Difference between revisions of "2015 AMC 8 Problems/Problem 1"
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== Solution 1 == | == Solution 1 == | ||
− | We need <math>12 | + | We need <math>12\cdot9</math> square feet of carpet to cover the floor. Since there are <math>9</math> square feet in a square yard, we divide this by <math>9</math> to get <math>\bold{\boxed{\textbf{(A) }12}}</math> square yards. |
== 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>3\cdot4=\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> yards, and <math>12</math> by <math>3</math> to get <math>4</math> yards. To find the area of the carpet, we then multiply these two values together to get <math>3\cdot4=\boxed{\textbf{(A) }12}</math> square yards. |
==Video Solution 1 (HOW TO THINK CRITICALLY)== | ==Video Solution 1 (HOW TO THINK CRITICALLY)== |
Latest revision as of 21:53, 16 January 2025
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.)
Solution 1
We need square feet of carpet to cover the floor. Since there are square feet in a square yard, we divide this by to get square yards.
Solution 2
Since there are feet in a yard, we divide by to get yards, and by to get yards. To find the area of the carpet, we then multiply these two values together to get square yards.
Video Solution 1 (HOW TO THINK CRITICALLY)
~Education, the Study of Everything
Video Solution 2
~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.