1965 AHSME Problems/Problem 18

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 $\boxed{\textbf{(B) }y^2}$

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1965 AHSME Problems/Problem 18

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