Difference between revisions of "1968 AHSME Problems/Problem 4"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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<math>4 \star 4 \star 4 = 4 \star 2 = \frac{4}{3}</math>
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Therefore, our answer is <math>\fbox{C}</math>
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Solution by VivekA
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1968|num-b=3|num-a=5}}   
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{{AHSME 35p box|year=1968|num-b=3|num-a=5}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:52, 16 August 2023

Problem

Define an operation $\star$ for positive real numbers as $a\star b=\frac{ab}{a+b}$. Then $4 \star (4 \star 4)$ equals:

$\text{(A) } \frac{3}{4}\quad \text{(B) } 1\quad \text{(C) } \frac{4}{3}\quad \text{(D) } 2\quad \text{(E )} \frac{16}{3}$


Solution

$4 \star 4 \star 4 = 4 \star 2 = \frac{4}{3}$

Therefore, our answer is $\fbox{C}$

Solution by VivekA

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

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