Difference between revisions of "2002 AMC 8 Problems/Problem 17"
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| − | We can try to guess and check to find the answer. If she got five right, her score would be <math>(5*5)-(5*2)=15</math>. If she got six right her score would be <math>(6*5)-(2*4)=22</math>. That's close, but it's still not right! If she got 7 right, her score would be <math>(7*5)-(2*3)=29</math>. Thus, our answer is <math>\boxed{C}</math> | + | We can try to guess and check to find the answer. If she got five right, her score would be <math>(5*5)-(5*2)=15</math>. If she got six right her score would be <math>(6*5)-(2*4)=22</math>. That's close, but it's still not right! If she got 7 right, her score would be <math>(7*5)-(2*3)=29</math>. Thus, our answer is <math>\boxed{\text{(C)}\ 7</math>. ~avamarora |
==Solution 2== | ==Solution 2== | ||
Revision as of 18:07, 19 October 2023
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 try to guess and check to find the answer. If she got five right, her score would be 
. If she got six right her score would be 
. That's close, but it's still not right! If she got 7 right, her score would be 
. Thus, our answer is $\boxed{\text{(C)}\ 7$ (Error compiling LaTeX. Unknown error_msg). ~avamarora
Solution 2
We can start with the full score, 50, and subtract not only 2 points for each incorrect 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 
.
Video Solution
https://youtu.be/8YXPMTjOyvM Soo, DRMS, NM
https://www.youtube.com/watch?v=aTeyOXo6-Uo ~David
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=1560
~pi_is_3.14
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. 
