Difference between revisions of "2002 AMC 10A Problems/Problem 6"

Latest revision as of 07:46, 18 February 2009

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-=Problem==
+#redirect [[2002 AMC 12A Problems/Problem 2]]
-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?
-<math>\text{(A)}\ 15 \qquad \text{(B)}\ 34 \qquad \text{(C)}\ 43 \qquad \text{(D)}\ 51 \qquad \text{(E)} 138</math>
-==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}=\boxed{15}</math>. Our answer is <math>\text{(A)}\ 15</math>.