Difference between revisions of "2015 AMC 8 Problems/Problem 1"
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==Solution 1== | ==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>. | 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>. | ||
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| + | ===Solution 1.1=== | ||
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| + | You could just leave it as <math>12\cdot9</math> and divide it by <math>9</math> to get <math>12</math> square yards to save some time. This gets us <math>\bold{\boxed{\textbf{(A)}~12}}</math>. | ||
==Solution 2== | ==Solution 2== | ||
Latest revision as of 06:21, 18 November 2024
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
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 1.1
You could just leave it as and divide it by to get square yards to save some time. This gets us .
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 
.
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Video Solution
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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
