Difference between revisions of "2002 AMC 8 Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
We can simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! If she got six right, her score would be (6*5)-(2*4)=22. That's close, but it's still not right! If she got 7 right, her score would be (7*5)-(2*3)=29. Which is what we need! Thus, our answer is \boxed{\text{(C)}\ 7} | We can simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! If she got six right, her score would be (6*5)-(2*4)=22. That's close, but it's still not right! If she got 7 right, her score would be (7*5)-(2*3)=29. Which is what we need! Thus, our answer is \boxed{\text{(C)}\ 7} | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 12:12, 25 November 2020
Contents
Problem
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
Solution 1
We can simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! If she got six right, her score would be (6*5)-(2*4)=22. That's close, but it's still not right! If she got 7 right, her score would be (7*5)-(2*3)=29. Which is what we need! Thus, our answer is \boxed{\text{(C)}\ 7}
Solution 2
We can start with the full score, 50, and subtract not only 2 points for each correct answer but also the 5 points we gave her credit for. This expression is equivalent to her score, 29.
Let 
be the number of questions she answers correctly. Then, we will represent the number incorrect by 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
