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Difference between revisions of "1952 AHSME Problems/Problem 28"

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== Solution ==
 
== Solution ==
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<math>\fbox{}</math>
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A simple method of solving the problem is trying each of the answer choices.
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One can plug in values of x and y for each, because many x-values and their single corresponding y-values are given.
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#Choice A works for (1,3) and (2,7) but fails to work on (3,13) because 3*4=12 and 12-1=11, not 13.
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#Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7.
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#Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31.
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Thus, the answer is <math>\boxed{\bf{(C)}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 01:10, 13 November 2023

Problem

In the table shown, the formula relating x and y is:

\[\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\ \hline y & 3 & 7 & 13 & 21 & 31\\ \hline\end{array}\]

$\text{(A) } y = 4x - 1 \qquad\quad \text{(B) } y = x^3 - x^2 + x + 2 \qquad\\ \text{(C) } y = x^2 + x + 1 \qquad \text{(D) } y = (x^2 + x + 1)(x - 1) \qquad\\ \text{(E) } \text{none of these}$

Solution

A simple method of solving the problem is trying each of the answer choices. One can plug in values of x and y for each, because many x-values and their single corresponding y-values are given.

  1. Choice A works for (1,3) and (2,7) but fails to work on (3,13) because 3*4=12 and 12-1=11, not 13.
  2. Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7.
  3. Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31.

Thus, the answer is $\boxed{\bf{(C)}}$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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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