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

(Solution 3)
(Problem)
 
(24 intermediate revisions by 16 users not shown)
Line 1: Line 1:
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.)
+
==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 <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>
  
==Solution==
+
==Solution 1==
 +
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>.
 +
 
 +
==Solution 2==
 +
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>.
  
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{\textbf{(A)}}</math>.
+
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/8mpjdpKcFs8
  
==Solution 2==
+
~Education, the Study of Everything
  
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>
+
==Video Solution==
 +
https://youtu.be/758_W_eK81g
  
===Solution 3===
+
~savannahsolver
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}}
 +
Placement:Easy Geometry

Latest revision as of 18:42, 30 November 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.)

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

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 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 (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 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