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==
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==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>
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==Solution 2 (Overkill)==
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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>
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<cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c} </cmath>
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<cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - ab}{a - b}}{\dfrac{a + b - ac + bc}{a - b}} </cmath>
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<cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac - ab}{a + b - ac + ab} </cmath>
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Now, substituting <math>a=6</math>, <math>b=4</math>, and <math>c=3</math>:
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<cmath> (6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12} </cmath>
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<cmath> (6\otimes 4)\otimes 3 = \dfrac{16}{4} </cmath>
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<cmath> (6\otimes 4)\otimes 3 = 4 </cmath>
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<math>\boxed{\text {(A)}</math>
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~megaboy6679
  
 
==Video Solution-Cooler Method==
 
==Video Solution-Cooler Method==

Revision as of 23:07, 19 December 2024

Problem

If $a\otimes b = \dfrac{a + b}{a - b}$, then $(6\otimes 4)\otimes 3 =$

$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

Solution 1

$6\otimes4=\frac{6+4}{6-4}=\frac{10}{2}=5$. $5\otimes3=\frac{5+3}{5-3}=\frac{8}{2}=\boxed{\textbf{(A)}\ 4}$

Solution 2 (Overkill)

When you expand the general form of $(a\otimes b)\otimes c$, you get \[(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}\] \[(a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c}\] \[(a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - ab}{a - b}}{\dfrac{a + b - ac + bc}{a - b}}\] \[(a\otimes b)\otimes c = \dfrac{a + b + ac - ab}{a + b - ac + ab}\]

Now, substituting $a=6$, $b=4$, and $c=3$:

\[(6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12}\] \[(6\otimes 4)\otimes 3 = \dfrac{16}{4}\] \[(6\otimes 4)\otimes 3 = 4\] $\boxed{\text {(A)}$ (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 (ProblemsAnswer KeyResources)
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

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