2002 AMC 12A Problems/Problem 2

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?

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

Solution

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

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

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2002 AMC 12A Problems/Problem 2

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