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

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==Problem 7==
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==Problem==
  
 
<math>100\times 19.98\times 1.998\times 1000=</math>
 
<math>100\times 19.98\times 1.998\times 1000=</math>

Revision as of 11:49, 23 December 2012

Problem

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

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

Solution

$19.98\times100=1998$

$1.998\times1000=1998$

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

See also

1998 AJHSME (ProblemsAnswer KeyResources)
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