Difference between revisions of "2004 AMC 12A Problems/Problem 2"

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==Solution==
 
==Solution==
 
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She gets <math>8*2.5=20</math> points for the problems she didn't answer. She must get <math>\left\lceil \frac{100-20}{6} \right\rceil =14 \Rightarrow \text {(C)}</math> problems right to score at least 100.
She gets <math>8*2.5=20</math> points for the problems she didn't answer. She must get <math>\lceil \frac{100-20}{6} \rceil =14 \Rightarrow \text {(C)}</math> problems right to score at least 100.
 
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2004|ab=A|num-b=1|num-a=3}}
  
{{AMC12 box|year=2004|ab=A|num-b=1|num-a=3}}
 
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:14, 3 July 2013

Problem

On the AMC 12, each correct answer is worth $6$ points, each incorrect answer is worth $0$ points, and each problem left unanswered is worth $2.5$ points. If Charlyn leaves $8$ of the $25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $100$?

$\text {(A)} 11 \qquad \text {(B)} 13 \qquad \text {(C)} 14 \qquad \text {(D)} 16\qquad \text {(E)}17$

Solution

She gets $8*2.5=20$ points for the problems she didn't answer. She must get $\left\lceil \frac{100-20}{6} \right\rceil =14 \Rightarrow \text {(C)}$ problems right to score at least 100.

See Also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AMC 12 Problems and Solutions

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