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

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How many square yards of carpet are required to cover a rectangular floor that is <math>12</math> feet long and <math>9</math> feet wide? (There are 3 feet in a yard.)
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==Problem==
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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 <math>12</math> feet long and <math>9</math> feet wide? (There are 3 feet in a yard.)
  
 
<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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==Solution==
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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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First, we multiply <math>12\cdot9</math> to get that you need <math>108</math> square feet of carpet you need to cover.  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>\boxed{(A)}</math>.
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===Solution 1.1===
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You could just leave it as <math>12\cdot9</math> and divide it by <math>9</math> to get <math>12</math> square yards to save some time. This gets us <math>\bold{\boxed{\textbf{(A)}~12}}</math>.
  
 
==Solution 2==
 
==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>.
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/8mpjdpKcFs8
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~Education, the Study of Everything
  
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First we should write an equation for it: <math>\frac{12*9}{9}</math> Then we simplify: <math>\frac{12}{1}</math> Then further simplify and get our answer: <math>12 \textbf{(A)}</math>
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==Video Solution==
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https://youtu.be/758_W_eK81g
  
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===Solution 3===
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~savannahsolver
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Since there are 3 feet in a yard, we divide 9 by 3 to get 3, and 12 by 3 to get 4. To find the area of the carpet, we then multiply these two values together to get 12 square yards, which yields <math>12 \textbf{(A)}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2015|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2015|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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Placement:Easy Geometry

Latest revision as of 06:21, 18 November 2024

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.)

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

Solution 1

First, we multiply $12\cdot9$. To get that, we need $108$ square feet of carpet to cover the room's floor. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{\textbf{(A)}~12}}$.


Solution 1.1

You could just leave it as $12\cdot9$ and divide it by $9$ to get $12$ square yards to save some time. This gets us $\bold{\boxed{\textbf{(A)}~12}}$.

Solution 2

Since there are $3$ feet in a yard, we divide $9$ by $3$ to get $3$, and $12$ by $3$ to get $4$. To find the area of the carpet, we then multiply these two values together to get $\boxed{\textbf{(A)}~12}$.

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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