Difference between revisions of "2018 AMC 10B Problems/Problem 1"
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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 | + | ==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? | |
− | + | <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 == | ||
+ | |||
+ | 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. | ||
== Solution 2 == | == Solution 2 == | ||
− | By dividing each of the dimensions by <math>2</math>, we get a <math>10\times9</math> grid | + | 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. |
==Video Solution== | ==Video Solution== |
Latest revision as of 11:09, 7 December 2021
- 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 -inch by -inch pan of cornbread. The cornbread is cut into pieces that measure inches by inches. How many pieces of cornbread does the pan contain?
Solution 1
The area of the pan is . Since the area of each piece is , there are pieces.
Solution 2
By dividing each of the dimensions by , we get a grid that makes pieces.
Video Solution
~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.