1987 AHSME Problems/Problem 26

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Problem

The amount $2.5$ is split into two nonnegative real numbers uniformly at random, for instance, into $2.143$ and $.357$, or into $\sqrt{3}$ and $2.5-\sqrt{3}$. Then each number is rounded to its nearest integer, for instance, $2$ and $0$ in the first case above, $2$ and $1$ in the second. What is the probability that the two integers sum to $3$?

$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{2}{5} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{3}{4}$

Solution

The two parts may round to $0$ and $2$; $1$ and $2$; $1$ and $1$; $2$ and $1$; or $2$ and $0$. By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from $0$ to $2.5$, in which one of the parts is found, into five equal ranges of $0$ to $0.5$, $0.5$ to $1$, $1$ to $1.5$, $1.5$ to $2$, and $2$ to $2.5$). As exactly two of the five cases give a sum of $1 + 2 = 3$, the answer is $\frac{2}{5}$, which is answer $\boxed{B}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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