Difference between revisions of "2018 AMC 10B Problems/Problem 1"
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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> | |
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| − | + | == 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 <math>\frac{360}{4} = 90</math> pieces. Thus, the answer is <math>\boxed{A}</math>. | ||
== Solution 2 == | == Solution 2 == | ||
Revision as of 14:34, 23 December 2020
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. Thus, the answer is 
.
Solution 2
By dividing each of the dimensions by , we get a 
grid which makes 
pieces. Thus, the answer is .
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. 
