Difference between revisions of "2011 AMC 10B Problems/Problem 3"

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== Solution ==
 
== Solution ==
  
The minimum dimensions of the rectangle are <math>1.5</math> inches by <math>2.5</math> inches. The minimum area is <math>1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}</math> inches.
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The minimum dimensions of the rectangle are <math>1.5</math> inches by <math>2.5</math> inches. The minimum area is <math>1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}</math> square inches.
  
 
== See Also==
 
== See Also==

Revision as of 21:30, 16 January 2016

Problem

At a store, when a length is reported as $x$ inches that means the length is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?

$\textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75$

Solution

The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}$ square inches.

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AMC 10 Problems and Solutions

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