Difference between revisions of "2011 AMC 10B Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | At a store, when a length | + | 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? |
<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 08:58, 13 January 2025
Problem
At a store, when a length is reported as 
inches that means the length 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
