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Difference between revisions of "1959 AHSME Problems/Problem 50"

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Astrophysics is the sub-discipline of [[Physics|physics]] that studies the behavior of astronomical objects such as stars and planets. In their studies, astrophysicists use theories like [[Relativity|General Relativity]], [[Chemistry|chemistry]], and [[Thermodynamics|thermodynamics]].
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A club with <math>x</math> members is organized into four committees in accordance with these two rules:
  
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== Current theories ==
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<cmath>\text{(1)}\ \textup{Each member belongs to two and only two committees}\qquad \\ \text{(2)}\ \textup{Each pair of committees has one and only one member in common}</cmath>
  
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From both theoretical expertise and observations, we have the following theories:
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Then <math>x</math>:
  
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*the Big Bang (estimated to be about <math>13.8</math> billion years ago), from which the entirety of the universe originated;
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<math>\textbf{(A)} \ \textup{cannot be determined} \qquad \\ \textbf{(B)} \ \textup{has a single value between 8 and 16} \qquad \\ \textbf{(C)} \ \textup{has two values between 8 and 16} \qquad \\ \textbf{(D)} \ \textup{has a single value between 4 and 8} \qquad \\ \textbf{(E)} \ \textup{has two values between 4 and 8}</math>
  
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*the existence of dark matter (otherwise unknown outside of gravitational interactions) and dark energy (otherwise unknown outside of speeding up inflation), thought to comprise 95% of the universe's total energy.
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== Solution ==
  
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{{Category:Stubs}}
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We can label each of the <math>x</math> members with the pair of committees that they belong to, which is clearly valid due to rule (1). Then, by rule (2), for each pair of committees, there is exactly one member labeled with that pair. But since we have four committees, there must be <math>x = \binom{4}{2} = 6</math> members in total. Thusly our choice is <math>\boxed{\textbf{(D)}}</math>, and we are done.
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{{Category:Physics}}
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== See also ==
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{{AHSME 50p box|year=1959|num-b=49|after=Last Problem}}
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{{MAA Notice}}
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[[Category: AHSME]][[Category:AHSME Problems]]

Latest revision as of 10:13, 20 July 2020

A club with $x$ members is organized into four committees in accordance with these two rules:

\[\text{(1)}\ \textup{Each member belongs to two and only two committees}\qquad \\ \text{(2)}\ \textup{Each pair of committees has one and only one member in common}\]

Then $x$:

$\textbf{(A)} \ \textup{cannot be determined} \qquad \\ \textbf{(B)} \ \textup{has a single value between 8 and 16} \qquad \\ \textbf{(C)} \ \textup{has two values between 8 and 16} \qquad \\ \textbf{(D)} \ \textup{has a single value between 4 and 8} \qquad \\ \textbf{(E)} \ \textup{has two values between 4 and 8}$

Solution

We can label each of the $x$ members with the pair of committees that they belong to, which is clearly valid due to rule (1). Then, by rule (2), for each pair of committees, there is exactly one member labeled with that pair. But since we have four committees, there must be $x = \binom{4}{2} = 6$ members in total. Thusly our choice is $\boxed{\textbf{(D)}}$, and we are done.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 49 		Followed by
Last Problem
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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