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 == | ||
| − | 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? | + | 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>? |
<math> | <math> | ||
| − | \ | + | \textbf{(A) }\ 1 \qquad |
| − | \ | + | \textbf{(B) }\ 2 \qquad |
| − | \ | + | \textbf{(C) }\ 3 \qquad |
| − | \ | + | \textbf{(D) }\ 4 \qquad |
| − | \ | + | \textbf{(E) }\ 5 |
</math> | </math> | ||
== Solution == | == Solution == | ||
| − | 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{2}</math> of them. | + | 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 == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=5|num-a=7}} | |
| + | {{AMC12 box|year=2005|ab=B|num-b=3|num-a=5}} | ||
| + | {{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 
on at least 
of her 
quizzes for the year. She earned an on 
of the first 
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 ?
Solution
Lisa's goal was to get an on 
quizzes. She already has 's on quizzes, so she needs to get 's on 
more. There are 
quizzes left, so she can afford to get less than an on 
of them.
Here, only the '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. 
