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

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== Problem ==
 
== Problem ==
  
At a store, when a length is reported as <math>x</math> inches that means the length is at least <math>x - 0.5</math> inches and at most <math>x + 0.5</math> inches. Suppose the dimensions of a rectangular tile are reported as <math>2</math> inches by <math>3</math> inches. In square inches, what is the minimum area for the rectangle?
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At a store, when a length or a width is reported as <math>x</math> inches that means it is at least <math>x - 0.5</math> inches and at most <math>x + 0.5</math> inches. Suppose the dimensions of a rectangular tile are reported as <math>2</math> inches by <math>3</math> inches. In square inches, what is the minimum area for the rectangle?
  
 
<math> \textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75 </math>
 
<math> \textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75 </math>

Latest revision as of 10:51, 21 November 2018

Problem

At a store, when a length or a width is reported as $x$ inches that means it 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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