1989 AMC 12 Problem 20

Question

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.

)

\frac2{25}

\frac{241}{2500}

\frac1{10}

\frac{96}{625}

1

Convert each floor condition into an interval for

x

The conditional probability is the ratio of the two relevant interval lengths

The given condition is

144\le x\lt169,

an interval of length

25.

The desired condition is

120^2\le100x\lt121^2,

or

144\le x\lt146.41.

Its length is

2.41=\frac{241}{100},

so the conditional probability is

\frac{\frac{241}{100}}{25}=\frac{241}{2500}.

Thus the correct answer is B.

Problem 20 in other years