Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 8 Problems/Problem 21"

(→‎Solution 3 (Cheap))
(→‎Solution 3 (Cheap))
Line 37: Line 37:
  
 
==Solution 3 (Cheap)==
 
==Solution 3 (Cheap)==
−
Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most shots candace
+
Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is
  
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==

Revision as of 20:58, 21 December 2023

Problem

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ 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] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

Solution 1 (Inequalities)

Let $x$ be the number of shots that Candace made in the first half, and let $y$ 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 $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x}{12}<\frac{15}{20} \implies x<9,\] and \[\frac{y}{18}<\frac{10}{10} \implies y<18.\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{\textbf{(C) } 9}.$

~wamofan

Solution 2 (Answer Choices)

Clearly, Steph made $15 + 10 = 25$ 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 $\boxed{\textbf{(C) } 9}.$

Solution 3 (Cheap)

Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=gkFZrlmVOdkOwwTT&t=3793

~Math-X

Video Solution (🚀58 seconds. Straightforward Explanation.🚀)

https://youtu.be/MKzyAvby_HQ

~Education, the Study of Everything

Video Solution

https://youtu.be/QF9iZEjQ4AI

Please like and subscribe!

Video Solution by OmegaLearn

https://youtu.be/DzQZtQvNDwA?t=460

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=IbsSecIq8FE

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1995

~Interstigation

Video Solution

https://www.youtube.com/watch?v=nXHHM1884Jo

~David

Video Solution

https://youtu.be/0orAAUaLIO0

~STEMbreezy

Video Solution

https://youtu.be/diWioLTXpYk

~savannahsolver

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_21&oldid=208077"