Difference between revisions of "2002 AMC 12A Problems/Problem 2"
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| + | {{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #2]] and [[2002 AMC 10A Problems|2002 AMC 10A #6]]}} | ||
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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== | ||
| − | {{ | + | We work backwards; the number that Cindy started with is <math>3(43)+9=138</math>. Now, the correct result is <math>\frac{138-3}{9}=\frac{135}{9}=15</math>. Our answer is <math>\boxed{\text{(A)}\ 15}</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}} | ||
| + | {{AMC10 box|year=2002|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 10:06, 4 July 2013
- The following problem is from both the 2002 AMC 12A #2 and 2002 AMC 10A #6, so both problems redirect to this page.
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
We work backwards; the number that Cindy started with is 
. Now, the correct result is 
. Our answer is 
.
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 | |
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
