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 "1986 AJHSME Problems/Problem 4"

(→‎Solution)
(→‎Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
The easiest way to do this problem is to estimate, since it says "closest to", and the multiple choices are pretty spread out.
+
<center>
 +
<math>(1.8)(40.37)\approx (1.8)(40)=72.</math>
 +
</center>
  
−
<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>.
+
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>).
−
Putting this in, we get <math>(2)(40 + 0) = 80</math>
+
<math>74</math> is the closest to <math>72</math>, so the answer is <math>\boxed{\text{C}}</math>.
−
 
 
−
<math>80</math> is about <math>84</math>
 
−
 
 
−
<math>\boxed{\text{D}}</math>
 
−
 
 
−
edit: the answer is C. The reason for this is that 1.8 and 2 are close, but when multiplied by 40, it's a big difference (80 and 72). So you could just do 1.8(40.37) which is basically 1.8*40=72. 74 is the closest to 72.
 
  
 
==See Also==
 
==See Also==
Line 22: Line 18:
 
{{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 01:17, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_4&oldid=88625"