1989 AHSME Problems/Problem 20: Difference between revisions

Latest revision as of 06:58, 22 October 2014

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 $\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.$

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

@@ Line 16: / Line 16: @@
 <cmath>\frac{2.41}{25}=\boxed{\frac{241}{2500}};\;\boxed{B}.</cmath>
 == See also ==
-{{AHSME box|year=1989|before=[[1988 AHSME]]|after=[[1990 AHSME]]}}
+{{AHSME box|year=1989|num-b=19|num-a=21}}
-* [[AHSME]]
+[[Category: Intermediate Algebra Problems]]
-* [[AHSME Problems and Solutions]]
-* [[Mathematics competition resources]]
 {{MAA Notice}}

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1989 AHSME Problems/Problem 20: Difference between revisions

Latest revision as of 06:58, 22 October 2014

Problem

Solution

See also