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Difference between revisions of "1989 AJHSME Problems"

(New page: == Problem 1 == Solution == Problem 2 == Solution == Problem 3 == Solution ==...)
 
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== Problem 1 ==
 
== Problem 1 ==
 +
 +
<math>(1+11+21+31+41)+(9+19+29+39+49)=</math>
 +
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<math>\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250</math>
  
 
[[1989 AJHSME Problems/Problem 1|Solution]]
 
[[1989 AJHSME Problems/Problem 1|Solution]]
  
 
== Problem 2 ==
 
== Problem 2 ==
 +
 +
<math>\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} =</math>
 +
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<math>\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246</math>
  
 
[[1989 AJHSME Problems/Problem 2|Solution]]
 
[[1989 AJHSME Problems/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
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 +
Which of the following numbers is the largest?
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 +
<math>\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009</math>
  
 
[[1989 AJHSME Problems/Problem 3|Solution]]
 
[[1989 AJHSME Problems/Problem 3|Solution]]
  
 
== Problem 4 ==
 
== Problem 4 ==
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Estimate to determine which of the following numbers is closest to <math>\frac{401}{.205}</math>.
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<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math>
  
 
[[1989 AJHSME Problems/Problem 4|Solution]]
 
[[1989 AJHSME Problems/Problem 4|Solution]]
  
 
== Problem 5 ==
 
== Problem 5 ==
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<math>-15+9\times (6\div 3) =</math>
 +
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<math>\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12</math>
  
 
[[1989 AJHSME Problems/Problem 5|Solution]]
 
[[1989 AJHSME Problems/Problem 5|Solution]]
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== Problem 7 ==
 
== Problem 7 ==
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If the value of <math>20</math> quarters and <math>10</math> dimes equals the value of <math>10</math> quarters and <math>n</math> dimes, then <math>n=</math>
 +
 +
<math>\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45</math>
  
 
[[1989 AJHSME Problems/Problem 7|Solution]]
 
[[1989 AJHSME Problems/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
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 +
<math>(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =</math>
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26</math>
  
 
[[1989 AJHSME Problems/Problem 8|Solution]]
 
[[1989 AJHSME Problems/Problem 8|Solution]]
  
 
== Problem 9 ==
 
== Problem 9 ==
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 +
There are <math>2</math> boys for every <math>3</math> girls in Ms. Johnson's math class.  If there are <math>30</math> students in her class, what percent of them are boys?
 +
 +
<math>\text{(A)}\ 12\% \qquad \text{(B)}\ 20\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 60\% \qquad \text{(E)}\ 66\frac{2}{3}\% </math>
  
 
[[1989 AJHSME Problems/Problem 9|Solution]]
 
[[1989 AJHSME Problems/Problem 9|Solution]]
  
 
== Problem 10 ==
 
== Problem 10 ==
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 +
What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
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 +
<math>\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ</math>
  
 
[[1989 AJHSME Problems/Problem 10|Solution]]
 
[[1989 AJHSME Problems/Problem 10|Solution]]
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== Problem 12 ==
 
== Problem 12 ==
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 +
<math>\frac{1-\frac{1}{3}}{1-\frac{1}{2}} =</math>
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<math>\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{3}{2} \qquad \text{(E)}\ \frac{4}{3}</math>
  
 
[[1989 AJHSME Problems/Problem 12|Solution]]
 
[[1989 AJHSME Problems/Problem 12|Solution]]
  
 
== Problem 13 ==
 
== Problem 13 ==
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 +
<math>\frac{9}{7\times 53} =</math>
 +
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<math>\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \text{(D)}\ \frac{.9}{7\times .53} \qquad \text{(E)}\ \frac{.09}{.07\times .53}</math>
  
 
[[1989 AJHSME Problems/Problem 13|Solution]]
 
[[1989 AJHSME Problems/Problem 13|Solution]]
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== Problem 16 ==
 
== Problem 16 ==
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In how many ways can <math>47</math> be written as the sum of two primes?
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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}</math>
  
 
[[1989 AJHSME Problems/Problem 16|Solution]]
 
[[1989 AJHSME Problems/Problem 16|Solution]]
  
 
== Problem 17 ==
 
== Problem 17 ==
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 +
The number <math>\text{N}</math> is between <math>9</math> and <math>17</math>.  The average of <math>6</math>, <math>10</math>, and <math>\text{N}</math> could be
 +
 +
<math>\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16</math>
  
 
[[1989 AJHSME Problems/Problem 17|Solution]]
 
[[1989 AJHSME Problems/Problem 17|Solution]]
  
 
== Problem 18 ==
 
== Problem 18 ==
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 +
Many calculators have a reciprocal key <math>\boxed{\frac{1}{x}}</math> that replaces the current number displayed with its reciprocal.  For example, if the display is <math>\boxed{00004}</math> and the <math>\boxed{\frac{1}{x}}</math> key is depressed, then the display becomes <math>\boxed{000.25}</math>.  If <math>\boxed{00032}</math> is currently displayed, what is the fewest number of times you must depress the <math>\boxed{\frac{1}{x}}</math> key so the display again reads <math>\boxed{00032}</math>?
 +
 +
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5</math>
  
 
[[1989 AJHSME Problems/Problem 18|Solution]]
 
[[1989 AJHSME Problems/Problem 18|Solution]]
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== Problem 21 ==
 
== Problem 21 ==
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Jack had a bag of <math>128</math> apples.  He sold <math>25\% </math> of them to Jill.  Next he sold <math>25\% </math> of those remaining to June.  Of those apples still in his bag, he gave the shiniest one to his teacher.  How many apples did Jack have then?
 +
 +
<math>\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111</math>
  
 
[[1989 AJHSME Problems/Problem 21|Solution]]
 
[[1989 AJHSME Problems/Problem 21|Solution]]

Revision as of 10:41, 24 April 2009

Problem 1

$(1+11+21+31+41)+(9+19+29+39+49)=$

$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$

Solution

Problem 2

$\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} =$

$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$

Solution

Problem 3

Which of the following numbers is the largest?

$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$

Solution

Problem 4

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$.

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution

Problem 5

$-15+9\times (6\div 3) =$

$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$

Solution

Problem 6

Solution

Problem 7

If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$

Solution

Problem 8

$(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution

Problem 9

There are $2$ boys for every $3$ girls in Ms. Johnson's math class. If there are $30$ students in her class, what percent of them are boys?

$\text{(A)}\ 12\% \qquad \text{(B)}\ 20\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 60\% \qquad \text{(E)}\ 66\frac{2}{3}\%$

Solution

Problem 10

What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?

$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$

Solution

Problem 11

Solution

Problem 12

$\frac{1-\frac{1}{3}}{1-\frac{1}{2}} =$

$\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{3}{2} \qquad \text{(E)}\ \frac{4}{3}$

Solution

Problem 13

$\frac{9}{7\times 53} =$

$\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \text{(D)}\ \frac{.9}{7\times .53} \qquad \text{(E)}\ \frac{.09}{.07\times .53}$

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

In how many ways can $47$ be written as the sum of two primes?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

Solution

Problem 17

The number $\text{N}$ is between $9$ and $17$. The average of $6$, $10$, and $\text{N}$ could be

$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution

Problem 18

Many calculators have a reciprocal key $\boxed{\frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $\boxed{00004}$ and the $\boxed{\frac{1}{x}}$ key is depressed, then the display becomes $\boxed{000.25}$. If $\boxed{00032}$ is currently displayed, what is the fewest number of times you must depress the $\boxed{\frac{1}{x}}$ key so the display again reads $\boxed{00032}$?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?

$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111$

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

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