2022 AMC 8 Problems/Problem 21
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?
![[asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label("12", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label("18", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label("20", (28,18)); label("15", (28,24)); draw((25,21)--(31,21)); label("10", (56,18)); label("10", (56,24)); draw((53,21)--(59,21)); label("First Half", (28,31)); label("Second Half", (56,31)); label("Candace", (2.35,6)); label("Steph", (0,21)); [/asy]](http://latex.artofproblemsolving.com/4/2/a/42aefbe42671b2aeb0b90d88e171e5e03a00b73e.png)
Solution 1 (Inequalities)
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: ![\[\frac{x}{12}<\frac{15}{20} \implies x<9,\]](http://latex.artofproblemsolving.com/7/1/6/716a42f88961e64e0d35c406547f428f6f675fe6.png)
and ![\[\frac{y}{18}<\frac{10}{10} \implies y<18.\]](http://latex.artofproblemsolving.com/a/3/3/a33cc629e050ccf31e08c13b068e5dd97ddd5290.png)
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 odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is 
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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 | ||
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