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2011 AMC 10B Problems/Problem 3: Difference between revisions

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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 2 inches by 3 inches. In square inches, what is the minimum area for the rectangle?
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?


(A) 3.75 (B) 4.5 (C) 5 (D) 6 (E) 8.75
<math> \textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75 </math>


== Solution ==
== Solution 1 ==


The minimum dimensions of the rectangle is 1.5 inches by 2.5 inches. The minimum area is (A) <math>1.5\times2.5=\boxed{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.
 
== Solution 2 (Utilize algebra) ==
 
Notice that the minimum area of a rectangle with dimensions <math>x</math> inches by <math>y</math> inches will result in <math>(x - 0.5)\times(y - 0.5)</math>. Simplify to get <math>xy - 0.5x - 0.5y + 0.25</math>. Plug in <math>2</math> and <math>3</math> for <math>x</math> and <math>y</math> to get <math>6 - 1 - 1.5 + 0.25 = \boxed{\text{(A)\ 3.75}}</math> square inches.
 
== See Also==
 
{{AMC10 box|year=2011|ab=B|num-b=2|num-a=4}}
{{MAA Notice}}

Latest revision as of 17:20, 22 October 2025

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 1

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.

Solution 2 (Utilize algebra)

Notice that the minimum area of a rectangle with dimensions $x$ inches by $y$ inches will result in $(x - 0.5)\times(y - 0.5)$. Simplify to get $xy - 0.5x - 0.5y + 0.25$. Plug in $2$ and $3$ for $x$ and $y$ to get $6 - 1 - 1.5 + 0.25 = \boxed{\text{(A)\ 3.75}}$ 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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