Difference between revisions of "1986 AJHSME Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | + | <center> | |
| + | <math>(1.8)(40.37)\approx (1.8)(40)=72.</math> | ||
| + | </center> | ||
| − | <math>1.8</math> is | + | Approximating <math>40.37</math> instead of <math>1.8</math> is more effective because larger numbers are less affected by absolute changes (e.g <math>1001</math> is much closer relatively to <math>1000</math> than <math>2</math> is to <math>1</math>). |
| − | + | <math>74</math> is the closest to <math>72</math>, so the answer is <math>\boxed{\text{C}}</math>. | |
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1986|num-b=3|num-a=5}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:17, 24 November 2017
Problem
The product 
is closest to
Solution
Approximating 
instead of 
is more effective because larger numbers are less affected by absolute changes (e.g 
is much closer relatively to 
than 
is to 
).
is the closest to 
, so the answer is 
.
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
