# 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 $x = \boxed{\text{(C)}\ 7}$.

## 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

~pi_is_3.14