Difference between revisions of "2022 AMC 8 Problems/Problem 21"
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Let <math>x</math> be the number of shots that Candace made in the first half, and let <math>y</math> 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 <math>x+y=10+15=25.</math> In addition, we have the following inequalities: <cmath>\frac{x}{12}<\frac{15}{20} \implies x<9,</cmath> and <cmath>\frac{y}{18}<\frac{10}{10} \implies y<18.</cmath> Pairing this up with <math>x+y=25</math> we see the ONLY possible solution is <math>(x,y)=(8,17),</math> for an answer of <math>17-8 = \boxed{\textbf{(C) }9}</math> | Let <math>x</math> be the number of shots that Candace made in the first half, and let <math>y</math> 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 <math>x+y=10+15=25.</math> In addition, we have the following inequalities: <cmath>\frac{x}{12}<\frac{15}{20} \implies x<9,</cmath> and <cmath>\frac{y}{18}<\frac{10}{10} \implies y<18.</cmath> Pairing this up with <math>x+y=25</math> we see the ONLY possible solution is <math>(x,y)=(8,17),</math> for an answer of <math>17-8 = \boxed{\textbf{(C) }9}</math> | ||
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| + | ~wamofan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=20|num-a=22}} | {{AMC8 box|year=2022|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:01, 28 January 2022
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
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
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. 
