1991 AJHSME Problems/Problem 3: Difference between revisions

Revision as of 23:06, 4 July 2013

Problem

Two hundred thousand times two hundred thousand equals

$\text{(A)}\ \text{four hundred thousand} \qquad \text{(B)}\ \text{four million} \qquad \text{(C)}\ \text{forty thousand} \qquad \text{(D)}\ \text{four hundred million} \qquad \text{(E)}\ \text{forty billion}$

Solution

Writing the numbers in scientific notation, we have $\begin{align*} (2\times 10^5)\times (2\times 10^5) &= 4\times 10^{10} \\ &= 40\times 10^9 \\ &= \text{forty billion} \rightarrow \boxed{\text{E}}. \end{align*}$

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

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 {{AJHSME box|year=1991|num-b=2|num-a=4}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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1991 AJHSME Problems/Problem 3: Difference between revisions

Revision as of 23:06, 4 July 2013

Problem

Solution

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