Difference between revisions of "2015 AMC 8 Problems/Problem 1"
(→Solution 2) |
(→Solution 3) |
||
Line 12: | Line 12: | ||
===Solution 3=== | ===Solution 3=== | ||
− | 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>12</math> square yards, which yields <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>12</math> square yards, which yields <math>\boxed{\textbf{(A)} 12}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|before=First Problem|num-a=2}} | {{AMC8 box|year=2015|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:37, 30 November 2015
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.)
Contents
Solution
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
First we should write an equation for it: Then we simplify:
Then further simplify and get our answer:
Solution 3
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
square yards, which yields
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.