1989 AHSME Problems/Problem 20: Difference between revisions
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== Problem == | |||
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> | ||
==Solution== | |||
Since <math>\lfloor\sqrt{x}\rfloor=12</math>, <math>12\leq\sqrt{x}<13</math> and thus <math>144\leq x<169</math>. | |||
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 | |||
<cmath>144\leq x<146.41</cmath> | |||
The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide. Thus, the probability is | |||
<cmath>\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.</cmath> | |||
Revision as of 13:24, 3 August 2011
Problem
Let 
be a real number selected uniformly at random between 100 and 200. If 
, find the probability that 
. (
means the greatest integer less than or equal to 
.)
Solution
Since 
, 
and thus 
.
The successful region is when 
in which case 
Thus, the successful region is when
![\[144\leq x<146.41\]](http://latex.artofproblemsolving.com/b/2/f/b2fa28342b49d4a51faf759f9997e7696b3c9554.png)
The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide. Thus, the probability is
![\[\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.\]](http://latex.artofproblemsolving.com/8/0/8/808f0ba392b5ef18f5bceff36869961a5cb66d84.png)
