Difference between revisions of "2007 AMC 12B Problems/Problem 5"
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The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attempt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points? | The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attempt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points? | ||
| − | <math>\mathrm {(A)} 13 | + | <math>\mathrm {(A)}\ 13 \qquad \mathrm {(B)}\ 14 \qquad \mathrm {(C)}\ 15 \qquad \mathrm {(D)}\ 16 \qquad \mathrm {(E)} 17</math> |
==Solution== | ==Solution== | ||
| − | She must get at least 100-4.5=95.5 points, and that can only be possible by answering at least <math>16 \Rightarrow \mathrm {(D)}</math> questions | + | She must get at least <math>100 - 4.5 = 95.5</math> points, and that can only be possible by answering at least <math>\lceil \frac{95.5}{6}\rceil = 16 \Rightarrow \mathrm {(D)}</math> questions correctly. |
| − | |||
==See Also== | ==See Also== | ||
| + | {{AMC12 box|year=2007|ab=B|num-b=4|num-a=6}} | ||
| − | {{ | + | [[Category:Introductory Algebra Problems]] |
| + | {{MAA Notice}} | ||
Latest revision as of 09:50, 4 July 2013
Problem
The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attempt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points?
Solution
She must get at least 
points, and that can only be possible by answering at least 
questions correctly.
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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. 
