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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
We can simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! 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. Which is what we need! Thus, our answer is \boxed{\text{(C)}\ 7}
+
We can simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! 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. Which is what we need! Thus, our answer is 7 or C.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 12:13, 25 November 2020

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 simply use the options. If she got five right, her score would be (5*5)-(5*2)=15. That's not right! 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. Which is what we need! Thus, our answer is 7 or C.

Solution 2

We can start with the full score, 50, and subtract not only 2 points for each correct 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*}

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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

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