

Problem 2

$90+91+92+93+94+95+96+97+98+99=$


$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

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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