Difference between revisions of "1965 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
| − | <math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2} | + | 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*} | ||
| − | <math> \ | + | Thus, our answer is <math>\boxed{\textbf{(B) }y^2}</math> |
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| − | |||
==See Also== | ==See Also== | ||
Revision as of 15:54, 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 | |
| 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 | ||
