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

(Problem)
Line 11: Line 11:
 
===Solution 2===
 
===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>.
 
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==
 +
https://youtu.be/758_W_eK81g
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 08:55, 26 March 2022

Problem

Onkon wants to cover his rooms 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.)

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

Solutions

Solution 1

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 $\bold{\boxed{\textbf{(A)}~12}}$.

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

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