Difference between revisions of "1989 AJHSME Problems/Problem 8"
5849206328x (talk | contribs) (New page: ==Problem== <math>(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =</math> <math>\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad ...) |
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{{AJHSME box|year=1989|num-b=7|num-a=9}} | {{AJHSME box|year=1989|num-b=7|num-a=9}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Revision as of 23:57, 4 July 2013
Problem
Solution
Solution 1
We use the distributive property to get ![\[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{\text{E}}\]](http://latex.artofproblemsolving.com/6/9/6/696057855d9d7437bb8c492ee09749c3a843647f.png)
Solution 2
Since 
, we have ![\[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\]](http://latex.artofproblemsolving.com/7/3/f/73fb09048546749e763867c37a9daf9b86e9a9df.png)
The only answer choice greater than 
is 
.
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
