Difference between revisions of "2002 AMC 8 Problems/Problem 17"
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We see that Olivia's score is odd. Since subtracting multiples of <math>2</math> (even numbers) does not change a number's parity (odd or even), Olivia's score from only her correct answers must be odd. Then, we test odd multiples of <math>5</math> greater than <math>29</math> to see which one works. The smallest odd multiple of <math>5</math> greater than <math>29</math> is <math>35</math>, meaning <math>7</math> answers were correct and the remaining <math>3</math> answers were wrong. We see that the final score is | We see that Olivia's score is odd. Since subtracting multiples of <math>2</math> (even numbers) does not change a number's parity (odd or even), Olivia's score from only her correct answers must be odd. Then, we test odd multiples of <math>5</math> greater than <math>29</math> to see which one works. The smallest odd multiple of <math>5</math> greater than <math>29</math> is <math>35</math>, meaning <math>7</math> answers were correct and the remaining <math>3</math> answers were wrong. We see that the final score is | ||
<cmath>7(5)-3(2)=35-6=29,</cmath> | <cmath>7(5)-3(2)=35-6=29,</cmath> | ||
| − | which is the score Olivia got. Thus, the answer is <math>\boxed{(C)7}</math>. | + | which is the score Olivia got. Thus, the answer is <math>\boxed{(C) 7}</math>. |
==Video Solution== | ==Video Solution== | ||
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~pi_is_3.14 | ~pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/9dl4iKzW6Tg | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=16|num-a=18}} | {{AMC8 box|year=2002|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:35, 29 October 2024
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 
. ~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 
.
Solution 3
Suppose she got questions right. Then she got 
questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve 
to get that 
.
Solution 4
We see that Olivia's score is odd. Since subtracting multiples of 
(even numbers) does not change a number's parity (odd or even), Olivia's score from only her correct answers must be odd. Then, we test odd multiples of 
greater than 
to see which one works. The smallest odd multiple of greater than is 
, meaning 
answers were correct and the remaining 
answers were wrong. We see that the final score is
![\[7(5)-3(2)=35-6=29,\]](http://latex.artofproblemsolving.com/e/1/4/e1493f5ea042a99d02e427deaf3a01ea9c11a65e.png)
which is the score Olivia got. Thus, the answer is 
.
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
Video Solution by WhyMath
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. 
