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

Revision as of 15:44, 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

$\frac{1}{1 + y} = \frac{1 - y}{1 - y^2}$

$\frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2$

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

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

@@ Line 16: / Line 16: @@
 <math> \boxed{(B) y^2}</math>
+==See Also==
+{{AHSME 40p box|year=1965|num-b=17|num-a=19}}
+[[Category:Introductory Algebra Problems]]

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

Revision as of 15:44, 18 July 2024

Problem

Solution

See Also