Difference between revisions of "2013 AMC 10B Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | + | Let <math>x</math> be the number of two point shots attempted and <math>y</math> the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, <math>0.5(2x)+0.4(3y)=54</math> or <math>x+1.2y=54</math>. Because the team attempted 50% more two point shots then threes, <math>x=1.5y</math>. Substituting <math>1.5y</math> for <math>x</math> in the first equation gives <math>1.5y+1.2y=54</math>, which equals <math>2.7y=54</math> so <math>y=</math> <math>\boxed{\textbf{(C) }20}</math> | |
== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2013|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:44, 25 December 2017
Problem
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
Solution
Let be the number of two point shots attempted and the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, or . Because the team attempted 50% more two point shots then threes, . Substituting for in the first equation gives , which equals so
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
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