Difference between revisions of "1965 AHSME Problems/Problem 18"
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− | + | == Problem == | |
− | <math> | + | If <math>1 - y</math> is used as an approximation to the value of <math>\frac {1}{1 + y}, |y| < 1</math>, the ratio of the error made to the correct value is: |
− | <math> \boxed{(B) y^2}</math> | + | <math>\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 </math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | 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*} \\ | ||
+ | \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 <math>\boxed{\textbf{(B) }y^2}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1965|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 15:57, 18 July 2024
Problem
If is used as an approximation to the value of , the ratio of the error made to the correct value is:
Solution
The error made in this approximation is , and the correct value is . 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
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AHSME Problems and Solutions |
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