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Difference between revisions of "2005 AMC 12B Problems/Problem 4"

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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #4]] and [[2005 AMC 10B Problems|2005 AMC 10B #6]]}}
 
{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #4]] and [[2005 AMC 10B Problems|2005 AMC 10B #6]]}}
 
== Problem ==
 
== Problem ==
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At the beginning of the school year, Trahee's goal was to earn an A on at least <math>80\%</math> of her <math>50</math> quizzes for the year.  She earned an A on <math>22</math> of the first <math>30</math> quizzes.  If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?
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At the beginning of the school year, Lisa's goal was to earn an A on at least <math>80\%</math> of her <math>50</math> quizzes for the year.  She earned an A on <math>22</math> of the first <math>30</math> quizzes.  If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?
  
 
<math>
 
<math>

Revision as of 06:33, 1 June 2021

The following problem is from both the 2005 AMC 12B #4 and 2005 AMC 10B #6, so both problems redirect to this page.

Problem

At the beginning of the school year, Lisa's goal was to earn an A on at least $80\%$ of her $50$ quizzes for the year. She earned an A on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$

Solution

Trahee's goal was to get an A on $80\% \cdot 50 = 40$ quizzes. She already has A's on $22$ quizzes, so she needs to get A's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an A on $20-18=\boxed{\text{(B)}2}$ of them. Here, only the A's matter... No complicated stuff!

See also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 10 Problems and Solutions
2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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