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Difference between revisions of "2018 AMC 10B Problems/Problem 1"

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Kate bakes 20-inch by 18-inch pan of cornbread. The cornbread is cut into pieces that measure 2 inches by 2 inches. How many pieces of cornbread does the pan contain?
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{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #1]] and [[2018 AMC 10B Problems|2018 AMC 10B #1]]}}
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==Problem==
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Kate bakes a <math>20</math>-inch by <math>18</math>-inch pan of cornbread. The cornbread is cut into pieces that measure <math>2</math> inches by <math>2</math> inches. How many pieces of cornbread does the pan contain?
  
 
<math>\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360</math>
 
<math>\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360</math>
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== Solution 1 ==
 
== Solution 1 ==
  
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The area of the pan is <math>20\cdot18</math> = <math>360</math>. Since the area of each piece is <math>4</math>, there are 360/40 = 90 pieces. Thus, the answer is <math>\boxed{A}</math>.
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The area of the pan is <math>20\cdot18=360</math>. Since the area of each piece is <math>2\cdot2=4</math>, there are <math>\frac{360}{4} = \boxed{\textbf{(A) } 90}</math> pieces.
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== Solution 2 ==
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By dividing each of the dimensions by <math>2</math>, we get a <math>10\times9</math> grid that makes <math>\boxed{\textbf{(A) } 90}</math> pieces.
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/uCRbTeTlbuU
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/o5MUHOmF1zo
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2018|ab=B|before=First Problem|num-a=2}}
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{{AMC12 box|year=2018|ab=B|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 01:38, 28 May 2023

The following problem is from both the 2018 AMC 12B #1 and 2018 AMC 10B #1, so both problems redirect to this page.

Problem

Kate bakes a $20$-inch by $18$-inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

$\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360$

Solution 1

The area of the pan is $20\cdot18=360$. Since the area of each piece is $2\cdot2=4$, there are $\frac{360}{4} = \boxed{\textbf{(A) } 90}$ pieces.

Solution 2

By dividing each of the dimensions by $2$, we get a $10\times9$ grid that makes $\boxed{\textbf{(A) } 90}$ pieces.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/uCRbTeTlbuU

~Education, the Study of Everything

Video Solution

https://youtu.be/o5MUHOmF1zo

~savannahsolver

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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
2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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