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Difference between revisions of "1987 AHSME Problems/Problem 19"

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== Solution ==
 
== Solution ==
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We have <math>\sqrt{65} > 8 > 7.5</math>. Also <math>7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63</math>, so <math>\sqrt{63} > 7.5</math>. Thus <math>\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15</math>. Now notice that <math>\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}</math>, so <math>\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...</math>, so the answer must be <math>A</math> or <math>B</math>. To determine which, we write <math>\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 32 \cdot 0.0078125 + 0.0078125^2 = 4096 - 0.25 + 0.0078125^2</math>, which is true. Hence as the expression is greater than <math>0.125</math>, and less than or equal to <math>0.13</math> (since we showed it is certainly less than <math>0.1333333...</math>), it is closest to <math>0.13</math>, which is answer <math>\boxed{\text{B}}</math>.
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We have <math>\sqrt{65} > 8 > 7.5</math>. Also <math>7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63</math>, so <math>\sqrt{63} > 7.5</math>. Thus <math>\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15</math>. Now notice that <math>\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}</math>, so <math>\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...</math>, so the answer must be <math>A</math> or <math>B</math>. To determine which, we write <math>\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2</math> which is true. Hence as the expression is greater than <math>0.125</math>, and less than or equal to <math>0.13</math> (since we showed it is certainly less than <math>0.1333333...</math>), it is closest to <math>0.13</math>, which is answer <math>\boxed{\text{B}}</math>.
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(<math>\sqrt{65} - \sqrt{63}</math> is approximately equal to <math>0.125003815</math>)
  
 
== See also ==
 
== See also ==

Latest revision as of 07:02, 12 July 2018

Problem

Which of the following is closest to $\sqrt{65}-\sqrt{63}$?

$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$

Solution

We have $\sqrt{65} > 8 > 7.5$. Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$, so $\sqrt{63} > 7.5$. Thus $\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15$. Now notice that $\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}$, so $\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...$, so the answer must be $A$ or $B$. To determine which, we write $\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$, and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$), it is closest to $0.13$, which is answer $\boxed{\text{B}}$.

($\sqrt{65} - \sqrt{63}$ is approximately equal to $0.125003815$)

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AHSME Problems and Solutions

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