Difference between revisions of "2015 AMC 10B Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | <math>1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}</math>, so <math>(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}</math>. Also, <math>2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}</math>, so <math>1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}</math>. Thus, <math>((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}</math> | + | <math>1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}</math>, so <math>(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}</math>. Also, <math>2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}</math>, so <math>1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}</math>. Thus, <math>((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\dfrac{1}{6}-\dfrac{2}{5}=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}</math> |
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/opghB-8aScI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/SkInNyZkxzA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=B|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:11, 2 August 2022
Problem
Consider the operation "minus the reciprocal of," defined by . What is ?
Solution
, so . Also, , so . Thus,
Video Solution 1
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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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