Difference between revisions of "2022 AMC 8 Problems/Problem 21"
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==Solution 2 (Answer Choices)== | ==Solution 2 (Answer Choices)== | ||
− | Clearly, Candace made <math>15 + 10 = 25</math> shots in total. | + | Clearly, Candace made <math>15 + 10 = 25</math> shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be off. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is <math>\boxed{\textbf{(C)}9}.</math> |
==Video Solution== | ==Video Solution== |
Revision as of 19:17, 20 July 2022
Contents
Problem
Steph scored baskets out of
attempts in the first half of a game, and
baskets out of
attempts in the second half. Candace took
attempts in the first half and
attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?
Solution
Let be the number of shots that Candace made in the first half, and let
be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have
In addition, we have the following inequalities:
and
Pairing this up with
we see the only possible solution is
for an answer of
~wamofan
Solution 2 (Answer Choices)
Clearly, Candace made shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be off. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is
Video Solution
https://www.youtube.com/watch?v=IbsSecIq8FE
~Mathematical Dexterity
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1995
~Interstigation
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.