Difference between revisions of "2002 AMC 10A Problems/Problem 6"
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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}= | + | 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== | ||
Revision as of 18:16, 26 December 2008
Problem=
From a starting number, Cindy was supposed to subtract 3, and then divide by 9, but instead, Cindy subtracted 9, then divided by 3, getting 43. If the correct instructions were followed, what would the result be?
Solution
We work backwards; the number that Cindy started with is 
. Now, the correct result is 
. Our answer is 
.
See Also
| 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 | ||
