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Difference between revisions of "1965 AHSME Problems/Problem 18"

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== Solution ==
 
== Solution ==
  
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<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math>
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The error made in this approximation is <math>\frac{1}{1+y}-(1-y)</math>, and the correct value is <math>\frac{1}{1+y}</math>. Taking the ratio of these two values, we have:
 +
\begin{align*} \\
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\frac{\frac{1}{1+y}-(1-y)}{\frac{1}{1+y}}&=\frac{[\frac{1}{1+y}-(1-y)][1+y]}{[\frac{1}{1+y}][1+y]} \\
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&=\frac{1-(1-y^2)}{1} \\
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&=y^2 \\
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\end{align*}
  
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<math> \frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2</math>
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Thus, our answer is <math>\boxed{\textbf{(B) }y^2}</math>
  
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<math> \boxed{(B) y^2}</math>
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==See Also==
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{{AHSME 40p box|year=1965|num-b=17|num-a=19}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 15:57, 18 July 2024

Problem

If $1 - y$ is used as an approximation to the value of $\frac {1}{1 + y}, |y| < 1$, the ratio of the error made to the correct value is:

$\textbf{(A)}\ y \qquad \textbf{(B) }\ y^2 \qquad \textbf{(C) }\ \frac {1}{1 + y} \qquad \textbf{(D) }\ \frac{y}{1+y}\qquad \textbf{(E) }\ \frac{y^2}{1+y}\qquad$

Solution

The error made in this approximation is $\frac{1}{1+y}-(1-y)$, and the correct value is $\frac{1}{1+y}$. Taking the ratio of these two values, we have: \begin{align*} \\ \frac{\frac{1}{1+y}-(1-y)}{\frac{1}{1+y}}&=\frac{[\frac{1}{1+y}-(1-y)][1+y]}{[\frac{1}{1+y}][1+y]} \\ &=\frac{1-(1-y^2)}{1} \\ &=y^2 \\ \end{align*}

Thus, our answer is $\boxed{\textbf{(B) }y^2}$

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
All AHSME Problems and Solutions

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