Difference between revisions of "2001 AMC 8 Problems/Problem 12"
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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72</math> | <math>\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72</math> | ||
− | ==Solution== | + | ==Solution 1== |
<math> 6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5 </math>. | <math> 6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5 </math>. | ||
<math> 5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4} </math> | <math> 5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4} </math> | ||
+ | |||
+ | ==Solution 2 (Overkill)== | ||
+ | When you expand the general form of <math>(a\otimes b)\otimes c</math>, you get <cmath>(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}</cmath> | ||
+ | <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c} </cmath> | ||
+ | <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - ab}{a - b}}{\dfrac{a + b - ac + bc}{a - b}} </cmath> | ||
+ | <cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac - ab}{a + b - ac + ab} </cmath> | ||
+ | |||
+ | Now, substituting <math>a=6</math>, <math>b=4</math>, and <math>c=3</math>: | ||
+ | |||
+ | <cmath> (6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12} </cmath> | ||
+ | <cmath> (6\otimes 4)\otimes 3 = \dfrac{16}{4} </cmath> | ||
+ | <cmath> (6\otimes 4)\otimes 3 = 4 </cmath> | ||
+ | <math>\boxed{\text {(A)}</math> | ||
+ | |||
+ | ~megaboy6679 | ||
==Video Solution-Cooler Method== | ==Video Solution-Cooler Method== |
Revision as of 23:07, 19 December 2024
Problem
If , then
Solution 1
.
Solution 2 (Overkill)
When you expand the general form of , you get
Now, substituting , , and :
$\boxed{\text {(A)}$ (Error compiling LaTeX. Unknown error_msg)
~megaboy6679
Video Solution-Cooler Method
https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.