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

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===Solution 3===
 
===Solution 3===
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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>12</math> square yards, which yields <math>12 \textbf{(A)}</math>
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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>12</math> square yards, which yields <math>\boxed{\textbf{(A)}12}</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}}

Revision as of 11:37, 30 November 2015

How many square yards of 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

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

Solution 2

First we should write an equation for it: $\frac{12*9}{9}$ Then we simplify: $\frac{12}{1}$ Then further simplify and get our answer: $12 \textbf{(A)}$

Solution 3

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 $\boxed{\textbf{(A)}12}$

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

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