1986 AJHSME Problems/Problem 20: Difference between revisions
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==Solution== | ==Solution== | ||
<cmath> \frac{(304)^5}{(29.7)(399)^4} \ | <cmath> \frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}</cmath> | ||
Which is closest to <math>3\rightarrow\boxed{\text{D}}</math>. | |||
(The original expression is approximately equal to <math>3.44921198</math>.) | (The original expression is approximately equal to <math>3.44921198</math>.) | ||
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==See Also== | ==See Also== | ||
[[ | {{AJHSME box|year=1986|num-b=19|num-a=21}} | ||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 20:28, 3 July 2013
Problem
The value of the expression 
is closest to
Solution
![\[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\]](http://latex.artofproblemsolving.com/6/a/3/6a3c99b6be27b8ea707b57c0bf34a0b1d74615ef.png)
Which is closest to 
.
(The original expression is approximately equal to 
.)
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
