2002 AMC 12A Problem 2

Attempt Problem 2 of the 2002 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

Cindy was asked by her teacher to subtract 33 from a certain number and then divide the result by 9.9. Instead, she subtracted 99 and then divided the result by 3,3, giving an answer of 43.43. What would her answer have been had she worked the problem correctly?
1515
3434
4343
5151
138138
Answer: A
Concepts:work backwardslinear equation
Small hint:
Her computation was x−93=43\dfrac{x-9}{3}=43
Big hint:
Solve for x,x, then evaluate x−39\dfrac{x-3}{9}
Solution:
Let xx be the number. Cindy computed x−93=43,\dfrac{x-9}{3} = 43, so x−9=129x - 9 = 129 and x=138.x = 138.
The correct computation gives 138−39=1359=15.\dfrac{138-3}{9} = \dfrac{135}{9} = 15.
Thus, the correct answer is A.
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