Difference between revisions of "2008 AMC 10B Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | The number of points could have been 10, 11, 12, 13, 14, or 15. Thus, the answer is <math>\mathrm{(E)}</math>. | + | The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is <math>\boxed{\mathrm{(E)}\ 6}</math>. |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|before=First Question|num-a=2}} | {{AMC10 box|year=2008|ab=B|before=First Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:54, 7 June 2021
Problem
A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
Solution
The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is 
.
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
