Difference between revisions of "2015 AMC 8 Problems/Problem 1"
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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.) | 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.) | ||
<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== | + | ==Solutions== |
− | + | ===Solution 1=== | |
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>\bold{\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>\bold{\boxed{\textbf{(A)}~12}}</math>. | ||
Revision as of 15:52, 16 January 2021
Problem
How many square yards of 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 you need
square feet of carpet you need to cover. 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
.
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