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Difference between revisions of "1987 AJHSME Problems"

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== Problem 11 ==
 
== Problem 11 ==
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The sum <math>2\frac17+3\frac12+5\frac{1}{19}</math> is between
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<math>\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\text{ and }11\frac12 \qquad \text{(D)}\ 11\frac12 \text{ and }12 \qquad \text{(E)}\ 12\text{ and }12\frac12</math>
  
 
[[1987 AJHSME Problems/Problem 11|Solution]]
 
[[1987 AJHSME Problems/Problem 11|Solution]]

Revision as of 08:01, 16 February 2009

Problem 1

$.4+.02+.006=$

$\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426$

Solution

Problem 2

$\frac{2}{25}=$

$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)} 1.25 \qquad \text{(E)}\ 12.5$

Solution

Problem 3

$2(81+83+85+87+89+91+93+95+97+99)=$

$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$

Solution

Problem 4

Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?

$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250$

Solution

Problem 5

The area of the rectangular region is

[asy] draw((0,0)--(4,0)--(4,2.2)--(0,2.2)--cycle,linewidth(.5 mm)); label(".22 m",(4,1.1),E); label(".4 m",(2,0),S); [/asy]

$\text{(A)}\ \text{.088 m}^2 \qquad \text{(B)}\ \text{.62 m}^2 \qquad \text{(C)}\ \text{.88 m}^2 \qquad \text{(D)}\ \text{1.24 m}^2 \qquad \text{(E)}\ \text{4.22 m}^2$

Solution

Problem 6

The smallest product one could obtain by multiplying two numbers in the set $\{ -7, -5, -1, 1, 3 \}$ is

$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$, the least common denominator used is

$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$

Solution

Problem 10

$4(299)+3(299)+2(299)+298=$

$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$

Solution

Problem 11

The sum $2\frac17+3\frac12+5\frac{1}{19}$ is between

$\text{(A)}\ 10\text{ and }10\frac12 \qquad \text{(B)}\ 10\frac12 \text{ and } 11 \qquad \text{(C)}\ 11\text{ and }11\frac12 \qquad \text{(D)}\ 11\frac12 \text{ and }12 \qquad \text{(E)}\ 12\text{ and }12\frac12$

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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