2018 AMC 10B Problems/Problem 1: Difference between revisions
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== Solution 1 == | == Solution 1 == | ||
The area of the pan is <math>20\cdot18=360</math>. Since the area of each piece is <math>4</math>, there are <math>\frac{360}{4} = \boxed{\textbf{(A) } 90}</math> pieces. | 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 that makes <math>\boxed{\textbf{(A) } 90}</math> pieces. | 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 (HOW TO THINK CRITICALLY!!!)== | |||
https://youtu.be/uCRbTeTlbuU | |||
~Education, the Study of Everything | |||
==Video Solution== | ==Video Solution== | ||
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 
-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 (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
