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

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==Problem==
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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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Suppose we write down the smallest (positive) <math>2</math>-digit, <math>3</math>-digit, and <math>4</math>-digit multiples of <math>8</math>.
 
 
<math>\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360</math>
 
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 19:43, 22 December 2020

Problem: Suppose we write down the smallest (positive) $2$-digit, $3$-digit, and $4$-digit multiples of $8$.

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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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