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Difference between revisions of "1989 AHSME Problems/Problem 20"

(Created page with "Let <math>x</math> be a real number selected uniformly at random between 100 and 200. If <math>\lfloor {\sqrt{x}} \rfloor = 12</math>, find the probability that <math>\lfloor {\s...")
 
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== Problem ==
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Let <math>x</math> be a real number selected uniformly at random between 100 and 200. If <math>\lfloor {\sqrt{x}} \rfloor = 12</math>, find the probability that <math>\lfloor {\sqrt{100x}} \rfloor = 120</math>. (<math>\lfloor {v} \rfloor</math> means the greatest integer less than or equal to <math>v</math>.)
 
Let <math>x</math> be a real number selected uniformly at random between 100 and 200. If <math>\lfloor {\sqrt{x}} \rfloor = 12</math>, find the probability that <math>\lfloor {\sqrt{100x}} \rfloor = 120</math>. (<math>\lfloor {v} \rfloor</math> means the greatest integer less than or equal to <math>v</math>.)
  
 
<math>\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1</math>
 
<math>\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1</math>
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==Solution==
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Since <math>\lfloor\sqrt{x}\rfloor=12</math>, <math>12\leq\sqrt{x}<13</math> and thus <math>144\leq x<169</math>.
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The successful region is when <math>120\leq10\sqrt{x}<121</math> in which case <math>12\leq\sqrt{x}<12.1</math>  Thus, the successful region is when
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<cmath>144\leq x<146.41</cmath>
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The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide.  Thus, the probability is
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<cmath>\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.</cmath>

Revision as of 13:24, 3 August 2011

Problem

Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor {\sqrt{x}} \rfloor = 12$, find the probability that $\lfloor {\sqrt{100x}} \rfloor = 120$. ($\lfloor {v} \rfloor$ means the greatest integer less than or equal to $v$.)

$\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1$

Solution

Since $\lfloor\sqrt{x}\rfloor=12$, $12\leq\sqrt{x}<13$ and thus $144\leq x<169$.

The successful region is when $120\leq10\sqrt{x}<121$ in which case $12\leq\sqrt{x}<12.1$ Thus, the successful region is when \[144\leq x<146.41\]

The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide. Thus, the probability is \[\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.\]

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