Difference between revisions of "2005 AMC 8 Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | Connie multiplies a number by 2 and gets 60 as her answer. However, she should | + | 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? |
| − | have divided the number by 2 to get the correct answer. What is the correct | ||
| − | answer? | ||
<math> \textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240 </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>. | + | 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>. |
| + | |||
| + | A second way to do it is to divide the number by <math>4</math>, as you multiplied by <math>2</math> when you were supposed to divide by <math>2</math>. So, <math>\dfrac{60}{4} = \boxed{\textbf{(B)}\ 15}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|before=First <br/> Question|num-a=2}} | {{AMC8 box|year=2005|before=First <br/> Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:10, 8 January 2024
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 
.
A second way to do it is to divide the number by 
, as you multiplied by when you were supposed to divide by . So, 
.
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. 
