Difference between revisions of "1998 AJHSME Problems/Problem 7"

Revision as of 12:49, 23 December 2012

$100\times 19.98\times 1.998\times 1000=$

$\text{(A)}\ (1.998)^2 \qquad \text{(B)}\ (19.98)^2 \qquad \text{(C)}\ (199.8)^2 \qquad \text{(D)}\ (1998)^2 \qquad \text{(E)}\ (19980)^2$

$19.98\times100=1998$

$1.998\times1000=1998$

$1998\times1998=(1998)^2=\boxed{D}$

1998 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

Revision as of 11:12, 31 July 2011 (view source) Talkinaway (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 12:49, 23 December 2012 (view source) Gina (talk \| contribs) m (→‎Problem 7) Newer edit →
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−	==Problem 7==	+	==Problem==

	<math>100\times 19.98\times 1.998\times 1000=</math>		<math>100\times 19.98\times 1.998\times 1000=</math>