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

Revision as of 19:41, 31 December 2020

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 $4$ , there are $\frac{360}{4} = 90$ pieces. Thus, the answer is $\boxed{A}$ .

Solution 2

By dividing each of the dimensions by $2$ , we get a $10\times9$ grid which makes $90$ pieces. Thus, the answer is $\boxed{A}$ .

Video Solution

https://youtu.be/o5MUHOmF1zo

~savannahsolver

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.

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+{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #1]] and [[2018 AMC 10B Problems|2018 AMC 10B #1]]}}
 ==Problem==
 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?

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