Difference between revisions of "1991 AJHSME Problems/Problem 7"

Revision as of 23:07, 4 July 2013

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 $\begin{align*} \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\ &= 500,000 \times 22,000 \\ &= 1.1\times 10^{10} \\ &\approx 10,000,000,000 \rightarrow \boxed{\text{D}}. \end{align*}$

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, as part of the American Mathematics Competitions.

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

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Difference between revisions of "1991 AJHSME Problems/Problem 7"

Revision as of 23:07, 4 July 2013

Problem

Solution

See Also