2002 AMC 12A Problems/Problem 2

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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?

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

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 $\boxed{\textbf{(A) }15}$.


Solution 2

Let the number be $x$. We transform the problem into an equation: $\frac{x-9}{3}=43$. Solve for $x$ gives us $x=138$. Therefore, the correct result is $\frac{138-3}{9}=\frac{135}{9}=\boxed{\textbf{(A) }15}$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 12 Problems and Solutions
2002 AMC 10A (ProblemsAnswer KeyResources)
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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