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Difference between revisions of "1986 AJHSME Problems"

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== Problem 4 ==
 
== Problem 4 ==
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The product <math>(1.8)(40.3+.07)</math> is closest to
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<math>\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737</math>
  
 
[[1986 AJHSME Problems/Problem 4|Solution]]
 
[[1986 AJHSME Problems/Problem 4|Solution]]

Revision as of 17:09, 15 January 2009

Problem 1

In July 1861, $366$ inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

$\text{(A)}\ \frac{366}{31\times 24}$

$\text{(B)}\ \frac{366\times 31}{24}$

$\text{(C)}\ \frac{366\times 24}{31}$

$\text{(D)}\ \frac{31\times 24}{366}$

$\text{(E)}\ 366\times 31\times 24$

Solution

Problem 2

Which of the following numbers has the largest reciprocal?

$\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{5} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 1986$

Solution

Problem 3

The smallest sum one could get by adding three different numbers from the set $\{ 7,25,-1,12,-3 \}$ is

$\text{(A)}\ -3 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 21$

Solution

Problem 4

The product $(1.8)(40.3+.07)$ is closest to

$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

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

Solution

See also

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