Difference between revisions of "1986 AJHSME Problems/Problem 4"
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==Solution== | ==Solution== | ||
| + | <center> | ||
| + | First, solve the sum of \( 40.3 \) and \(.07\) | ||
| + | |||
| + | \( \Rightarrow \) \( 40.3 + .07 = 40.37 \) | ||
| + | |||
| + | Multiply it with \( 1.8 \) | ||
| + | |||
| + | \( \Rightarrow \) \( (1.8)(40.37) \) | ||
| + | |||
| + | Evalutate the result | ||
| + | |||
| + | \( \Rightarrow \) \( 72.666 \) | ||
| + | |||
| + | </center> | ||
| + | |||
| + | Since we found the answer to be \( 72.666 \), the closest option is <math>\boxed{\text{C) 74}}</math>. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\text{C}}</math>. | ||
| + | |||
| + | ~ Solution by abcde26 | ||
| + | ==Better Solution== | ||
<center> | <center> | ||
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</center> | </center> | ||
| + | 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>. | <math>74</math> is the closest to <math>72</math>, so the answer is <math>\boxed{\text{C}}</math>. | ||
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{{AJHSME box|year=1986|num-b=3|num-a=5}} | {{AJHSME box|year=1986|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:53, 23 October 2025
Contents
Problem
The product 
is closest to
Solution
First, solve the sum of \( 40.3 \) and \(.07\)
\( \Rightarrow \) \( 40.3 + .07 = 40.37 \)
Multiply it with \( 1.8 \)
\( \Rightarrow \) \( (1.8)(40.37) \)
Evalutate the result
\( \Rightarrow \) \( 72.666 \)
Since we found the answer to be \( 72.666 \), the closest option is 
.
Therefore, the answer is 
.
~ Solution by abcde26
Better 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
