2002 AMC 12A Problems/Problem 2

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?

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 34\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 51\qquad \mathrm{(E) \ } 138$

Solution

$\dfrac{x-9}{3}=43$

$x=43*3+9=138$

$\dfrac{x-3}{9}=15 \Rightarrow \mathrm {(A)}$

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2002 AMC 12A Problems/Problem 2

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