Difference between revisions of "2002 AMC 12A Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | {{ | + | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly? |
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| + | <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 34\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 51\qquad \mathrm{(E) \ } 138 </math> | ||
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==Solution== | ==Solution== | ||
| − | {{ | + | <math>\dfrac{x-9}{3}=43</math> |
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| + | <math>x=43*3+9=138</math> | ||
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| + | <math>\dfrac{x-3}{9}=15 \Rightarrow \mathrm {(A)}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2002|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2002|ab=A|num-b=1|num-a=3}} | ||
Revision as of 14:35, 17 October 2007
Problem
Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
Solution
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 AMC 12 Problems and Solutions | |
