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Difference between revisions of "1986 AJHSME Problems/Problem 4"

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==Solution==
 
==Solution==
  
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The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.
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<center>
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<math>(1.8)(40.37)\approx (1.8)(40)=72.</math>
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</center>
  
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<math>1.8</math> is about <math>2</math>, <math>40.3</math> is about <math>40</math>, and <math>.07</math> is about <math>0</math>.
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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>).
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Putting this in, we get <math>(2)(40 + 0) = 80</math>
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<math>74</math> is the closest to <math>72</math>, so the answer is <math>\boxed{\text{C}}</math>.
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<math>80</math> is about <math>84</math>
 
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<math>\boxed{\text{D}}</math>
 
  
 
==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=3|num-a=5}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 02:17, 24 November 2017

Problem

The product $(1.8)(40.3+.07)$ is closest to

$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

Solution

$(1.8)(40.37)\approx (1.8)(40)=72.$

Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$). $74$ is the closest to $72$, so the answer is $\boxed{\text{C}}$.

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

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