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1991 AJHSME Problems/Problem 7: Difference between revisions

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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>
~*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==
==See Also==


{{AJHSME box|year=1991|num-b=6|num-a=8}}
{{AJHSME box|year=1991|num-b=6|num-a=8}}
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 19:53, 14 January 2025

Problem

The value of $\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}$ is closest to

$\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000$

Solution

We can make the approximations \begin{align*} 487,000 &\approx 500,000 \\ 12,027,300 &\approx 12,000,000 \\ 9,621,001 &\approx 10,000,000 \\ 19,367 &\approx 20,000. \end{align*}

Using these instead of the original numbers for an estimate, we have \begin{align*} \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\ &= 500,000 \times 22,000 \\ &= 1.1\times 10^{10} \\ &\approx 10,000,000,000 \rightarrow \boxed{\text{D}}. \end{align*}


~*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

These problems are copyrighted © by the Mathematical Association of America.

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