1951 AHSME Problems/Problem 41

Revision as of 10:14, 19 April 2014 by Hukilau17 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png