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?

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

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 $\frac{138-3}{9}=\frac{135}{9}=\boxed{\textbf{(A) }15}$ .

Video Solution by Daily Dose of Math

https://youtu.be/ZgEkIB1qmRw?si=LpBb-qb3OaUbQ8_W

~Thesmartgreekmathdude

2002 AMC 12A (Problems • Answer Key • Resources)
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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 2

Contents

Problem

Solution

Solution 2

Video Solution by Daily Dose of Math

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