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Difference between revisions of "2015 AMC 8 Problems/Problem 1"

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<math>\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972</math>
 
<math>\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972</math>
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==Solutions==
 
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===Solution 1===
 
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First, we multiply <math>12\cdot9</math>. To get that, we need <math>108</math> square feet of carpet to cover the room's floor.  Since there are <math>9</math> square feet in a square yard, you divide <math>108</math> by <math>9</math> to get <math>12</math> square yards, so our answer is <math>\bold{\boxed{\textbf{(A)}~12}}</math>.
 
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===Solution 2===
 
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Since there are <math>3</math> feet in a yard, we divide <math>9</math> by <math>3</math> to get <math>3</math>, and <math>12</math> by <math>3</math> to get <math>4</math>. To find the area of the carpet, we then multiply these two values together to get <math>\boxed{\textbf{(A)}~12}</math>.
 
  
 
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 
==Video Solution (HOW TO THINK CRITICALLY!!!)==

Revision as of 00:19, 19 October 2023

Problem

Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)

$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/8mpjdpKcFs8

~Education, the Study of Everything

Video Solution

https://youtu.be/758_W_eK81g

~savannahsolver

See Also

2015 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Placement:Easy Geometry

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