Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AJHSME Problems"

(New page: == Problem 1 == The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of {{image}} <math>\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad...)
 
Line 3: Line 3:
 
The diagram shows part of a scale of a measuring device.  The arrow indicates an approximate reading of
 
The diagram shows part of a scale of a measuring device.  The arrow indicates an approximate reading of
  
−
{{image}}
+
<asy>
 +
draw((-3,0)..(0,3)..(3,0));
 +
draw((-3.5,0)--(-2.5,0));
 +
draw((0,2.5)--(0,3.5));
 +
draw((2.5,0)--(3.5,0));
 +
draw((1.8,1.8)--(2.5,2.5));
 +
draw((-1.8,1.8)--(-2.5,2.5));
 +
draw((0,0)--3*dir(120),EndArrow);
 +
label("$10$",(-2.6,0),E);
 +
label("$11$",(2.6,0),W);
 +
</asy>
  
 
<math>\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6</math>
 
<math>\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6</math>
Line 10: Line 20:
  
 
== Problem 2 ==
 
== Problem 2 ==
 +
 +
The product <math>8\times .25\times 2\times .125 =</math>
 +
 +
<math>\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2</math>
  
 
[[1988 AJHSME Problems/Problem 2|Solution]]
 
[[1988 AJHSME Problems/Problem 2|Solution]]
  
 
== Problem 3 ==
 
== Problem 3 ==
 +
 +
<math>\frac{1}{10}+\frac{2}{20}+\frac{3}{30} = </math>
 +
 +
<math>\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6</math>
  
 
[[1988 AJHSME Problems/Problem 3|Solution]]
 
[[1988 AJHSME Problems/Problem 3|Solution]]
Line 26: Line 44:
  
 
== Problem 6 ==
 
== Problem 6 ==
 +
 +
<math>\frac{(.2)^3}{(.02)^2} =</math>
 +
 +
<math>\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20</math>
  
 
[[1988 AJHSME Problems/Problem 6|Solution]]
 
[[1988 AJHSME Problems/Problem 6|Solution]]
  
 
== Problem 7 ==
 
== Problem 7 ==
 +
 +
<math>2.46\times 8.163\times (5.17+4.829)</math> is closest to
 +
 +
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500</math>
  
 
[[1988 AJHSME Problems/Problem 7|Solution]]
 
[[1988 AJHSME Problems/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
 +
 +
Betty used a calculator to find the product <math>0.075 \times 2.56</math>.  She forgot to enter the decimal points.  The calculator showed <math>19200</math>.  If Betty had entered the decimal points correctly, the answer would have been
 +
 +
<math>\text{(A)}\ .0192 \qquad \text{(B)}\ .192 \qquad \text{(C)}\ 1.92 \qquad \text{(D)}\ 19.2 \qquad \text{(E)}\ 192</math>
  
 
[[1988 AJHSME Problems/Problem 8|Solution]]
 
[[1988 AJHSME Problems/Problem 8|Solution]]
Line 42: Line 72:
  
 
== Problem 10 ==
 
== Problem 10 ==
 +
 +
Chris' birthday is on a Thursday this year.  What day of the week will it be <math>60</math> days after her birthday?
 +
 +
<math>\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Thursday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}</math>
  
 
[[1988 AJHSME Problems/Problem 10|Solution]]
 
[[1988 AJHSME Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
 +
 +
<math>\sqrt{164}</math> is
 +
 +
<math>\text{(A)}\ 42 \qquad \text{(B)}\ \text{less than }10 \qquad \text{(C)}\ \text{between }10\text{ and }11 \qquad \text{(D)}\ \text{between }11\text{ and }12 \qquad \text{(E)}\ \text{between }12\text{ and }13</math>
  
 
[[1988 AJHSME Problems/Problem 11|Solution]]
 
[[1988 AJHSME Problems/Problem 11|Solution]]
  
 
== Problem 12 ==
 
== Problem 12 ==
 +
 +
Suppose the estimated <math>20</math> billion dollar cost to send a person to the planet Mars is shared equally by the <math>250</math> million people in the U.S. Then each person's share is
 +
 +
<math>\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}</math>
  
 
[[1988 AJHSME Problems/Problem 12|Solution]]
 
[[1988 AJHSME Problems/Problem 12|Solution]]

Revision as of 10:38, 15 March 2009

Problem 1

The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of

[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2.6,0),E); label("$11$",(2.6,0),W); [/asy]

$\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6$

Solution

Problem 2

The product $8\times .25\times 2\times .125 =$

$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution

Problem 3

$\frac{1}{10}+\frac{2}{20}+\frac{3}{30} =$

$\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6$

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

$\frac{(.2)^3}{(.02)^2} =$

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$

Solution

Problem 7

$2.46\times 8.163\times (5.17+4.829)$ is closest to

$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

Solution

Problem 8

Betty used a calculator to find the product $0.075 \times 2.56$. She forgot to enter the decimal points. The calculator showed $19200$. If Betty had entered the decimal points correctly, the answer would have been

$\text{(A)}\ .0192 \qquad \text{(B)}\ .192 \qquad \text{(C)}\ 1.92 \qquad \text{(D)}\ 19.2 \qquad \text{(E)}\ 192$

Solution

Problem 9

Solution

Problem 10

Chris' birthday is on a Thursday this year. What day of the week will it be $60$ days after her birthday?

$\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Thursday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}$

Solution

Problem 11

$\sqrt{164}$ is

$\text{(A)}\ 42 \qquad \text{(B)}\ \text{less than }10 \qquad \text{(C)}\ \text{between }10\text{ and }11 \qquad \text{(D)}\ \text{between }11\text{ and }12 \qquad \text{(E)}\ \text{between }12\text{ and }13$

Solution

Problem 12

Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is

$\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}$

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AJHSME_Problems&oldid=30726"