Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 8 Problems/Problem 17"

(Created page with "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 sco...")
 
 
(28 intermediate revisions by 13 users not shown)
Line 1: Line 1:
 +
==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?
 
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?
  
 
<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 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*}
 +
50-7(10-x)&=29\\
 +
50-70+7x&=29\\
 +
7x&=49\\
 +
x&=\boxed{\text{(C)}\ 7}
 +
\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]
 +
 +
==Solution 4==
 +
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>
 +
which is the score Olivia got. Thus, the answer is <math>\boxed{(C) 7}</math>.
 +
 +
==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==
 +
https://youtu.be/9dl4iKzW6Tg
 +
 +
==See Also==
 +
{{AMC8 box|year=2002|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 13:35, 29 October 2024

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?

$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution 1

We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$. 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 $(7*5)-(2*3)=29$. Thus, our answer is $\boxed{\text{(C)}\ 7}$. ~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 $x$ be the number of questions she answers correctly. Then, we will represent the number incorrect by $10-x$.

\begin{align*} 50-7(10-x)&=29\\ 50-70+7x&=29\\ 7x&=49\\ x&=\boxed{\text{(C)}\ 7} \end{align*}

Solution 3

Suppose she got $x$ questions right. Then she got $10 - x$ questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve $5x - 2(10 - x) = 29$ to get that $x = \boxed{\text{(C)}\ 7}$.

~cxsmi

Solution 4

We see that Olivia's score is odd. Since subtracting multiples of $2$ (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 $5$ greater than $29$ to see which one works. The smallest odd multiple of $5$ greater than $29$ is $35$, meaning $7$ answers were correct and the remaining $3$ answers were wrong. We see that the final score is \[7(5)-3(2)=35-6=29,\] which is the score Olivia got. Thus, the answer is $\boxed{(C) 7}$.

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

https://youtu.be/9dl4iKzW6Tg

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_17&oldid=230851"