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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]]}}
 
== Problem ==
 
== Problem ==
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At the beginning of the school year, Lisa's goal was to earn an <math>A</math> on at least <math>80\%</math> of her <math>50</math> quizzes for the year.  She earned an <math>A</math> 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 <math>A</math>?
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 +
<math>
 +
\textbf{(A) }\ 1      \qquad
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\textbf{(B) }\ 2      \qquad
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\textbf{(C) }\ 3      \qquad
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\textbf{(D) }\ 4      \qquad
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\textbf{(E) }\ 5
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</math>
  
 
== Solution ==
 
== Solution ==
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Lisa's goal was to get an <math>A</math> on <math>80\% \cdot 50 = 40</math> quizzes.  She already has <math>A</math>'s on <math>22</math> quizzes, so she needs to get <math>A</math>'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 <math>A</math> on <math>20-18=\boxed{\textbf{(B) }2}</math> of them.
 +
 +
Here, only the <math>A</math>'s matter... No complicated stuff!
  
 
== See also ==
 
== See also ==
−
* [[2005 AMC 12B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=5|num-a=7}}
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{{AMC12 box|year=2005|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 13:16, 14 December 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 $A$?

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

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 $20-18=\boxed{\textbf{(B) }2}$ 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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