Difference between revisions of "1989 AHSME Problems/Problem 20"
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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 | ||
+ | <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 | ||
+ | <cmath>\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.</cmath> | ||
+ | == See also == | ||
+ | {{AHSME box|year=1989|num-b=19|num-a=21}} | ||
+ | |||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:58, 22 October 2014
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
The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide. Thus, the probability is
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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