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

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

Latest 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

Lisa'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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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