Difference between revisions of "2005 AMC 8 Problems/Problem 1"
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− | ==Problem | + | ==Problem== |
− | + | Connie multiplies a number by <math>2</math> and gets <math>60</math> as her answer. However, she should have divided the number by <math>2</math> to get the correct answer. What is the correct answer? | |
− | <math>\ | + | <math> \textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240 </math> |
==Solution== | ==Solution== | ||
− | + | If <math>x</math> is the number, then <math>2x=60</math> and <math>x=30</math>. Dividing the number by <math>2</math> yields <math>\dfrac{30}{2} = \boxed{\textbf{(B)}\ 15}</math>. | |
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2005|before=First <br/> Question|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 22:06, 11 February 2020
Problem
Connie multiplies a number by and gets as her answer. However, she should have divided the number by to get the correct answer. What is the correct answer?
Solution
If is the number, then and . Dividing the number by yields .
See Also
2005 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Question |
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.