Difference between revisions of "2021 Mock AMC 8 Problems/Problem 2"
Arcticturn (talk | contribs) (→Solution 1) |
Arcticturn (talk | contribs) (→Solution 1) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 2: | Line 2: | ||
Aaron has a rectangular yard measuring <math>4</math> feet by <math>10</math> feet. How many <math>2</math> inch by <math>5</math> inch rectangular bricks can he fit in his yard? | Aaron has a rectangular yard measuring <math>4</math> feet by <math>10</math> feet. How many <math>2</math> inch by <math>5</math> inch rectangular bricks can he fit in his yard? | ||
− | <math>\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ | + | <math>\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 72 \qquad\mathrm{(C)}\ 144 \qquad\mathrm{(D)}\ 288 \qquad\mathrm{(E)}\ 576</math> |
==Solution 1== | ==Solution 1== | ||
− | Since there are 12 inches in a feet, | + | Since there are <math>12</math> inches in a feet, there are <math>2 \cdot 12</math> = <math>24</math> blocks every row. There are also <math>2 \cdot 12</math> = <math>24</math> blocks in each column. Therefore, we have <math>24 \cdot 24 = \boxed{\textbf{(E) } 576} \qquad</math> |
Latest revision as of 15:55, 16 October 2021
Problem
Aaron has a rectangular yard measuring feet by feet. How many inch by inch rectangular bricks can he fit in his yard?
Solution 1
Since there are inches in a feet, there are = blocks every row. There are also = blocks in each column. Therefore, we have