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

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Problem:
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{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #1]] and [[2018 AMC 10B Problems|2018 AMC 10B #1]]}}
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>.
 
  
Problem:
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==Problem==
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>.
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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?
  
What is the sum of these three numbers?
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<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=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 which makes <math>90</math> pieces. Thus, the answer is <math>\boxed{A}</math>.
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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 $20$-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 $2\cdot2=4$, there are $\frac{360}{4} = \boxed{\textbf{(A) } 90}$ pieces.

Solution 2

By dividing each of the dimensions by $2$, we get a $10\times9$ grid that makes $\boxed{\textbf{(A) } 90}$ pieces.

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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