Difference between revisions of "2002 AMC 8 Problems/Problem 17"
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<math> \text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9 </math> | <math> \text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | 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== | ||
| + | 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 <math>x</math> be the number of questions she answers correctly. Then, we will represent the number incorrect by <math>10-x</math>. | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | + | 50-7(10-x)&=29\\ | |
| − | + | 50-70+7x&=29\\ | |
| − | + | 7x&=49\\ | |
| − | + | x&=\boxed{\text{(C)}\ 7} | |
| − | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | |||
| + | ==Solution 3== | ||
| + | Suppose she got <math>x</math> questions right. Then she got <math>10 - x</math> questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve <math>5x - 2(10 - x) = 29</math> to get that <math>x = \boxed{\text{(C)}\ 7}</math>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
| + | |||
| + | ==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== | ==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 18:31, 20 January 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 
.
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. 
