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 | + | 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> | ||
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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. | + | 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== | ||
{{AMC10 box|year=2011|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2011|ab=B|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:51, 21 November 2018
Problem
At a store, when a length or a width is reported as inches that means it is at least inches and at most inches. Suppose the dimensions of a rectangular tile are reported as inches by inches. In square inches, what is the minimum area for the rectangle?
Solution
The minimum dimensions of the rectangle are inches by inches. The minimum area is square inches.
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
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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