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?

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

Solution

We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $\frac{138-3}{9}=\frac{135}{9}=\boxed{15}$ . Our answer is $\text{(A)}\ 15$ .

2002 AMC 10A (Problems • Answer Key • Resources)
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2002 AMC 10A Problems/Problem 6

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