Difference between revisions of "1951 AHSME Problems/Problem 41"

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==Problem==
 
==Problem==
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The formula expressing the relationship between <math>x</math> and <math>y</math> in the table is:
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<cmath> Unknown environment 'tabular' </cmath>
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<math> \textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x</math>
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<math> \textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4 </math>
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==Solution==
 
==Solution==
Unsolved
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Just plug the <math>x,y</math> pair <math>(6,20)</math> into each of the 5 answer choices:
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(A): <math>2(6)-4=8\ne20</math>
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(B): <math>6^2-3(6)+2=20</math>
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(C): <math>6^3-3(6^2)+2(6)=120\ne20</math>
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(D): <math>6^2-4(6)=12\ne20</math>
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(E): <math>6^2-4=32\ne20</math>
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The only one that works is <math>\boxed{\textbf{(B)}}</math>.
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=40|num-a=42}}  
 
{{AHSME 50p box|year=1951|num-b=40|num-a=42}}  

Latest revision as of 10:14, 19 April 2014

Problem

The formula expressing the relationship between $x$ and $y$ in the table is: \[\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}\]

$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$ $\textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4$

Solution

Just plug the $x,y$ pair $(6,20)$ into each of the 5 answer choices:

(A): $2(6)-4=8\ne20$

(B): $6^2-3(6)+2=20$

(C): $6^3-3(6^2)+2(6)=120\ne20$

(D): $6^2-4(6)=12\ne20$

(E): $6^2-4=32\ne20$

The only one that works is $\boxed{\textbf{(B)}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 40
Followed by
Problem 42
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All AHSME Problems and Solutions

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