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

Revision as of 18:47, 29 August 2019

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*~

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 18:00, 29 August 2019 (view source) Qkddud (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 18:47, 29 August 2019 (view source) Qkddud (talk \| contribs) (→‎Solution) Newer edit →
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−	~If you typed this into a calculator or actually got the answer, it would be AROUND 10 ~~million~~.~	+	~If you typed this into a calculator or actually got the answer, it would be AROUND 10 billion.~

	~Calypso~		~Calypso~

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

Revision as of 18:47, 29 August 2019

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