Difference between revisions of "2015 AMC 8 Problems/Problem 1"
m |
(Undo revision 199815 by Cherry blossom (talk)) (Tag: Undo) |
||
Line 4: | Line 4: | ||
<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> | ||
+ | |||
+ | ==Solutions== | ||
+ | ===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>. | ||
==Video Solution (HOW TO THINK CRITICALLY!!!)== | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
Revision as of 00:05, 19 October 2023
Contents
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 feet long and feet wide? (There are 3 feet in a yard.)
Solutions
Solution 1
First, we multiply . To get that, we need square feet of carpet to cover the room's floor. Since there are square feet in a square yard, you divide by to get square yards, so our answer is .
Solution 2
Since there are feet in a yard, we divide by to get , and by to get . To find the area of the carpet, we then multiply these two values together to get .
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~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.
Placement:Easy Geometry