Difference between revisions of "1991 AJHSME Problems/Problem 7"
5849206328x (talk | contribs) (Created page with '==Problem== The value of <math>\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}</math> is closest to <math>\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000…') |
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&\approx 10,000,000,000 \rightarrow \boxed{\text{D}}. | &\approx 10,000,000,000 \rightarrow \boxed{\text{D}}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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+ | ~*If you typed this into a calculator or actually got the answer, it would be AROUND 10 billion.*~ | ||
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+ | ~*Calypso*~ | ||
+ | |||
+ | "Google says the answer is 10,887,305,816.1" ~ xxsc | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1991|num-b=6|num-a=8}} | {{AJHSME box|year=1991|num-b=6|num-a=8}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 15:50, 12 November 2019
Problem
The value of is closest to
Solution
We can make the approximations
Using these instead of the original numbers for an estimate, we have
~*If you typed this into a calculator or actually got the answer, it would be AROUND 10 billion.*~
~*Calypso*~
"Google says the answer is 10,887,305,816.1" ~ xxsc
See Also
1991 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.