2002 AMC 8 Problems/Problem 17

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

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

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.

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